What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads?Applied Probability- Bayes theoremWhat is the probability that the coin drawn is fairFinding probability in the case of a biased coinat least one coin has occured then the probability that coin produced different result = (1)2P1p2p1+p2−2p1p2Probability of exactly one head from two coin toss (biased)Biased coin toss probabilityWhat is the Probability that coin is tossed three times1 Biased Coin and 1 Fair Coin, probability of 3rd Head given first 2 tosses are head?What is p(biased coin given heads) in 2 Fair coin, 1 biased coin experimentOne coin chosen between a biased coin and a fair coin, and is tossed n times. Find probability of having gotten the biased coin.How to find the number of tosses a biased coin goes through until a tail occurs?
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What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads?
Applied Probability- Bayes theoremWhat is the probability that the coin drawn is fairFinding probability in the case of a biased coinat least one coin has occured then the probability that coin produced different result = (1)2P1p2p1+p2−2p1p2Probability of exactly one head from two coin toss (biased)Biased coin toss probabilityWhat is the Probability that coin is tossed three times1 Biased Coin and 1 Fair Coin, probability of 3rd Head given first 2 tosses are head?What is p(biased coin given heads) in 2 Fair coin, 1 biased coin experimentOne coin chosen between a biased coin and a fair coin, and is tossed n times. Find probability of having gotten the biased coin.How to find the number of tosses a biased coin goes through until a tail occurs?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
A biased coin is tossed.
Probability of Head - $frac18$
Probability of Tail - $frac78$
A liar watches the coin toss. Probability of his lying is $frac34$ and telling the truth is $frac14$. He says that that the outcome is Head. What is the probability that the coin has truly turned Head?
My Attempt:
I used the formula: $$P(A mid B) = fracP(Acap B)P(B)$$
$ $P(it is head GIVEN liar said it's head) = P(it's head AND liar said it's head) / P(liar said it's head)
or, P(it is head GIVEN liar said it's head) = $frac frac18 frac14 frac78frac34 + frac18frac14 $ [using a probability tree will be helpful here]
or, P(it is head GIVEN liar said it's head) = $frac122$
The Question: Is the method I used wrong in any way? Some others I have talked to are saying the answer will be $frac14$. Their reasoning is this: since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4. So who is right? What will be the answer?
probability conditional-probability
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
$endgroup$
|
show 1 more comment
$begingroup$
A biased coin is tossed.
Probability of Head - $frac18$
Probability of Tail - $frac78$
A liar watches the coin toss. Probability of his lying is $frac34$ and telling the truth is $frac14$. He says that that the outcome is Head. What is the probability that the coin has truly turned Head?
My Attempt:
I used the formula: $$P(A mid B) = fracP(Acap B)P(B)$$
$ $P(it is head GIVEN liar said it's head) = P(it's head AND liar said it's head) / P(liar said it's head)
or, P(it is head GIVEN liar said it's head) = $frac frac18 frac14 frac78frac34 + frac18frac14 $ [using a probability tree will be helpful here]
or, P(it is head GIVEN liar said it's head) = $frac122$
The Question: Is the method I used wrong in any way? Some others I have talked to are saying the answer will be $frac14$. Their reasoning is this: since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4. So who is right? What will be the answer?
probability conditional-probability
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
$endgroup$
2
$begingroup$
Is the probability of the liar lying 3/4, or is the probability of the liar lying in this case 3/4? I.e. are the result of the cointoss and the liar's response correlated?
$endgroup$
– bukwyrm
Aug 6 at 9:38
1
$begingroup$
To convince yourself that "others" are wrong let the probability on head be $0$ and on tail $1$. Then again they will argue that the answer is $frac14$ which is evidently wrong.
$endgroup$
– drhab
Aug 6 at 10:27
3
$begingroup$
Before the liar spoke, and before we see how the coin actually fell, we should say there is a $frac18$ probability of heads. Now someone says it is heads, and it is known that this person usually lies--does that make it more likely that the coin is heads? All those "others" are saying that it does!
$endgroup$
– David K
Aug 6 at 11:21
2
$begingroup$
FYI, your friends who say 1/4 are committing the base rate fallacy.
$endgroup$
– Bridgeburners
Aug 6 at 20:23
2
$begingroup$
This is a variation on the MontyHall problem. The important thing is that the liar knows the true state of the coin, so if it's heads, 3/4 of the time he'll say "tails" and so on
$endgroup$
– Carl Witthoft
Aug 7 at 13:29
|
show 1 more comment
$begingroup$
A biased coin is tossed.
Probability of Head - $frac18$
Probability of Tail - $frac78$
A liar watches the coin toss. Probability of his lying is $frac34$ and telling the truth is $frac14$. He says that that the outcome is Head. What is the probability that the coin has truly turned Head?
My Attempt:
I used the formula: $$P(A mid B) = fracP(Acap B)P(B)$$
$ $P(it is head GIVEN liar said it's head) = P(it's head AND liar said it's head) / P(liar said it's head)
or, P(it is head GIVEN liar said it's head) = $frac frac18 frac14 frac78frac34 + frac18frac14 $ [using a probability tree will be helpful here]
or, P(it is head GIVEN liar said it's head) = $frac122$
The Question: Is the method I used wrong in any way? Some others I have talked to are saying the answer will be $frac14$. Their reasoning is this: since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4. So who is right? What will be the answer?
probability conditional-probability
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
$endgroup$
A biased coin is tossed.
Probability of Head - $frac18$
Probability of Tail - $frac78$
A liar watches the coin toss. Probability of his lying is $frac34$ and telling the truth is $frac14$. He says that that the outcome is Head. What is the probability that the coin has truly turned Head?
My Attempt:
I used the formula: $$P(A mid B) = fracP(Acap B)P(B)$$
$ $P(it is head GIVEN liar said it's head) = P(it's head AND liar said it's head) / P(liar said it's head)
or, P(it is head GIVEN liar said it's head) = $frac frac18 frac14 frac78frac34 + frac18frac14 $ [using a probability tree will be helpful here]
or, P(it is head GIVEN liar said it's head) = $frac122$
The Question: Is the method I used wrong in any way? Some others I have talked to are saying the answer will be $frac14$. Their reasoning is this: since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4. So who is right? What will be the answer?
probability conditional-probability
probability conditional-probability
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
edited Aug 6 at 22:05
NolantheNerd
155 bronze badges
155 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
asked Aug 6 at 8:45
HoqueHoque
1326 bronze badges
1326 bronze badges
2
$begingroup$
Is the probability of the liar lying 3/4, or is the probability of the liar lying in this case 3/4? I.e. are the result of the cointoss and the liar's response correlated?
$endgroup$
– bukwyrm
Aug 6 at 9:38
1
$begingroup$
To convince yourself that "others" are wrong let the probability on head be $0$ and on tail $1$. Then again they will argue that the answer is $frac14$ which is evidently wrong.
$endgroup$
– drhab
Aug 6 at 10:27
3
$begingroup$
Before the liar spoke, and before we see how the coin actually fell, we should say there is a $frac18$ probability of heads. Now someone says it is heads, and it is known that this person usually lies--does that make it more likely that the coin is heads? All those "others" are saying that it does!
$endgroup$
– David K
Aug 6 at 11:21
2
$begingroup$
FYI, your friends who say 1/4 are committing the base rate fallacy.
$endgroup$
– Bridgeburners
Aug 6 at 20:23
2
$begingroup$
This is a variation on the MontyHall problem. The important thing is that the liar knows the true state of the coin, so if it's heads, 3/4 of the time he'll say "tails" and so on
$endgroup$
– Carl Witthoft
Aug 7 at 13:29
|
show 1 more comment
2
$begingroup$
Is the probability of the liar lying 3/4, or is the probability of the liar lying in this case 3/4? I.e. are the result of the cointoss and the liar's response correlated?
$endgroup$
– bukwyrm
Aug 6 at 9:38
1
$begingroup$
To convince yourself that "others" are wrong let the probability on head be $0$ and on tail $1$. Then again they will argue that the answer is $frac14$ which is evidently wrong.
$endgroup$
– drhab
Aug 6 at 10:27
3
$begingroup$
Before the liar spoke, and before we see how the coin actually fell, we should say there is a $frac18$ probability of heads. Now someone says it is heads, and it is known that this person usually lies--does that make it more likely that the coin is heads? All those "others" are saying that it does!
$endgroup$
– David K
Aug 6 at 11:21
2
$begingroup$
FYI, your friends who say 1/4 are committing the base rate fallacy.
$endgroup$
– Bridgeburners
Aug 6 at 20:23
2
$begingroup$
This is a variation on the MontyHall problem. The important thing is that the liar knows the true state of the coin, so if it's heads, 3/4 of the time he'll say "tails" and so on
$endgroup$
– Carl Witthoft
Aug 7 at 13:29
2
2
$begingroup$
Is the probability of the liar lying 3/4, or is the probability of the liar lying in this case 3/4? I.e. are the result of the cointoss and the liar's response correlated?
$endgroup$
– bukwyrm
Aug 6 at 9:38
$begingroup$
Is the probability of the liar lying 3/4, or is the probability of the liar lying in this case 3/4? I.e. are the result of the cointoss and the liar's response correlated?
$endgroup$
– bukwyrm
Aug 6 at 9:38
1
1
$begingroup$
To convince yourself that "others" are wrong let the probability on head be $0$ and on tail $1$. Then again they will argue that the answer is $frac14$ which is evidently wrong.
$endgroup$
– drhab
Aug 6 at 10:27
$begingroup$
To convince yourself that "others" are wrong let the probability on head be $0$ and on tail $1$. Then again they will argue that the answer is $frac14$ which is evidently wrong.
$endgroup$
– drhab
Aug 6 at 10:27
3
3
$begingroup$
Before the liar spoke, and before we see how the coin actually fell, we should say there is a $frac18$ probability of heads. Now someone says it is heads, and it is known that this person usually lies--does that make it more likely that the coin is heads? All those "others" are saying that it does!
$endgroup$
– David K
Aug 6 at 11:21
$begingroup$
Before the liar spoke, and before we see how the coin actually fell, we should say there is a $frac18$ probability of heads. Now someone says it is heads, and it is known that this person usually lies--does that make it more likely that the coin is heads? All those "others" are saying that it does!
$endgroup$
– David K
Aug 6 at 11:21
2
2
$begingroup$
FYI, your friends who say 1/4 are committing the base rate fallacy.
$endgroup$
– Bridgeburners
Aug 6 at 20:23
$begingroup$
FYI, your friends who say 1/4 are committing the base rate fallacy.
$endgroup$
– Bridgeburners
Aug 6 at 20:23
2
2
$begingroup$
This is a variation on the MontyHall problem. The important thing is that the liar knows the true state of the coin, so if it's heads, 3/4 of the time he'll say "tails" and so on
$endgroup$
– Carl Witthoft
Aug 7 at 13:29
$begingroup$
This is a variation on the MontyHall problem. The important thing is that the liar knows the true state of the coin, so if it's heads, 3/4 of the time he'll say "tails" and so on
$endgroup$
– Carl Witthoft
Aug 7 at 13:29
|
show 1 more comment
5 Answers
5
active
oldest
votes
$begingroup$
You're correct, an easy way to check this is by writing out the possible outcomes. In a perfect world, if we flip the coin 32 times the following will happen:
- 21 times it lands tails and the liar said head.
- 7 times it lands tails and the liar said tails.
- 3 times it lands head and the liar said tails.
- 1 time it lands head and the liar said head.
Since it is given that the liar said head there are 22 options left over, only one of which has the coin actually landing head.
I know this is not the proper way of solving this, but I always found it useful to write things out like this when I was getting confused about conditional probability.
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
$endgroup$
22
$begingroup$
This is not improper. It is absolutely the right thing to do, before trying to remember which "magic probability formula" should be applied to the question.
$endgroup$
– Martin Kochanski
Aug 6 at 9:29
$begingroup$
+1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same.
$endgroup$
– drhab
Aug 6 at 9:36
$begingroup$
+1 As someone new to probability, I also find this useful.
$endgroup$
– Toby Mak
Aug 6 at 9:38
4
$begingroup$
+1 Your method even has a name. You are using a contingency table built with natural frequencies. math.stackexchange.com/questions/2279851/…
$endgroup$
– Ethan Bolker
Aug 6 at 20:21
add a comment |
$begingroup$
The Question: Is the method I used wrong in any way?
Nope, your solution is accurate. The coin is rarely heads and truthfully said to be so. The coin is much more often tails yet said to be heads. So when the result is said to be heads, the coin is quite unlikely to truly be heads.$$tfractfrac 18tfrac 14tfrac 18tfrac 14+tfrac 78tfrac 34=dfrac 122$$
Some others I have talked to are saying the answer will be $1/4$. Their reasoning is this: since the liar lies $3$ times out of $4$ and he said it is head, then the probability of it being head is $1/4$. So who is right? What will be the answer?
Consider using another coin, one with two tails - so the result cannot truly be heads - while the reporter lies with the same probability as above. So if this coin is flipped and said to be heads, what is the (conditional) probability that the result is truly heads?
Your method says: $0$, while their method says $1/4$.
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
$endgroup$
$begingroup$
I see now that in my comment I repeated what you already said here.
$endgroup$
– drhab
Aug 6 at 10:28
1
$begingroup$
I came to comment the same thing - if I have an "intuitive" answer to a probability problem, I try to reason if it still holds at the extremes. So one check would be a coin only showing head once on a million tries (or never like you proposed) the other check would be a liar who lies with 99.999% probability. Both will show an answer has to include both factors to hold at the extremes.
$endgroup$
– Falco
Aug 7 at 12:53
add a comment |
$begingroup$
I find a nice way to think about the problem is a table. First let's declare probabilities
P(Heads) = 1/8
P(Tails) = 7/8
P(Lie) = 3/4
P(Truth) = 1/4
With this information we can make the following table
+-------+------------+------------+
| | Says Heads | Says Tails |
+-------+------------+------------+
| Heads | (1/8)(1/4) | (1/8)(3/4) |
+-------+------------+------------+
| Tails | (7/8)(3/4) | (7/8)(1/4) |
+-------+------------+------------+
Since the liar already told us heads we can ignore the right column. So the Probability of $P(Heads)P(Truth)$ would be $P(Heads)P(Truth)$ over the sum of all other possibilities where the liar says heads, or
$$
dfrac P(Heads)P(Truth)
P(Heads)P(Truth) + P(Tails)P(Lie)
= 1/22
$$
Your approach is correct but this is another way to justify the number you arrive to.
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
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add a comment |
$begingroup$
Your application of the conditional probability is correct.
since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4.
One must remember he can say a truth or a lie. If it were indeed heads, then the probability of him telling the truth is 1/4 according to the given condition. But there is 3/4 probability he is lying, that is, it was tails and he is calling it heads.
In fact, with the information "he said it is head", the sample space consists of the two outcomes: (it was heads and he told truth) or (it was tails and he lied). The question is asking the probability of the first outcome happening. Hence, it is the probability of the first out of total probability of the two outcomes, which you calculated correctly.
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
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I agree with the other answers here regarding $frac122$ being the "correct" answer to this question.
However, I also agree with your friends. The correct answer to this question is also $frac14$ depending on how you interpret the question.
I would also have agreed with the assertion that in fact the answer to this question is $frac18$ on the basis that the liar's claim couldn't possibly have had anything to do with the coin-toss outcome itself.
The point being, the question "What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads?" is a bad question, because it can be interpreted in a number of equally valid ways.
In other words, the key problem here is the importance of mathematical rigorousness when stating a problem. The use of the word "given" is a particular pet peeve of mine, since it can be interpreted (linguistically) in many different ways. The mathematical community has informally agreed that it should imply only one of these (the conditional formulation), but the fact that this statement is linguistically ambiguous is not helping, especially when attempting to discuss such problems in a conversational context.
A better and more mathematically precise way to formulate your question is "what is the probability of the coin-toss having resulted in heads, conditioned on the fact that the liar claimed it was heads".
Your friends have effectively interpreted the question as "based on what we know about the liar, what is the probability of the coin having "in fact" resulted in heads (i.e. the liar having told the truth), "given" (i.e. "when") we know the liar lies with a certain fixed probability". Why, $frac14$ of course.
An equally valid interpretation could have been "How is the coin toss affected by the liar's statement?" When "given" the liar's statement, does this affect the coin toss in any way? No. Therefore, "what is the probability of a coin actually landing heads, 'given' (i.e. factoring in the casual influence to the outcome from) what the liar said?". Why, $frac18$ of course!
In other words, like the vast majority of arguments in this world, your disagreement with your friends wasn't a disagreement based on facts, but one based on definitions, masquerading as an argument about facts. The question you ask is twice as important as the answer you give. To paraphrase John Tukey: "I would prefer an approximate answer to an exact problem a great deal more than an exact answer to an approximate problem".
This may sound like a pedantic point, but in more subtle cases, it is in fact a very big problem when working with probabilities in formal problems. Read about the "Monty Hall" problem for a famous example illustrating this nicely.
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
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$begingroup$
"given" was only used as a summary version of the problem in the title. The actual statement of the problem in the post does not use it.
$endgroup$
– Paul Sinclair
Aug 7 at 16:49
$begingroup$
@PaulSinclair indeed, so the correct rebuttal to his friends is that they're answering a different question.
$endgroup$
– Tasos Papastylianou
Aug 7 at 17:29
add a comment |
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5 Answers
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active
oldest
votes
5 Answers
5
active
oldest
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active
oldest
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active
oldest
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$begingroup$
You're correct, an easy way to check this is by writing out the possible outcomes. In a perfect world, if we flip the coin 32 times the following will happen:
- 21 times it lands tails and the liar said head.
- 7 times it lands tails and the liar said tails.
- 3 times it lands head and the liar said tails.
- 1 time it lands head and the liar said head.
Since it is given that the liar said head there are 22 options left over, only one of which has the coin actually landing head.
I know this is not the proper way of solving this, but I always found it useful to write things out like this when I was getting confused about conditional probability.
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
$endgroup$
22
$begingroup$
This is not improper. It is absolutely the right thing to do, before trying to remember which "magic probability formula" should be applied to the question.
$endgroup$
– Martin Kochanski
Aug 6 at 9:29
$begingroup$
+1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same.
$endgroup$
– drhab
Aug 6 at 9:36
$begingroup$
+1 As someone new to probability, I also find this useful.
$endgroup$
– Toby Mak
Aug 6 at 9:38
4
$begingroup$
+1 Your method even has a name. You are using a contingency table built with natural frequencies. math.stackexchange.com/questions/2279851/…
$endgroup$
– Ethan Bolker
Aug 6 at 20:21
add a comment |
$begingroup$
You're correct, an easy way to check this is by writing out the possible outcomes. In a perfect world, if we flip the coin 32 times the following will happen:
- 21 times it lands tails and the liar said head.
- 7 times it lands tails and the liar said tails.
- 3 times it lands head and the liar said tails.
- 1 time it lands head and the liar said head.
Since it is given that the liar said head there are 22 options left over, only one of which has the coin actually landing head.
I know this is not the proper way of solving this, but I always found it useful to write things out like this when I was getting confused about conditional probability.
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
$endgroup$
22
$begingroup$
This is not improper. It is absolutely the right thing to do, before trying to remember which "magic probability formula" should be applied to the question.
$endgroup$
– Martin Kochanski
Aug 6 at 9:29
$begingroup$
+1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same.
$endgroup$
– drhab
Aug 6 at 9:36
$begingroup$
+1 As someone new to probability, I also find this useful.
$endgroup$
– Toby Mak
Aug 6 at 9:38
4
$begingroup$
+1 Your method even has a name. You are using a contingency table built with natural frequencies. math.stackexchange.com/questions/2279851/…
$endgroup$
– Ethan Bolker
Aug 6 at 20:21
add a comment |
$begingroup$
You're correct, an easy way to check this is by writing out the possible outcomes. In a perfect world, if we flip the coin 32 times the following will happen:
- 21 times it lands tails and the liar said head.
- 7 times it lands tails and the liar said tails.
- 3 times it lands head and the liar said tails.
- 1 time it lands head and the liar said head.
Since it is given that the liar said head there are 22 options left over, only one of which has the coin actually landing head.
I know this is not the proper way of solving this, but I always found it useful to write things out like this when I was getting confused about conditional probability.
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
$endgroup$
You're correct, an easy way to check this is by writing out the possible outcomes. In a perfect world, if we flip the coin 32 times the following will happen:
- 21 times it lands tails and the liar said head.
- 7 times it lands tails and the liar said tails.
- 3 times it lands head and the liar said tails.
- 1 time it lands head and the liar said head.
Since it is given that the liar said head there are 22 options left over, only one of which has the coin actually landing head.
I know this is not the proper way of solving this, but I always found it useful to write things out like this when I was getting confused about conditional probability.
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
answered Aug 6 at 9:27
Floris ClaassensFloris Claassens
3,2481 gold badge4 silver badges17 bronze badges
3,2481 gold badge4 silver badges17 bronze badges
22
$begingroup$
This is not improper. It is absolutely the right thing to do, before trying to remember which "magic probability formula" should be applied to the question.
$endgroup$
– Martin Kochanski
Aug 6 at 9:29
$begingroup$
+1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same.
$endgroup$
– drhab
Aug 6 at 9:36
$begingroup$
+1 As someone new to probability, I also find this useful.
$endgroup$
– Toby Mak
Aug 6 at 9:38
4
$begingroup$
+1 Your method even has a name. You are using a contingency table built with natural frequencies. math.stackexchange.com/questions/2279851/…
$endgroup$
– Ethan Bolker
Aug 6 at 20:21
add a comment |
22
$begingroup$
This is not improper. It is absolutely the right thing to do, before trying to remember which "magic probability formula" should be applied to the question.
$endgroup$
– Martin Kochanski
Aug 6 at 9:29
$begingroup$
+1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same.
$endgroup$
– drhab
Aug 6 at 9:36
$begingroup$
+1 As someone new to probability, I also find this useful.
$endgroup$
– Toby Mak
Aug 6 at 9:38
4
$begingroup$
+1 Your method even has a name. You are using a contingency table built with natural frequencies. math.stackexchange.com/questions/2279851/…
$endgroup$
– Ethan Bolker
Aug 6 at 20:21
22
22
$begingroup$
This is not improper. It is absolutely the right thing to do, before trying to remember which "magic probability formula" should be applied to the question.
$endgroup$
– Martin Kochanski
Aug 6 at 9:29
$begingroup$
This is not improper. It is absolutely the right thing to do, before trying to remember which "magic probability formula" should be applied to the question.
$endgroup$
– Martin Kochanski
Aug 6 at 9:29
$begingroup$
+1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same.
$endgroup$
– drhab
Aug 6 at 9:36
$begingroup$
+1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same.
$endgroup$
– drhab
Aug 6 at 9:36
$begingroup$
+1 As someone new to probability, I also find this useful.
$endgroup$
– Toby Mak
Aug 6 at 9:38
$begingroup$
+1 As someone new to probability, I also find this useful.
$endgroup$
– Toby Mak
Aug 6 at 9:38
4
4
$begingroup$
+1 Your method even has a name. You are using a contingency table built with natural frequencies. math.stackexchange.com/questions/2279851/…
$endgroup$
– Ethan Bolker
Aug 6 at 20:21
$begingroup$
+1 Your method even has a name. You are using a contingency table built with natural frequencies. math.stackexchange.com/questions/2279851/…
$endgroup$
– Ethan Bolker
Aug 6 at 20:21
add a comment |
$begingroup$
The Question: Is the method I used wrong in any way?
Nope, your solution is accurate. The coin is rarely heads and truthfully said to be so. The coin is much more often tails yet said to be heads. So when the result is said to be heads, the coin is quite unlikely to truly be heads.$$tfractfrac 18tfrac 14tfrac 18tfrac 14+tfrac 78tfrac 34=dfrac 122$$
Some others I have talked to are saying the answer will be $1/4$. Their reasoning is this: since the liar lies $3$ times out of $4$ and he said it is head, then the probability of it being head is $1/4$. So who is right? What will be the answer?
Consider using another coin, one with two tails - so the result cannot truly be heads - while the reporter lies with the same probability as above. So if this coin is flipped and said to be heads, what is the (conditional) probability that the result is truly heads?
Your method says: $0$, while their method says $1/4$.
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
$endgroup$
$begingroup$
I see now that in my comment I repeated what you already said here.
$endgroup$
– drhab
Aug 6 at 10:28
1
$begingroup$
I came to comment the same thing - if I have an "intuitive" answer to a probability problem, I try to reason if it still holds at the extremes. So one check would be a coin only showing head once on a million tries (or never like you proposed) the other check would be a liar who lies with 99.999% probability. Both will show an answer has to include both factors to hold at the extremes.
$endgroup$
– Falco
Aug 7 at 12:53
add a comment |
$begingroup$
The Question: Is the method I used wrong in any way?
Nope, your solution is accurate. The coin is rarely heads and truthfully said to be so. The coin is much more often tails yet said to be heads. So when the result is said to be heads, the coin is quite unlikely to truly be heads.$$tfractfrac 18tfrac 14tfrac 18tfrac 14+tfrac 78tfrac 34=dfrac 122$$
Some others I have talked to are saying the answer will be $1/4$. Their reasoning is this: since the liar lies $3$ times out of $4$ and he said it is head, then the probability of it being head is $1/4$. So who is right? What will be the answer?
Consider using another coin, one with two tails - so the result cannot truly be heads - while the reporter lies with the same probability as above. So if this coin is flipped and said to be heads, what is the (conditional) probability that the result is truly heads?
Your method says: $0$, while their method says $1/4$.
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
$endgroup$
$begingroup$
I see now that in my comment I repeated what you already said here.
$endgroup$
– drhab
Aug 6 at 10:28
1
$begingroup$
I came to comment the same thing - if I have an "intuitive" answer to a probability problem, I try to reason if it still holds at the extremes. So one check would be a coin only showing head once on a million tries (or never like you proposed) the other check would be a liar who lies with 99.999% probability. Both will show an answer has to include both factors to hold at the extremes.
$endgroup$
– Falco
Aug 7 at 12:53
add a comment |
$begingroup$
The Question: Is the method I used wrong in any way?
Nope, your solution is accurate. The coin is rarely heads and truthfully said to be so. The coin is much more often tails yet said to be heads. So when the result is said to be heads, the coin is quite unlikely to truly be heads.$$tfractfrac 18tfrac 14tfrac 18tfrac 14+tfrac 78tfrac 34=dfrac 122$$
Some others I have talked to are saying the answer will be $1/4$. Their reasoning is this: since the liar lies $3$ times out of $4$ and he said it is head, then the probability of it being head is $1/4$. So who is right? What will be the answer?
Consider using another coin, one with two tails - so the result cannot truly be heads - while the reporter lies with the same probability as above. So if this coin is flipped and said to be heads, what is the (conditional) probability that the result is truly heads?
Your method says: $0$, while their method says $1/4$.
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
$endgroup$
The Question: Is the method I used wrong in any way?
Nope, your solution is accurate. The coin is rarely heads and truthfully said to be so. The coin is much more often tails yet said to be heads. So when the result is said to be heads, the coin is quite unlikely to truly be heads.$$tfractfrac 18tfrac 14tfrac 18tfrac 14+tfrac 78tfrac 34=dfrac 122$$
Some others I have talked to are saying the answer will be $1/4$. Their reasoning is this: since the liar lies $3$ times out of $4$ and he said it is head, then the probability of it being head is $1/4$. So who is right? What will be the answer?
Consider using another coin, one with two tails - so the result cannot truly be heads - while the reporter lies with the same probability as above. So if this coin is flipped and said to be heads, what is the (conditional) probability that the result is truly heads?
Your method says: $0$, while their method says $1/4$.
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
answered Aug 6 at 10:13
Graham KempGraham Kemp
92.2k4 gold badges36 silver badges83 bronze badges
92.2k4 gold badges36 silver badges83 bronze badges
$begingroup$
I see now that in my comment I repeated what you already said here.
$endgroup$
– drhab
Aug 6 at 10:28
1
$begingroup$
I came to comment the same thing - if I have an "intuitive" answer to a probability problem, I try to reason if it still holds at the extremes. So one check would be a coin only showing head once on a million tries (or never like you proposed) the other check would be a liar who lies with 99.999% probability. Both will show an answer has to include both factors to hold at the extremes.
$endgroup$
– Falco
Aug 7 at 12:53
add a comment |
$begingroup$
I see now that in my comment I repeated what you already said here.
$endgroup$
– drhab
Aug 6 at 10:28
1
$begingroup$
I came to comment the same thing - if I have an "intuitive" answer to a probability problem, I try to reason if it still holds at the extremes. So one check would be a coin only showing head once on a million tries (or never like you proposed) the other check would be a liar who lies with 99.999% probability. Both will show an answer has to include both factors to hold at the extremes.
$endgroup$
– Falco
Aug 7 at 12:53
$begingroup$
I see now that in my comment I repeated what you already said here.
$endgroup$
– drhab
Aug 6 at 10:28
$begingroup$
I see now that in my comment I repeated what you already said here.
$endgroup$
– drhab
Aug 6 at 10:28
1
1
$begingroup$
I came to comment the same thing - if I have an "intuitive" answer to a probability problem, I try to reason if it still holds at the extremes. So one check would be a coin only showing head once on a million tries (or never like you proposed) the other check would be a liar who lies with 99.999% probability. Both will show an answer has to include both factors to hold at the extremes.
$endgroup$
– Falco
Aug 7 at 12:53
$begingroup$
I came to comment the same thing - if I have an "intuitive" answer to a probability problem, I try to reason if it still holds at the extremes. So one check would be a coin only showing head once on a million tries (or never like you proposed) the other check would be a liar who lies with 99.999% probability. Both will show an answer has to include both factors to hold at the extremes.
$endgroup$
– Falco
Aug 7 at 12:53
add a comment |
$begingroup$
I find a nice way to think about the problem is a table. First let's declare probabilities
P(Heads) = 1/8
P(Tails) = 7/8
P(Lie) = 3/4
P(Truth) = 1/4
With this information we can make the following table
+-------+------------+------------+
| | Says Heads | Says Tails |
+-------+------------+------------+
| Heads | (1/8)(1/4) | (1/8)(3/4) |
+-------+------------+------------+
| Tails | (7/8)(3/4) | (7/8)(1/4) |
+-------+------------+------------+
Since the liar already told us heads we can ignore the right column. So the Probability of $P(Heads)P(Truth)$ would be $P(Heads)P(Truth)$ over the sum of all other possibilities where the liar says heads, or
$$
dfrac P(Heads)P(Truth)
P(Heads)P(Truth) + P(Tails)P(Lie)
= 1/22
$$
Your approach is correct but this is another way to justify the number you arrive to.
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
$endgroup$
add a comment |
$begingroup$
I find a nice way to think about the problem is a table. First let's declare probabilities
P(Heads) = 1/8
P(Tails) = 7/8
P(Lie) = 3/4
P(Truth) = 1/4
With this information we can make the following table
+-------+------------+------------+
| | Says Heads | Says Tails |
+-------+------------+------------+
| Heads | (1/8)(1/4) | (1/8)(3/4) |
+-------+------------+------------+
| Tails | (7/8)(3/4) | (7/8)(1/4) |
+-------+------------+------------+
Since the liar already told us heads we can ignore the right column. So the Probability of $P(Heads)P(Truth)$ would be $P(Heads)P(Truth)$ over the sum of all other possibilities where the liar says heads, or
$$
dfrac P(Heads)P(Truth)
P(Heads)P(Truth) + P(Tails)P(Lie)
= 1/22
$$
Your approach is correct but this is another way to justify the number you arrive to.
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
$endgroup$
add a comment |
$begingroup$
I find a nice way to think about the problem is a table. First let's declare probabilities
P(Heads) = 1/8
P(Tails) = 7/8
P(Lie) = 3/4
P(Truth) = 1/4
With this information we can make the following table
+-------+------------+------------+
| | Says Heads | Says Tails |
+-------+------------+------------+
| Heads | (1/8)(1/4) | (1/8)(3/4) |
+-------+------------+------------+
| Tails | (7/8)(3/4) | (7/8)(1/4) |
+-------+------------+------------+
Since the liar already told us heads we can ignore the right column. So the Probability of $P(Heads)P(Truth)$ would be $P(Heads)P(Truth)$ over the sum of all other possibilities where the liar says heads, or
$$
dfrac P(Heads)P(Truth)
P(Heads)P(Truth) + P(Tails)P(Lie)
= 1/22
$$
Your approach is correct but this is another way to justify the number you arrive to.
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
$endgroup$
I find a nice way to think about the problem is a table. First let's declare probabilities
P(Heads) = 1/8
P(Tails) = 7/8
P(Lie) = 3/4
P(Truth) = 1/4
With this information we can make the following table
+-------+------------+------------+
| | Says Heads | Says Tails |
+-------+------------+------------+
| Heads | (1/8)(1/4) | (1/8)(3/4) |
+-------+------------+------------+
| Tails | (7/8)(3/4) | (7/8)(1/4) |
+-------+------------+------------+
Since the liar already told us heads we can ignore the right column. So the Probability of $P(Heads)P(Truth)$ would be $P(Heads)P(Truth)$ over the sum of all other possibilities where the liar says heads, or
$$
dfrac P(Heads)P(Truth)
P(Heads)P(Truth) + P(Tails)P(Lie)
= 1/22
$$
Your approach is correct but this is another way to justify the number you arrive to.
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
edited Aug 7 at 7:58
chi
1,4668 silver badges13 bronze badges
1,4668 silver badges13 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
answered Aug 6 at 20:36
Mitchel PaulinMitchel Paulin
1466 bronze badges
1466 bronze badges
add a comment |
add a comment |
$begingroup$
Your application of the conditional probability is correct.
since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4.
One must remember he can say a truth or a lie. If it were indeed heads, then the probability of him telling the truth is 1/4 according to the given condition. But there is 3/4 probability he is lying, that is, it was tails and he is calling it heads.
In fact, with the information "he said it is head", the sample space consists of the two outcomes: (it was heads and he told truth) or (it was tails and he lied). The question is asking the probability of the first outcome happening. Hence, it is the probability of the first out of total probability of the two outcomes, which you calculated correctly.
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
$endgroup$
add a comment |
$begingroup$
Your application of the conditional probability is correct.
since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4.
One must remember he can say a truth or a lie. If it were indeed heads, then the probability of him telling the truth is 1/4 according to the given condition. But there is 3/4 probability he is lying, that is, it was tails and he is calling it heads.
In fact, with the information "he said it is head", the sample space consists of the two outcomes: (it was heads and he told truth) or (it was tails and he lied). The question is asking the probability of the first outcome happening. Hence, it is the probability of the first out of total probability of the two outcomes, which you calculated correctly.
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
$endgroup$
add a comment |
$begingroup$
Your application of the conditional probability is correct.
since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4.
One must remember he can say a truth or a lie. If it were indeed heads, then the probability of him telling the truth is 1/4 according to the given condition. But there is 3/4 probability he is lying, that is, it was tails and he is calling it heads.
In fact, with the information "he said it is head", the sample space consists of the two outcomes: (it was heads and he told truth) or (it was tails and he lied). The question is asking the probability of the first outcome happening. Hence, it is the probability of the first out of total probability of the two outcomes, which you calculated correctly.
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
$endgroup$
Your application of the conditional probability is correct.
since the liar lies 3 times out of 4 and he said it is head, then the probability of it being head is 1/4.
One must remember he can say a truth or a lie. If it were indeed heads, then the probability of him telling the truth is 1/4 according to the given condition. But there is 3/4 probability he is lying, that is, it was tails and he is calling it heads.
In fact, with the information "he said it is head", the sample space consists of the two outcomes: (it was heads and he told truth) or (it was tails and he lied). The question is asking the probability of the first outcome happening. Hence, it is the probability of the first out of total probability of the two outcomes, which you calculated correctly.
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
answered Aug 6 at 11:08
farruhotafarruhota
24.5k2 gold badges9 silver badges46 bronze badges
24.5k2 gold badges9 silver badges46 bronze badges
add a comment |
add a comment |
$begingroup$
I agree with the other answers here regarding $frac122$ being the "correct" answer to this question.
However, I also agree with your friends. The correct answer to this question is also $frac14$ depending on how you interpret the question.
I would also have agreed with the assertion that in fact the answer to this question is $frac18$ on the basis that the liar's claim couldn't possibly have had anything to do with the coin-toss outcome itself.
The point being, the question "What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads?" is a bad question, because it can be interpreted in a number of equally valid ways.
In other words, the key problem here is the importance of mathematical rigorousness when stating a problem. The use of the word "given" is a particular pet peeve of mine, since it can be interpreted (linguistically) in many different ways. The mathematical community has informally agreed that it should imply only one of these (the conditional formulation), but the fact that this statement is linguistically ambiguous is not helping, especially when attempting to discuss such problems in a conversational context.
A better and more mathematically precise way to formulate your question is "what is the probability of the coin-toss having resulted in heads, conditioned on the fact that the liar claimed it was heads".
Your friends have effectively interpreted the question as "based on what we know about the liar, what is the probability of the coin having "in fact" resulted in heads (i.e. the liar having told the truth), "given" (i.e. "when") we know the liar lies with a certain fixed probability". Why, $frac14$ of course.
An equally valid interpretation could have been "How is the coin toss affected by the liar's statement?" When "given" the liar's statement, does this affect the coin toss in any way? No. Therefore, "what is the probability of a coin actually landing heads, 'given' (i.e. factoring in the casual influence to the outcome from) what the liar said?". Why, $frac18$ of course!
In other words, like the vast majority of arguments in this world, your disagreement with your friends wasn't a disagreement based on facts, but one based on definitions, masquerading as an argument about facts. The question you ask is twice as important as the answer you give. To paraphrase John Tukey: "I would prefer an approximate answer to an exact problem a great deal more than an exact answer to an approximate problem".
This may sound like a pedantic point, but in more subtle cases, it is in fact a very big problem when working with probabilities in formal problems. Read about the "Monty Hall" problem for a famous example illustrating this nicely.
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
$endgroup$
$begingroup$
"given" was only used as a summary version of the problem in the title. The actual statement of the problem in the post does not use it.
$endgroup$
– Paul Sinclair
Aug 7 at 16:49
$begingroup$
@PaulSinclair indeed, so the correct rebuttal to his friends is that they're answering a different question.
$endgroup$
– Tasos Papastylianou
Aug 7 at 17:29
add a comment |
$begingroup$
I agree with the other answers here regarding $frac122$ being the "correct" answer to this question.
However, I also agree with your friends. The correct answer to this question is also $frac14$ depending on how you interpret the question.
I would also have agreed with the assertion that in fact the answer to this question is $frac18$ on the basis that the liar's claim couldn't possibly have had anything to do with the coin-toss outcome itself.
The point being, the question "What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads?" is a bad question, because it can be interpreted in a number of equally valid ways.
In other words, the key problem here is the importance of mathematical rigorousness when stating a problem. The use of the word "given" is a particular pet peeve of mine, since it can be interpreted (linguistically) in many different ways. The mathematical community has informally agreed that it should imply only one of these (the conditional formulation), but the fact that this statement is linguistically ambiguous is not helping, especially when attempting to discuss such problems in a conversational context.
A better and more mathematically precise way to formulate your question is "what is the probability of the coin-toss having resulted in heads, conditioned on the fact that the liar claimed it was heads".
Your friends have effectively interpreted the question as "based on what we know about the liar, what is the probability of the coin having "in fact" resulted in heads (i.e. the liar having told the truth), "given" (i.e. "when") we know the liar lies with a certain fixed probability". Why, $frac14$ of course.
An equally valid interpretation could have been "How is the coin toss affected by the liar's statement?" When "given" the liar's statement, does this affect the coin toss in any way? No. Therefore, "what is the probability of a coin actually landing heads, 'given' (i.e. factoring in the casual influence to the outcome from) what the liar said?". Why, $frac18$ of course!
In other words, like the vast majority of arguments in this world, your disagreement with your friends wasn't a disagreement based on facts, but one based on definitions, masquerading as an argument about facts. The question you ask is twice as important as the answer you give. To paraphrase John Tukey: "I would prefer an approximate answer to an exact problem a great deal more than an exact answer to an approximate problem".
This may sound like a pedantic point, but in more subtle cases, it is in fact a very big problem when working with probabilities in formal problems. Read about the "Monty Hall" problem for a famous example illustrating this nicely.
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
$endgroup$
$begingroup$
"given" was only used as a summary version of the problem in the title. The actual statement of the problem in the post does not use it.
$endgroup$
– Paul Sinclair
Aug 7 at 16:49
$begingroup$
@PaulSinclair indeed, so the correct rebuttal to his friends is that they're answering a different question.
$endgroup$
– Tasos Papastylianou
Aug 7 at 17:29
add a comment |
$begingroup$
I agree with the other answers here regarding $frac122$ being the "correct" answer to this question.
However, I also agree with your friends. The correct answer to this question is also $frac14$ depending on how you interpret the question.
I would also have agreed with the assertion that in fact the answer to this question is $frac18$ on the basis that the liar's claim couldn't possibly have had anything to do with the coin-toss outcome itself.
The point being, the question "What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads?" is a bad question, because it can be interpreted in a number of equally valid ways.
In other words, the key problem here is the importance of mathematical rigorousness when stating a problem. The use of the word "given" is a particular pet peeve of mine, since it can be interpreted (linguistically) in many different ways. The mathematical community has informally agreed that it should imply only one of these (the conditional formulation), but the fact that this statement is linguistically ambiguous is not helping, especially when attempting to discuss such problems in a conversational context.
A better and more mathematically precise way to formulate your question is "what is the probability of the coin-toss having resulted in heads, conditioned on the fact that the liar claimed it was heads".
Your friends have effectively interpreted the question as "based on what we know about the liar, what is the probability of the coin having "in fact" resulted in heads (i.e. the liar having told the truth), "given" (i.e. "when") we know the liar lies with a certain fixed probability". Why, $frac14$ of course.
An equally valid interpretation could have been "How is the coin toss affected by the liar's statement?" When "given" the liar's statement, does this affect the coin toss in any way? No. Therefore, "what is the probability of a coin actually landing heads, 'given' (i.e. factoring in the casual influence to the outcome from) what the liar said?". Why, $frac18$ of course!
In other words, like the vast majority of arguments in this world, your disagreement with your friends wasn't a disagreement based on facts, but one based on definitions, masquerading as an argument about facts. The question you ask is twice as important as the answer you give. To paraphrase John Tukey: "I would prefer an approximate answer to an exact problem a great deal more than an exact answer to an approximate problem".
This may sound like a pedantic point, but in more subtle cases, it is in fact a very big problem when working with probabilities in formal problems. Read about the "Monty Hall" problem for a famous example illustrating this nicely.
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
$endgroup$
I agree with the other answers here regarding $frac122$ being the "correct" answer to this question.
However, I also agree with your friends. The correct answer to this question is also $frac14$ depending on how you interpret the question.
I would also have agreed with the assertion that in fact the answer to this question is $frac18$ on the basis that the liar's claim couldn't possibly have had anything to do with the coin-toss outcome itself.
The point being, the question "What is the probability of a biased coin coming up heads given that a liar is claiming that the coin came up heads?" is a bad question, because it can be interpreted in a number of equally valid ways.
In other words, the key problem here is the importance of mathematical rigorousness when stating a problem. The use of the word "given" is a particular pet peeve of mine, since it can be interpreted (linguistically) in many different ways. The mathematical community has informally agreed that it should imply only one of these (the conditional formulation), but the fact that this statement is linguistically ambiguous is not helping, especially when attempting to discuss such problems in a conversational context.
A better and more mathematically precise way to formulate your question is "what is the probability of the coin-toss having resulted in heads, conditioned on the fact that the liar claimed it was heads".
Your friends have effectively interpreted the question as "based on what we know about the liar, what is the probability of the coin having "in fact" resulted in heads (i.e. the liar having told the truth), "given" (i.e. "when") we know the liar lies with a certain fixed probability". Why, $frac14$ of course.
An equally valid interpretation could have been "How is the coin toss affected by the liar's statement?" When "given" the liar's statement, does this affect the coin toss in any way? No. Therefore, "what is the probability of a coin actually landing heads, 'given' (i.e. factoring in the casual influence to the outcome from) what the liar said?". Why, $frac18$ of course!
In other words, like the vast majority of arguments in this world, your disagreement with your friends wasn't a disagreement based on facts, but one based on definitions, masquerading as an argument about facts. The question you ask is twice as important as the answer you give. To paraphrase John Tukey: "I would prefer an approximate answer to an exact problem a great deal more than an exact answer to an approximate problem".
This may sound like a pedantic point, but in more subtle cases, it is in fact a very big problem when working with probabilities in formal problems. Read about the "Monty Hall" problem for a famous example illustrating this nicely.
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
edited Aug 7 at 16:17
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
answered Aug 7 at 16:11
Tasos PapastylianouTasos Papastylianou
2322 silver badges8 bronze badges
2322 silver badges8 bronze badges
$begingroup$
"given" was only used as a summary version of the problem in the title. The actual statement of the problem in the post does not use it.
$endgroup$
– Paul Sinclair
Aug 7 at 16:49
$begingroup$
@PaulSinclair indeed, so the correct rebuttal to his friends is that they're answering a different question.
$endgroup$
– Tasos Papastylianou
Aug 7 at 17:29
add a comment |
$begingroup$
"given" was only used as a summary version of the problem in the title. The actual statement of the problem in the post does not use it.
$endgroup$
– Paul Sinclair
Aug 7 at 16:49
$begingroup$
@PaulSinclair indeed, so the correct rebuttal to his friends is that they're answering a different question.
$endgroup$
– Tasos Papastylianou
Aug 7 at 17:29
$begingroup$
"given" was only used as a summary version of the problem in the title. The actual statement of the problem in the post does not use it.
$endgroup$
– Paul Sinclair
Aug 7 at 16:49
$begingroup$
"given" was only used as a summary version of the problem in the title. The actual statement of the problem in the post does not use it.
$endgroup$
– Paul Sinclair
Aug 7 at 16:49
$begingroup$
@PaulSinclair indeed, so the correct rebuttal to his friends is that they're answering a different question.
$endgroup$
– Tasos Papastylianou
Aug 7 at 17:29
$begingroup$
@PaulSinclair indeed, so the correct rebuttal to his friends is that they're answering a different question.
$endgroup$
– Tasos Papastylianou
Aug 7 at 17:29
add a comment |
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$begingroup$
Is the probability of the liar lying 3/4, or is the probability of the liar lying in this case 3/4? I.e. are the result of the cointoss and the liar's response correlated?
$endgroup$
– bukwyrm
Aug 6 at 9:38
1
$begingroup$
To convince yourself that "others" are wrong let the probability on head be $0$ and on tail $1$. Then again they will argue that the answer is $frac14$ which is evidently wrong.
$endgroup$
– drhab
Aug 6 at 10:27
3
$begingroup$
Before the liar spoke, and before we see how the coin actually fell, we should say there is a $frac18$ probability of heads. Now someone says it is heads, and it is known that this person usually lies--does that make it more likely that the coin is heads? All those "others" are saying that it does!
$endgroup$
– David K
Aug 6 at 11:21
2
$begingroup$
FYI, your friends who say 1/4 are committing the base rate fallacy.
$endgroup$
– Bridgeburners
Aug 6 at 20:23
2
$begingroup$
This is a variation on the MontyHall problem. The important thing is that the liar knows the true state of the coin, so if it's heads, 3/4 of the time he'll say "tails" and so on
$endgroup$
– Carl Witthoft
Aug 7 at 13:29