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A biased coin is tossed.

Probability of Head - $\frac{1}{8}$

Probability of Tail - $\frac{7}{8}$

A liar watches the coin toss. Probability of his lying is $\frac{3}{4}$ and telling the truth is $\frac{1}{4}$. 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: $$P(A \mid B) = \frac{P(A\cap B)}{P(B)}$$

$\ \ \ \ \ \ $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) = $\frac{ \frac{1}{8} \frac{1}{4} }{ \frac{7}{8}\frac{3}{4} + \frac{1}{8}\frac{1}{4} }$ [using a probability tree will be helpful here]

or, P(it is head GIVEN liar said it's head) = $\frac{1}{22}$

The Question: Is the method I used wrong in any way? Some others I have talked to are saying the answer will be $\frac{1}{4}$. 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?

NolantheNerd
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Hoque
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  • To continue with your method, you need to add $P(\text{it is head given liar says tail)}$. – Toby Mak Aug 06 '19 at 09:01
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    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? – bukwyrm Aug 06 '19 at 09:38
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    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. – drhab Aug 06 '19 at 10:27
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    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! – David K Aug 06 '19 at 11:21
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    FYI, your friends who say 1/4 are committing the base rate fallacy. – Bridgeburners Aug 06 '19 at 20:23
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    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 – Carl Witthoft Aug 07 '19 at 13:29
  • Not sure it's relevant, but... what is the liar's interest in the coin toss? – Robert Soupe Sep 17 '19 at 00:22

5 Answers5

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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.

Floris Claassens
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    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. – Martin Kochanski Aug 06 '19 at 09:29
  • +1) Nice answer! Yes, I am praising myself too ;-). I will delete my answer though because it is exactly the same. – drhab Aug 06 '19 at 09:36
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    +1 As someone new to probability, I also find this useful. – Toby Mak Aug 06 '19 at 09:38
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    +1 Your method even has a name. You are using a contingency table built with natural frequencies. https://math.stackexchange.com/questions/2279851/applied-probability-bayes-theorem/2279888#2279888 – Ethan Bolker Aug 06 '19 at 20:21
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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.$$\tfrac{\tfrac 18\tfrac 14}{\tfrac 18\tfrac 14+\tfrac 78\tfrac 34}=\dfrac 1{22}$$

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 $1/4$.

Graham Kemp
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  • I see now that in my comment I repeated what you already said here. – drhab Aug 06 '19 at 10:28
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    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. – Falco Aug 07 '19 at 12:53
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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.

chi
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Mitch
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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.

farruhota
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I agree with the other answers here regarding $\frac{1}{22}$ being the "correct" answer to this question.

However, I also agree with your friends. The correct answer to this question is also $\frac{1}{4}$ depending on how you interpret the question.

I would also have agreed with the assertion that in fact the answer to this question is $\frac{1}{8}$ 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, $\frac{1}{4}$ 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, $\frac{1}{8}$ 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.

Tasos Papastylianou
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