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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?

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.

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 $1/4$.

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

Your approach is correct but this is another way to justify the number you arrive to

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.