Chained Conditional Probability of winning

Question

A and B both tell the truth with probability 3/4 and lie with probability 1/4. A watches a cricket match and talks to B about the outcome (a cricket match can have 3 outcomes namely: win, lose & draw). B, in turn, tells C, "A told me that India won". What probability should C assign to India's win?

I applied Baye's formula as:

W = Event that India won according to C

X = Event that A told B that India didn't win

Y = Event that B lied to C

So,

X' = Event that A didn't tell B that India didn't win = Event that A told B that India won

Y' = Event that B didn't lie to C = Event that B told the truth to C

For C :

P(W) = P(Y'|X').P(X') + P(Y|X).P(X)

    = 3/4 . P(X') + 1/4 . P(X) [B's telling lie(or truth) to C is independent of A's telling lie(or truth) to B] ....... (1)

Now Let :

I = Event that India actually won (Probability is 1/3)

A = Event that A tells the truth

P(X') = P(A|I).P(I) + P(A'|I').P(I')

    = 3/4 . 1/3 + 1/4 . 2/3 [A's telling lie(or truth) is independent of India's win(or lose)]

    = 5/12 .......... (2)

And

P(X) = P(A'|I).P(I) + P(A|I').P(I')

   = 1/4 . 1/3 + 3/4 . 2/3

   = 7/12 ..........(3)

So, From Equation 1 we get :

P(W) = 3/4 . 5/12 + 1/4 . 7/12

   = 11/24

But the answer is 10/16.

Answer 1

There are eight cases (for simplicity, the wording is ungrammatical):

(1) Won, A truth, B truth.

(2) Won, A truth, B lie.

(3) Won, A lie, B truth.

(4) Won, A lie, B lie.

(5) Lost, A truth, B truth.

(6) Lost, A truth, B lie.

(7) Lost, A lie, B truth.

(8) Lost, A lie, B lie.

Now, you can easily check that B's statement that “A said India won” is compatible only with (1), (4), (6), and (7). The cumulative probability of these events (assuming that the probability of winning is $1/2$) is $$\left(\frac{1}{2}\times\frac{3}{4}\times\frac{3}{4}\right)+\left(\frac{1}{2}\times\frac{1}{4}\times\frac{1}{4}\right)+\left(\frac{1}{2}\times\frac{3}{4}\times\frac{1}{4}\right)+\left(\frac{1}{2}\times\frac{1}{4}\times\frac{3}{4}\right)=\frac{1}{2}.$$ Out of these cases, the cumulative probability of those involving winning is $$\left(\frac{1}{2}\times\frac{3}{4}\times\frac{3}{4}\right)+\left(\frac{1}{2}\times\frac{1}{4}\times\frac{1}{4}\right)=\frac{5}{16}.$$ Hence, the conditional probability you are looking for is $$\frac{5/16}{1/2}=\frac{10}{16}.$$

Note: if you insist that the probability of winning is $1/3$ and that of lost/draw is $2/3$, then the answer must be $5/11$ (by analogous computations).