# conditional probability on Truth and lies

Mike tells truth with probability 1/3 and lies with probability 2/3. Independently, David tells truth with probability 3/4 and lies with probability 1/4. Both watch a soccer match. David tells you that Spain won, Mike tells you that Spain lost. What probability will you assign to Spain's win?

Suppose there are 10 people instead of 2. Now, with each of them telling truth or lie, the probability of win will decrease? Or in the same way if i calculate P(loss) it will be 1-P(win)?

how should i approach for this question?

• Who were the opponents? The Dutch? Then Spain lost a.s.! – drhab Oct 7 '15 at 9:54
• There's not enough information to answer the question, since we don't know the prior probability of Spain winning. Perhaps you meant to imply that the prior probability is $\frac12$? – joriki Oct 7 '15 at 9:54
• Do games always end in a win for one team and loss for the other, without the possibility of tie? – kviiri Oct 7 '15 at 10:11
• @joriki No, there's enough information to obtain an exact answer for the first part, and make a qualitative assessment of the second; if we assume the given probabilities of a truthful answer is independent of the nature of the truth. – Graham Kemp Oct 7 '15 at 12:41
• @GrahamKemp: See my comment underneath your answer. – joriki Oct 7 '15 at 16:53

## 1 Answer

Mike tells truth with probability 1/3 and lies with probability 2/3. Independently, David tells truth with probability 3/4 and lies with probability 1/4. Both watch a soccer match. David tells you that Spain won, Mike tells you that Spain lost.

Edited: Full credit is due to joriki for this amended answer.

Correction: We're looking for Spain winning given what they said.

Letting $D_S, M_S$ be the events "David said Spain won" and "Mike said Spain won". These will be a lie or truth depending on what the truth is.

\begin{align}\mathsf P(S \mid D_S, \neg M_S) & = \frac{\mathsf P(S, D_S, \neg M_S)}{\mathsf P(S, D_S, \neg M_S)+\mathsf P(\neg S, D_S, \neg M_S)} \\ & = \frac{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)}{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)+\mathsf P(\neg S)\mathsf P(D_S\mid \neg S)\mathsf P(\neg M_S\mid \neg S)} \end{align}

We have   $\mathsf P(\neg M_S\mid S)=2/3 \\ \mathsf P(\neg M_S\mid\neg S)=1/3 \\ \mathsf P(D_S\mid S)=3/4 \\ \mathsf P(D_S\mid\neg S)=1/4$ .

What we don't have is: $\mathsf P(S)$, the prior probability that Spain won.

\begin{align}\mathsf P(S \mid D_S, \neg M_S) & = \frac{3\mathsf P(S)}{3\mathsf P(S)+2\mathsf P(\neg S)} \end{align}

• do we need to consider the case of match tie if both are lying. – umang Oct 7 '15 at 14:49
• We don't want $\mathsf P(S)$ but $\mathsf P(S\mid D_S\cap \neg M_S)$ (where $D_S$ and $M_S$ are the events that David and Mike, respectively, say that Spain won). This is $$\mathsf P(S\mid D_S\cap\neg M_S)=\frac{\mathsf P(S\cap D_S\cap\neg M_S)}{\mathsf P(D_S\cap\neg M_S)}=\frac{\mathsf P(S)\mathsf P(D)\mathsf P(\neg M)}{\mathsf P(S)\mathsf P(D)\mathsf P(\neg M)+\mathsf P(\neg S)\mathsf P(\neg D)\mathsf P(M)}\;,$$ where I followed both your notation for the events $D$ and $M$ and your reasonable assumption that the truth-telling is independent of the outcome of the game. – joriki Oct 7 '15 at 16:40
• This can't be simplified further without knowing the prior probability $\mathsf P(S)$. Your approach corresponds to assuming $\mathsf P(S)=\mathsf P(\neg S)$, which, as I suggested in a comment, is what the OP may have meant to imply but didn't, so the answer should at least mention this missing (and rather unrealistic) assumption. – joriki Oct 7 '15 at 16:40
• yah i think we have to assume P(S)=1/2. – umang Oct 7 '15 at 17:09
• mmm... Let's say Spain is having a lucky streak and always wins, then.... I see. Yes. Excellent point. @jonki – Graham Kemp Oct 7 '15 at 22:55