7

I recently saw a question somewhere where I got confused between what exactly I should do about it.

Q. Imagine person A speaks truth 9 out of 10 times and another person B speaks truth 8 out of 10 times. A random card is picked from Jack, Queen and Kings (12 cards total). If both A and B say the random card is Jack of Clubs, what is the probability that the Jack of Clubs was not the picked card?

A. In the answer the questioner said, the answer is supposed to be 1/144 because both are having 12 possibilities of saying something. I thought it was either 2/100 since then both have lied or 1/37 since if both say same card, then either both are lying or both are truthful and hence 2/(2+72).

Please tell me which is the correct answer and also please explain why. I am getting confused because of the questioners answer ignoring the truthfulness of A and B's claim.

Vaibhav D S
  • 71
  • You are correct. The questioner's idea that it is simply $1$ divided by the number of possibilities is nonsense. – saulspatz Mar 28 '21 at 17:00
  • Both answers so far make the assumptions that if a person lies about the card that was picked, they're equally likely to say any of the 11 cards that weren't picked, and that if both people lie, their choices of which card to falsely claim are independent. If these assumptions are correct, can you add them to your question? – Joseph Sible-Reinstate Monica Mar 29 '21 at 01:18

2 Answers2

3

Since they both affirm it is a JC, then either

  1. the card is actually a JC, i.e. both tell the truth, or
  2. the card is not a JC, i.e. both are lying, and in lying they both select JC.

The space of all possible events corresponds to the following Venn diagram

2_liars_1

and is $$ U = {1 \over {12}}\left( {\left( {{{8 + 2} \over {10}}} \right)\left( {{{9 + 1} \over {10}}} \right)} \right) + {{11} \over {12}}\left( {\left( {{{8 + 2} \over {10}}} \right)\left( {{{9 + 1} \over {10}}} \right)} \right) = 1 $$

The space of the event $X$ , corresponding to A and B affirming that the card is JC, whether it is actually so or not is $$ P(X) = {1 \over {12}}{9 \over {10}}{8 \over {10}} + {{11} \over {12}}{1 \over {10}}{2 \over {10}} {1 \over {11^{\,2} }} = {{11 \cdot 9 \cdot 8 + 2} \over {12 \cdot 100 \cdot 11}} $$ and the sought probability is therefore $$ P = {{{2 \over {12 \cdot 100 \cdot 11}}} \over {{{11 \cdot 9 \cdot 8 + 2} \over {12 \cdot 100 \cdot 11}}}} = {2 \over {794}} = {1 \over {397}} $$

G Cab
  • 35,272
  • I think your calculations are correct as conditional probabilities, but I would not describe them as involving "equiprobable events" – Henry Mar 28 '21 at 20:28
  • @Henry: you are right, thanks. I recasted and reworded my answer to render it a bit more rigorous. – G Cab Mar 28 '21 at 21:36
0

$\frac 1{144}$ is not correct because although there are $12$ things each can say they aren't equally likely. Yours... I don't understand your numbers.

There are two ways this can happen.

  1. It could be that that the card is the Jack of Clubs and both player A and B are telling the truth.

This has a probability of $\frac 1{12}\times \frac 9{10}\times \frac 8{10}=\frac {72}{1200}$ (or we could think of it as $72$ out of $1200$ times.)

  1. Or it could be that the card is not the Jack of Clubs and both player A and B are lying and of the 11 lies they could tell they both choose the Jack of Clubs.

This has a probability of $\frac {11}{12} \times (\frac 1{10}\times\frac 1{11})\times (\frac 1{10}\times \frac 1{11})=\frac {22}{1200\times 121}$. (or we could think of it as $22$ out of $1200\times 121$ times.).

So if we did this experiment $1200 \times 121$ times. and every probability occurred the number of times we expect.

Then the probability that the are both telling the truth would be $\frac{72}{1200}$ so this will occur $72\times 121=8712$ times.

And the probability that the are both lying and telling the same lie will is $\frac {22}{1200}$ and that will occur $22$ times.

So the number of times they both say "Jack of Clubs" will be $22 + 8712 = 8734$ times.

The remaining $136,416$ something else happens.

But of the $8734$ times they both say "Jack of clubs" $8712$ times it will be true. SO the probability it is not the Jack of clubs (given that they both said it was the jack of clubs) is $\frac {22}{8734}=\frac {22}{72\times 121 + 22}=\frac {1}{36\times 11 + 1} = \frac {1}{397}$.

This is an application of Bayes Theorem

$P(A|B) = \frac {P(B|A) \cdot P(A)}{P(B)}$ (where $M|N$ the probability om $M$ occurring given $N$ occured).

So $P($not jack of clubs| both said it is jack of clubs$)=\frac {P(\text{both saying it is jack of clubs given that it is not jack of clubs})\cdot P(\text{not jack of clubs})}{P(\text{both saying it is jack of clubs})}=$

$\frac {P(\text{both telling lying and saying it the same given card})\cdot \frac 1{12}}{P(\text{it is the jack of clubs and they both tell the truth}) + P(\text{it isn't the jack of clubs and they both lie and say it is})}$

$\frac {\frac 1{10}\cdot\frac 1{11}\cdot \frac {1}{10}\cdot \frac 1{11}}{\frac 1{12}\cdot \frac 8{10} \cdot \frac 9{10} + \frac {11}{12}\cdot \frac1{10}\cdot \frac1{11}\cdot \frac2{10}\cdot \frac1{11})}=$

$\frac {22}{72 + 22\frac 1{121}}=$

$\frac {2}{72\times 121 + 22}=$

$\frac {22}{8734}=\frac {1}{397}$.

fleablood
  • 124,253
  • This answer makes some assumptions, but some assumptions must be made in order to make the original question into a math problem, and these assumptions seem reasonable to me. Moreover I agree with the calculations. (In one place $\frac {22}{1200}$ is written instead of $\frac {22}{1200\times121}$, but the subsequent results are based on the correct value.) Too bad the downvoters did not explain why they objected. – David K Mar 28 '21 at 18:19
  • In the question, they worded Jack of Clubs part differently though, as in the question implies they both saw the card and then said which card it was. I thought it was 1/37 because someone would toss a coin and see the result and then say the result, if they said heads, then probability that it is heads becomes the probability of them speaking the truth? – Vaibhav D S Mar 29 '21 at 02:50
  • I was assuming it was common knowledge to everyone in the world involved knew what the card was. I'm not actually sure I'm understanding your issue. And I have no idea where the number $37$ is coming from. – fleablood Mar 29 '21 at 04:39