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Your friend places 100 cards in row, one of which contains jack of clubs that you need to guess. You pick card at position 15.The host removes all cards except card chosen by you at 15th position and another card at 87th position.He now says"one of the two cards is the jack of clubs". What is probability that card is at position 87?

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    $\begingroup$ Have you tried using Baye's Rule? $\endgroup$ Commented Aug 4, 2019 at 2:21
  • $\begingroup$ Is your friend trying to help you, or prevent you, or choosing at random? Each of those questions has a different answer. $\endgroup$ Commented Aug 4, 2019 at 2:24
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    $\begingroup$ @EthanBolker How is it matter? I thought this problem is similar to the two goats and one car (three doors) problem $\endgroup$ Commented Aug 4, 2019 at 2:26
  • $\begingroup$ @BaroqueFreak I started with Bayes rule but i am unable to find out conditions for bayes. $\endgroup$ Commented Aug 4, 2019 at 2:30
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    $\begingroup$ Answer for the problem is 0.99 given,not getting how it came $\endgroup$ Commented Aug 4, 2019 at 2:32

2 Answers 2

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We must assume that your friend would always do this (remove all cards but your chosen one and one other, and truthfully say that the jack of clubs is one of the two cards).

If your initial guess was correct, it is still the jack of clubs. If your initial guess was not correct, the jack of clubs is the other card. The probability that your initial guess was not correct is $99/100$.

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Let $A$ be the event that the player ended up with the jack of clubs $\textbf{after changes his choice}$, then let $H$ be the event that the player initially choose the right card, then $$P(A)=P(A|H)P(H)+P(A|H^C)P(H^C)=0\times 0.01+1\times 0.99=\boxed{0.99}.$$ Notice that $P(A)$ is the probability you want, but in a more general sense, since the numbers $15$ and $87$ are just dummy numbers.

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