Probability - Revealed/ Collapsed Options

Question

A small puzzle.

There is a deck of $52$ cards. There is a $70\%$ chance a card at random has been replaced with the joker, and $30\%$ chance that a card has not. $51$ of the cards are flipped over, and the joker is not among them. You are handed the last card, what is the chance that you have the joker?

My analysis:

I originally thought that the probability would be $0.7 \times \frac{1}{52}$, but since the other $51$ cards have been turned over is it not just $70\%$?

Answer 1

This problem, like so many others related to it, is approached by Bayes' Theorem which is a direct consequence of the definition of conditional probability, found by just rearranging the formula in the definition a bit. From $\Pr(A\mid B)\Pr(B)=\Pr(A\cap B)$ and similarly $\Pr(B\mid A)\Pr(A)=\Pr(A\cap B)$, setting these equal and rearranging a bit we get:

$$\Pr(A\mid B)=\dfrac{\Pr(B\mid A)\Pr(A)}{\Pr(B)}$$

As an intermediate step, it is common in these problems to have to rely on the law of total probability. In this context implying that $\Pr(B)=\Pr(A\cap B)+\Pr(A^c\cap B) = \Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)$.

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Now, as to your specific problem... let us begin by naming events and interpreting the problem statement.

Let $A$ be the event that there was in fact a joker mixed into the deck. Then $A^c$ is the event that there was not a joker mixed into the deck. We were told in the problem statement that $\Pr(A)=0.7$

Let $B$ be the event that the first $51$ cards were not jokers.

Recognize the probability you are after "What is the probability that you will be handed a joker as the 52nd card given that the first 51 cards are all not jokers" is precisely the probability $\Pr(A\mid B)$ since if there was a joker and the first 51 were not, the last card must be a joker, while if there was not a joker shuffled into the deck then clearly the 52nd card wouldn't be a joker either. (We could have been more careful and used a third event name talking about the 52nd card, but I don't feel that is necessary here)

Now, we should be able to intuit the values of $\Pr(B\mid A)$ and $\Pr(B\mid A^c)$, the probabilities that we in fact would have had the first 51 cards all not be the joker in the event that there had or had not been a joker added to the deck respectively in the first place.

$\Pr(B\mid A) = \dfrac{1}{52}$ since if there was a joker, it would have needed to be shuffled into the last spot. $\Pr(B\mid A^c)=1$ since if there was no joker, it is guaranteed we would not have found any jokers.

Next, we should have enough information to find $\Pr(B)$.

$\Pr(B) = \Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)=0.7\cdot \dfrac{1}{52}+0.3$ by the law of total probability and using the values found earlier

Finally, finish the problem by Bayes' Theorem to find $\Pr(A\mid B)$.

$\Pr(A\mid B) = \dfrac{\Pr(B\mid A)\Pr(A)}{\Pr(B)} = \dfrac{0.7\cdot \frac{1}{52}}{0.7\cdot \frac{1}{52}+0.3}\approx 0.04294\dots$