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Consider a card game with $100$ playing cards, of which exactly two cards are Jokers. After a good mix, the cards are divided into two piles of $50$ cards each. These are referred to as $Pile1$ and $Pile2$.

Suppose we get the information that at least one of the two jokers is in $Pile1$. What is the probability that the other joker is also in $Pile1$?

We will call the jokers now $Joker1$ and $Joker2$. What is the probability that $Joker2$ is in $Pile1$ if we already know that $Joker1$ is in $Pile1$?

So for the first question we know that one joker was already drawn so we use the Hyper-geometric distribution to find the probability of drawing the second one: $Hyp(99,1,49)(\{1\})$ so we are drawing $49$ from $99$ cards of which only $1$ is a joker and we calculate the probability of one success. And we get $0.25242$. But I have no idea why naming the jokers would make a difference in the second question.

Would appreciate any help.

Zntzozt
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  • Your result of $0.25242$ is way off. The correct answers to these two questions are approximately $\frac13$ and $\frac12$ respectively. Use Bayes' theorem if you want exact values. – TonyK Nov 25 '21 at 16:09
  • What events have you used in the Bayes theorem to get those numbers? (What is $A$ and what is $B$) – Zntzozt Nov 25 '21 at 16:17
  • For example, in the first question, $A$ is "two jokers in Pile$1$" and $B$ is "at least one joker in Pile$1$". – TonyK Nov 25 '21 at 17:19

1 Answers1

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Perhaps it helps to consider the $4$ a priori cases. We will label the cases $(i,j)$ for $i,j\in \{1,2\}$, where, say, $(1,2)$ means that $J_1\in \text{Pile}_1$ and $J_2\in \text{Pile}_2$ while $(1,1)$ means that both jokers are in $\text{Pile}_1$.

The probability of each case is:

$P_{1,1}=P_{2,2}= \binom {98}{48}\big / \binom {100}{50}=.2\overline {47}$

$P_{1,2}=P_{2,1}=\binom {98}{49}\big/ \binom {100}{50}=.\overline {25}$

Sanity checks: Note that these add to $1$ as they should, and note that we expected each case to be around $\frac 14$.

Now, your first question asks: What is the probability that you are in case $(1,1)$ given that you are not in $(2,2)$?

The answer is, of course, $$\frac {P_{1,1}}{1-P_{2,2}}=\frac {P_{1,1}}{P_{1,1}+P_{1,2}+P_{2,1}}=.328859$$

The second question asks: What is the probability that you are in $(1,1)$ given that you are either in $(1,1)$ or $(1,2)$?

The answer, of course, is: $$\frac {P_{1,1}}{P_{1,1}+P_{1,2}}=.\overline {49}$$

Note that these answers are easy to justify intuitively: if we ignore the slight dependence on the placement of the jokers, each case would have probability exactly $\frac 14$, so the answer to the first question would be $\frac 13$ while the answer to the second would be $\frac 12$. As expected, we see that the (very minor) dependence changes the intuitive results only slightly.

lulu
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  • So how would it look like if I wanted to get the exact numbers using Bayes formula? – Zntzozt Nov 25 '21 at 16:23
  • I have given you the exact numbers using Bayes' Theorem. Well, the $.328859$ is rounded down a little. – lulu Nov 25 '21 at 16:27
  • sorry, I wasn't able to see that. How did you get the probabilities for the cases? – Zntzozt Nov 25 '21 at 16:32
  • I wrote out the computations exactly. Since each shuffle is equally probable, it's just a matter of counting. Note that it suffices to look at a single Pile since the contents of one Pile determine the contents of the other. – lulu Nov 25 '21 at 16:33
  • For intuition you should consider the "simpler" problem: Suppose you toss a fair penny and a fair dime. $#1:$ What's the probability that they both come up $H$ given that at least one of them does? $#2:$ What's the probability that they both come up $H$ given that the dime does? – lulu Nov 25 '21 at 16:36
  • The answer is $1/2$ for $2$ – Zntzozt Nov 25 '21 at 16:38
  • Yes, the answer to $#2$ is $\frac 12$. – lulu Nov 25 '21 at 16:40
  • I really can't think about the first question without overwhelming my self. – Zntzozt Nov 25 '21 at 16:42
  • Where do the questions $1$ and $2$ even differ? – Zntzozt Nov 25 '21 at 16:44
  • Just use the same steps I followed. For the simpler problem, it is easier to compute the probabilities of each of the $4$ cases. – lulu Nov 25 '21 at 16:44
  • this question might help with intuition. Note too: it's easy to do the experiment to test the computations. All you need is two coins (or, really, just one since you can throw it twice). – lulu Nov 25 '21 at 16:45
  • So we have the $4$ cases ${(H,H),(H,T),(T,H),(T,T)}$ with prob. of $1/4$ each. In $#1$ we take out the case that don't have $H$ in them and calculate the prob. of getting another $H$ which is $1/3$ – Zntzozt Nov 25 '21 at 16:50
  • In $#2$ we have $\frac{1/4}{1/4+1/4}=1/2$ – Zntzozt Nov 25 '21 at 16:55
  • I'm still a bit fuzzy on how you got $0.\overline{25}$ for the probability of cases $2$ and $3$. – Zntzozt Nov 25 '21 at 17:08
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    If $J_1$ is in pile $1$ then we just need to choose $49$ other cards from the $98$ non-Jokers in order to complete Pile $1$. – lulu Nov 25 '21 at 17:11
  • Thx, really helped me out a lot :D – Zntzozt Nov 25 '21 at 17:15
  • I just thought of a different approach for the first question considering we don't differentiate between $joker_1$ and $joker_2$. We have exactly three outcomes which are $0$, $1$ or $2$ jokers in $Pile_1$. and given that we know that we already have a joker in $Pile_1$ that leaves us with two cases, of which only one has two jokers giving us a probability of $1/2$ . – Zntzozt Nov 25 '21 at 17:54
  • @Zntzozt Well, those three cases are not equally probable (that's why I used four cases). But, yes. This is the same idea. Note that it only works exactly in the case where you can disregard the dependence. – lulu Nov 25 '21 at 17:57
  • But the distinction between the jokers happens only in the second question. – Zntzozt Nov 25 '21 at 18:03