-1
$\begingroup$

Suppose we have a standard 52-card deck (4 sets of cards from number 1 to 13). We pick the cards without putting them back. We stop when two cards of the same value are drawn (not necesarily to be consective draws). What is the mathematical expectation of the number of cards we need to pick?

$\endgroup$

closed as off-topic by Shaun, NCh, Tpofofn, Saad, Leucippus Apr 5 '18 at 0:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, NCh, Saad, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome to stackexchange. What have you tried? Can you answer your question for a smaller deck? $\endgroup$ – Ethan Bolker Apr 4 '18 at 12:49
  • $\begingroup$ In order to help you, we also need to know how well acquainted you are with the principles of probability. Do you for instance know how $E[X]$ can be determined, given that $P(X=x)$ is known for all integer values $x$? $\endgroup$ – jvdhooft Apr 4 '18 at 13:03
  • $\begingroup$ @jvdhooft Quite often that route (finding expectation by means of distribution) is very cumbersome to go and can fortunately be left aside. $\endgroup$ – drhab Apr 4 '18 at 13:32
  • $\begingroup$ @drhab I agree, which is why I upvoted your answer. Even so, a good number of calculations are required when solving this one by hand. $\endgroup$ – jvdhooft Apr 4 '18 at 13:57
1
$\begingroup$

Hint:

For $n=0,1,2,\dots,13$ let $\mu_n$ denote the expectation of the number of cards that still must be picked if exactly $n$ cards with distinct values have been picked already.

Then $\mu_{13}=1$ and to be found is $\mu_0$.

Further for $n=0,1,2,\dots,12$ there is an expression for $\mu_n$ in $\mu_{n+1}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.