1
$\begingroup$

This question already has an answer here:

Suppose we have well shuffled standard deck of 52 cards. What would be the expected number of cards turned over without replacement until a spade is shown? The problem is easy if replacement is allowed, but I don't know how to setup a trial to take into account the conditionality of the probability. Solving the expected value problem recursively also doesn't work since there isn't replacement. Any tips on solving this kind of problem?

$\endgroup$

marked as duplicate by Lord_Farin, Amzoti, Martin, Julian Kuelshammer, Start wearing purple Jun 14 '13 at 8:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ please see this: math.stackexchange.com/questions/245354/expected-value-of-sums $\endgroup$ – jay-sun Nov 27 '12 at 18:44
  • 2
    $\begingroup$ I have given a very detailed answer to a very similar problem yesterday with full justification. You will just have to replace the $4$ (for $4$ Aces) by $13$, and the $48$ by $39$. $\endgroup$ – André Nicolas Nov 27 '12 at 18:48
  • $\begingroup$ Thank you! It took me a few minutes to work through your solution and understand it, but it makes sense. Pretty much we're summing up the probability of one card being before any spades plus two cards being before any spades and so on. I was thinking of it that way, but was getting the probability wrong as $\frac{13}{52}$. Could this be boiled down in terms of a Bernoulli trial where $n=39$ and $p=\frac{1}{14}$ in the formula $E[X]=np$? $\endgroup$ – kamikazekent Nov 27 '12 at 19:07
  • $\begingroup$ @kamikazekent the actual value is $53/14$. So you maybe bit off with a Bernoulli assumption. $\endgroup$ – jay-sun Nov 27 '12 at 19:35
  • $\begingroup$ @kamikazekent: I believe that (apart from its wordiness!) there is no really faster way to give a fully detailed solution, using only simple ideas. One can compute the probabilities that the ransom variable takes on various values, but then obtaining a simple expression for the mean is quite messy. $\endgroup$ – André Nicolas Nov 27 '12 at 20:35

Browse other questions tagged or ask your own question.