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?
Expected number of cards turned over until a spade is drawn.
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probability
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1please see this: http://math.stackexchange.com/questions/245354/expected-value-of-sums/ – 2012-11-27
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2I have given a very detailed answer to a very similar problem [yesterday](http://math.stackexchange.com/questions/245354/expected-value-of-sums) with full justification. You will just have to replace the $4$ (for $4$ Aces) by $13$, and the $48$ by $39$. – 2012-11-27
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0Thank 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 $E[X]=np$? – 2012-11-27
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0@kamikazekent the actual value is $53/14$. So you maybe bit off with a Bernoulli assumption. – 2012-11-27
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0@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. – 2012-11-27
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0@kamikazekent: There is a somewhat messy way to compute, using probabilities, but not $\Pr(X=k)$. For a non-negative random variable $X$ that takes on only integer values, $E(X)=\sum \Pr(X\ge k)$. These probabilities are easy to calculate. The sum is still a bit messy. – 2012-11-27
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0@AndréNicolas: Actually the expected value has a simple expression. I have posted an answer (http://math.stackexchange.com/a/245846/45813) on a similar question yesterday (http://math.stackexchange.com/questions/245354/expected-value-of-sums/). Of course, you provided an answer to that as well. – 2012-11-27
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0Oh, there is a simple expression all right, it can be obtained also by the method I used, exactly same proof. Can extend that also to expected number until the $r$-th success. But the evaluation of the combinatorial sum takes a couple of lines if done in detail. The expectation of a sum argument bypasses that. – 2012-11-27
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0@André: Would you consider getting this question off the Unanswered list by posting your first comment as an answer; I’d be happy to upvote it. – 2013-06-08