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Cards are drawn one after another from a standard 52 card deck until the first spade is drawn. Let the number of necessary draws be represented by X. What is the mean of X?

My professor said that this question looks like a Negative Binomial but is not. I can't think of what kind of distribution it could be. Any hints?

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  • $\begingroup$ It's not a negative binomial because you draw without replacement, so the odds change for each card. Specifically, the probability of having to draw more than $40$ cards is $0$. $\endgroup$ – Arthur Oct 18 '17 at 20:56
  • $\begingroup$ It's not binomial because the trials are not independent (cards are not put back in the deck). The probability $ p_X(x) $ that $ X = x $ is not hard to compute (hint: $ p_X(1) = 13/52 $, $ p_X(2) = 39/52 \cdot 13/51 $, etc.). Then you can use this knowledge to find $ \mathbb{E}[X] $. $\endgroup$ – derpy Oct 18 '17 at 20:57
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Imagine you have the following setup:

$\square\;{\spadesuit}1\;\square\;{\spadesuit}2\; \square .... \square\;{\spadesuit} 13\; \square $

On an average, each spade will be separated out evenly and we are interested in the pile that's before ${\spadesuit}1$. You have $52 - 13 = 39$ cards left, and $ \dfrac {39}{14}$ cards for each pile. So you would expect to turn $\dfrac{39}{14}$ cards $+ 1$ to get the first spade

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  • $\begingroup$ Very good. In other words, the number of cards drawn is $1$ (the first spade) plus the number of non-spades drawn before the first spade. There are $39$ non-spades, and each non-spade has a $\frac1{14}$ probability of being drawn ahead of all the spades. $\endgroup$ – bof Oct 19 '17 at 5:19
  • $\begingroup$ Why does each non-spade have a probability of $1/14$? $\endgroup$ – user478136 Nov 16 '17 at 4:09
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Let $E_n$ be the mean when starting with a deck of 13 spade and $n$ non-spade cards. Clearly, $E_0=1$ and $E_{n+1}=1+\frac{n}{13+n}E_n$. Conclude that $E_n=1+\frac n{14}$. You want $E_{39}$.

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    $\begingroup$ Can you explain how you got the equation for $E_{n+1}$? $\endgroup$ – user478136 Oct 18 '17 at 21:27
  • $\begingroup$ @user478136 Expectation / mean follows a law similar to that of total probability. If $Y$ is a random variable, and $B$ is an event, then $$E(Y)=E(Y\mid A)\cdot P(A)+E(Y\mid \overline A)\cdot P(\overline A)$$In this case, we have the random variable "The number of cars drawn from a deck with $n$ non-spades" and the event "The next card is a spade" $\endgroup$ – Arthur Oct 19 '17 at 6:04

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