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The question: What is the expected number of cards required to be drawn in order to draw 5 spades.

What I have:

Let $X=X_1+\cdots+X_{43}$ (43 because we're examining the case when 4 spades have been drawn and we're waiting for the 5th) with $X_i=1$ iff card $i$ is drawn before the fifth spade. Then $X=$ the number of cards drawn before the 5th spade. Let $Y=X+1$. Then $Y=$ the number of cards drawn before and including the 5th spade.

$P\left\{X_i=1\right\}=$ the probability that card $i$ is drawn before the 5th spade $=\frac{9!}{10!}=\frac{1}{10}$ (because there are $9!$ ways of ordering the $i$th card at the front of 9 spades and $10$ ways of ordering 10 cards).

Then $E\left[Y\right]=\frac{43}{10}+1=5.3$. But this is far too small. What went wrong?

Thanks.

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4 Answers 4

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There is a distribution to cover these waiting time problems without replacement. It is called the Negative Hypergeometric Distribution. It has an expected value of

$ \frac{N+1}{M+1} \cdot k = \frac{265}{14}$

where N ( 52 cards ) is the total number, M ( 13 spades ) is the total number of the target type and k ( 5 spades needed ) is the number wanted from M.

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Label the non-spades $1$ to $39$. Let $X_i=1$ if non-spade with label $i$ is drawn before the fifth spade, and $0$ otherwise.

Then the total number $Y$ of cards drawn up to an including the fifth spade is given by $$Y=5+X_1+X_2+\cdots+X_{39}.$$ I think you know how to find $E(X_i)$. The number of relevant cards is $14$.

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  • $\begingroup$ $P\left\{X_i=1\right\}=$ the probability that card $i$ is drawn before a spade = $\frac{13!}{14!}=\frac{1}{14}$. $E\left[Y\right]=5+E\left[X\right]=5+\frac{39}{14}=7.79$. 8 cards still seems too low, no? $\endgroup$ Dec 7, 2013 at 21:10
  • $\begingroup$ The probability $i$ is before the fifth spade is $\frac{5}{14}$. So we get $5 +(39)(5/14)$. This is about $13.93$. $\endgroup$ Dec 7, 2013 at 21:15
  • $\begingroup$ Andre, Out of curiosity, where do you live, I guess in the US $\endgroup$ Dec 7, 2013 at 22:05
  • $\begingroup$ It is Canada. ${}{}{}{}$ $\endgroup$ Dec 7, 2013 at 22:08
  • $\begingroup$ Thanks, looking forward to learning from you. $\endgroup$ Dec 7, 2013 at 22:08
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Fifty two cards are labeled with the numbers 1 through 52 . Note that there are 13 spades amongst the cards. For the first spade, We may view these as separating the remaining 39 cards into 14 groups of non-spades - those appearing before the first spade, between the first and second, etc. Each of these groups is equally likely to appear first, so 39/14 non-spades are drawn on average. Similarly for the second spade, (39-Avg Drawn for the 1st Spade)/13 non-spades are drawn on average and (39 - Average drawn for 1st and 2nd Spade)/12, (39-Average drawn for 1st, 2nd and 3rd Spade)/11, and (39-average drawn for 1st, 2nd, 3rd and 4th Spades)/10 non spades are drawn on average for 3rd, 4th and 5th spades

Thus the Expected number of cards drawn to have the first five spades

= $ 5+\frac{39}{14}$(2.785714)

+$\frac{39-2.785714}{13}$ (36.21429/13)

+$\frac{33.4287}{12}$

+$\frac{30.64286}{11}$

+$\frac{27.85714}{10}$

= 5+13.92857

= 18.9257 This equates to what Andre has suggested. I am sorry, I thought about it but was not sure.

Thanks

Satish

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Just so that the answer is here and clear.

Label the non-spades $1\cdots39$. Let $X_i=1$ iff card $i$ is drawn before the fifth spade (and $X_i=0$ iff card $i$ is drawn after the fifth spade). Then $X=\sum\limits_{i=1}^{39}X_i=$ the number of non-spade cards drawn before the fifth spade. Let $Y=5+X$. Then $Y=$ the number of cards drawn in order to obtain the fifth spade.

$P\left\{X_i=1\right\}=$ the probability that card $i$ is drawn before the fifth spade = $\frac{5}{14}$.

Then $E\left[Y\right]=5+E\left[X\right]=5+39*\frac{5}{14}=18.9$

Finally, given any shuffled standard deck of 52 cards, we expect the fifth spade to appear in about 19 draws.

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