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Suppose there are thirteen slips in a bag, labelled 1-13. If I draw a 10, 11, 12, or 13 then I stop adding to the sum and return the sum. If not, I add the number I draw to the current sum and place the slip back in the bag.

What is the expected value of the sum?

My Attempt:

S: total sum

Xi: the ith drawn slip

Probability of drawing a slip from 1-9: 9/13

Have a 9/13 chance of getting # from 1-9 and expect to add average of 5 and trying again...

Ending with a 10, 11, 12, or 13 each has 1/4 chance $$ E[S] = \frac{1}{4}(\frac{9}{13}(5+E[S]) +10) + \frac{1}{4}(\frac{9}{13}(5+E[S]) + 11) +... +\frac{1}{4}(\frac{9}{13}(5+E[S])+13) $$

which gives E[S] = 48.625

I wasn't sure how to factor in the four different ways that we could end the drawing (i.e. ending with a 10, 11, 12, 13). The answer doesn't seem right to me.

Additionally, let Y be the number of times a number of a particular number is drawn. What is E[Y]? I have no idea how to start with this with this part.

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2 Answers 2

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We have $$E(S)=\sum_{k=1}^{13}E(S\mid\hbox{first draw is $k$})P(\hbox{first draw is $k$})\ .$$ But if the first draw is $k=1,\ldots,9$ then the expected sum is $k$ plus the overall expected sum; if the first draw is $k=10,\ldots,13$ then the expected sum is just $k$. So $$E(S)=\frac{1}{13}\sum_{k=1}^9(k+E(S))+\frac{1}{13}\sum_{k=10}^{13}k\ ,\tag{$*$}$$ and you can solve this to find $E(S)$. Specifically, $$E(S)=\frac{45+9E(S)+46}{13}$$ and so $$E(S)=\frac{91}{4}\ .$$

Comment. I'm not quite clear from your question whether a $10,11,12$ or $13$ gets added to the sum and then stops the count, or if it just stops the count. I have assumed it is added. If not, remove the second sum in $(*)$: the answer will be $E(S)=\frac{45}{4}$.

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  • $\begingroup$ and so E(S) = 91/4. $\endgroup$
    – Ned
    Commented Jul 30, 2014 at 0:36
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Your sum is not correct. Your $E[S]$ includes the $\frac14$ component applied recursively, which doesn't make sense.

EDIT: David's answer (which appeared during this one) is correct. I'll address your second question instead.

The $E[Y]$ for the four numbers which end the drawing is obviously $\frac14$. :) Hopefully you can figure out why. For the remaining numbers, you need to consider each separately. The odds of a number being pulled is $\frac1{13}$, but it gets this chance as long as the four top numbers are not pulled. Think about this as though one slip reads "add 1", and the rest either read "do nothing" (there are eight of these) or "stop" (there are four of these). Let me know if that helps you reframe the problem.

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  • $\begingroup$ would I want to try to model this with some sort of discrete probability distribution with the above information? $\endgroup$ Commented Jul 30, 2014 at 1:03
  • $\begingroup$ Yes, but it's recursive. The expectation is 1/13 +1, 4/13 +0 and stop, and 8/13 the expectation again. To put it very informally. $\endgroup$ Commented Jul 30, 2014 at 13:10
  • $\begingroup$ I'm sorry, I'm not really sure what to do. $\endgroup$ Commented Jul 30, 2014 at 20:58
  • $\begingroup$ $E[S] = \frac 14 * 0 + \frac 1{13} * (1 + E[S]) + \frac 8{13} * E[S]$. Does that make sense? One-quarter of the time you get nothing, one-thirteenth of the time you get +1 and repeat, and eight-thirteenths of the time you just repeat. Solve for $E[S]$ and you're good to go. $\endgroup$ Commented Jul 31, 2014 at 0:08

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