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I'm working on an Expected Value problem and I have arrived at the following infinite sum:

$$2\sum_{n=1}^{1} \left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{n-1}\left(\frac{5}{6}\right)^{1-n} + 3\sum_{n=1}^{2} \left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{n-1}\left(\frac{5}{6}\right)^{2-n} + 4\sum_{n=1}^{3} \left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{n-1}\left(\frac{5}{6}\right)^{3-n}+\cdot\cdot\cdot$$

I'm seeking a way to simplify and eventually evaluate this sum. I've looked at Brian M. Scott's Answer to something similar and I propose the following, however, it is flawed:

$$\prod_{n=2}^{\infty}n\sum_{i=1}^{n-1}\left(\frac{5}{72}\right)\left(\frac{5}{9}\right)^{i-1}\left(\frac{5}{6}\right)^{n-2}$$

I think my understanding of the nested product/sum notation is incorrect, since Wolfram Alpha does not appear to give something reasonable. Any help is appreciated!

RobPratt
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mathlete_xD
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  • What was the motivating expected value problem? – RobPratt Sep 16 '21 at 01:59
  • It's based on a game called Tenzi. Be the first player to get all ten of your dice to show the same number. After the first roll, the player should ideally leave the most common face, and continue rolling the remaining dice. I'm trying to answer what the expected number of rolls is to finish the game. 10 dice is hard, so I started at 3. Assuming I did it correctly, 3 dice should be 63/8. – mathlete_xD Sep 16 '21 at 02:12
  • In that case, see https://puzzling.stackexchange.com/questions/99181/how-many-rolls-on-average-to-get-n-dice-to-all-show-the-same-value, which confirms your expected value of $63/8$ for $3$ dice. – RobPratt Sep 16 '21 at 03:00
  • Neat, thanks! Looks like it's just as hard as I imagined. – mathlete_xD Sep 16 '21 at 03:10

1 Answers1

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Here is a straightforward computation of your double sum: \begin{align} \sum_{k=2}^\infty k \sum_{n=1}^{k-1} \frac{5}{72} \left(\frac{5}{9}\right)^{n-1} \left(\frac{5}{6}\right)^{k-1-n} &= \frac{1}{8} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} \sum_{n=1}^{k-1} \left(\frac{2}{3}\right)^n \\ &= \frac{1}{8} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} \frac{2/3-(2/3)^k}{1-2/3} \\ &= \frac{1}{4} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} - \frac{1}{4} \sum_{k=2}^\infty k \left(\frac{5}{9}\right)^{k-1} \\ &= \frac{1}{4} \cdot \frac{(2-5/6)(5/6)}{(1-5/6)^2} - \frac{1}{4} \cdot \frac{(2-5/9)(5/9)}{(1-5/9)^2} \\ &= \frac{495}{64} \end{align}

Another approach is to interchange the order of summation: \begin{align} \sum_{k=2}^\infty k \sum_{n=1}^{k-1} \frac{5}{72} \left(\frac{5}{9}\right)^{n-1} \left(\frac{5}{6}\right)^{k-1-n} &= \frac{1}{8} \sum_{k=2}^\infty k \left(\frac{5}{6}\right)^{k-1} \sum_{n=1}^{k-1} \left(\frac{2}{3}\right)^n \\ &= \frac{1}{8} \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n \sum_{k=n+1}^\infty k \left(\frac{5}{6}\right)^{k-1} \\ &= \frac{1}{8} \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n 6(n+6)\left(\frac{5}{6}\right)^n \\ &= \frac{3}{4} \sum_{n=1}^\infty n\left(\frac{5}{9}\right)^n + \frac{9}{2} \sum_{n=1}^\infty \left(\frac{5}{9}\right)^n \\ &= \frac{3}{4} \cdot \frac{5/9}{(1-5/9)^2} + \frac{9}{2} \cdot \frac{5/9}{1-5/9} \\ &= \frac{495}{64} \end{align}

RobPratt
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  • Thanks a bunch! Guess I didn't see it as a double sum. I'll have to look more into that to see what went wrong in my thinking. – mathlete_xD Sep 16 '21 at 01:31