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You keep on throwing a dice and add the digit that appears to a sum. You stop when sum $\ge 100$. What’s the most frequently appearing digit in all such cases? $1$ or $6$?

I believe the probability of $1$ and $6$ should be equal as the whatever the number of rolls, the probability of getting a number should not be affected. However I don't have a formal proof for it and am not sure if this is right.

Kenta S
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oshhh
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    How about working it out for a much smaller number than 100, such as 5? Then 6... – Steve Kass Jul 08 '20 at 18:54
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    ...or consider: what is the largest number of $1$s that can ever fit your criterion? what is the largest number of $6$s that can ever fit your criterion? – David G. Stork Jul 08 '20 at 18:55
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    You may find some relevant discussion at this rather older question, which @SteveKass may remember (he made a comment on it). – Brian Tung Jul 08 '20 at 19:02
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    Suppose that when the sum reaches 100 or more, you stop the game, yell “Hooray!”, then start the game again with the same die. You do this for years on end. If the die is fair, how can yelling “Hooray!” now and then make the die unfair? – Steve Kass Jul 08 '20 at 20:26
  • @SteveKass Are you answering the same question as @BrianTung? It seems like you are finding the expected number of 1s whereas he is finding the number of 1s among all sequences of rolls which terminate once they reach 100? Why are these the same? I would be grateful if you could make your approach a little more formal or precise. – user293794 Jul 08 '20 at 21:21
  • I’m not saying anything about the expected number if 1s. I’m trying to make it seem intuitive that the stopping-at-100 criterion can’t affect the probability that the fair die will come up 1 (or any other number). – Steve Kass Jul 09 '20 at 00:58
  • @SteveKass that makes sense. But how exactly does that relate to the question? It's not obvious to me that what you said implies that 1s are as frequent as 6s among all possible strings of rolls which terminate at 100. – user293794 Jul 09 '20 at 13:09
  • I suggested a way to create a string of millions of die rolls by playing the game repeatedly. If 1s consistently appeared more frequently than 6s in each group of rolls representing a single game, then 1s would appear more frequently than 6s in the millions of rolls obtained by playing successive single games. But the coin is fair. You can’t magically make it unfair by saying you’re playing games. – Steve Kass Jul 09 '20 at 20:46
  • @SteveKass Thank you for bearing with me. I'm still a little confused though. I think you are saying something about the expected number of 1s vs 6s because your argument has to do with what happens over many trials in the long run. But I don't see why this is the same as showing #1s in all admissible strings = #6s since how does the latter see the probability (namely that longer strings are less probable). Also, it seems like the nature of the game matters since if it was "stop when the ratio of 6s to 1s is 3" then the arugment wouldn't work. So what exactly about this game are you using? – user293794 Jul 09 '20 at 21:48
  • Your original question doesn’t mention “admissible strings.” Maybe I’m not understanding what you meant by “all such cases.” If you play this game over and over, it’s impossible for 1 to consistently come up more (or less) often than 6. There is a very different question you might be thinking about: Given all distinct sequences of numbers 1 through 6 that add up to at least 100 (but add to less than 100 if the last term is omitted), the total number of 1s vs. 6s within all those distinct sequences not the same. But these sequences are unequally likely to arise from playing the game. – Steve Kass Jul 10 '20 at 13:59

2 Answers2

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Basic approach. Imagine drawing a tree, with a root labelled $0$. The running count of each node is the label on that node, plus the sum of the labels of all of its direct ancestors. We build on the tree as follows: Under any node whose running count is not yet $100$, we add six more nodes, labelled $1$ through $6$. We repeat until there are no nodes left whose running count is less than $100$.

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At the end of this process, we obviously have a finite tree. How many $1$s are there? How many $6$s? Was there any time when we added a $1$ but not a $6$, or vice versa?

Brian Tung
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The expected number of $1$'s is the same as the expected number of $6$'s. Let $n_j(k)$ denote the expected number of digit $j\in \{1,\ldots,6\}$ appearing in the sequence until the running sum reaches $k$ (in your case $k=100$). Then $$ n_j(k)=\frac{1}{6}(1+n_j(k-j))+\frac{1}{6}\sum_{i\ne j} n_j(k-i)=\frac{1}{6}+\frac{1}{6}\sum_{i\in \{1,\ldots,6\}} n_j(k-i) $$ with $n_j(1-i)=0$ for $i=1,\ldots,6$. This recurrence relation is the same for all $j$ and so its solution, $n_j(k)$, is the same for all $j$.

  • I don't believe your conclusion is correct. Although the recurrence is the same for all $\ j\ $, the initial conditions aren't. For $\ k=1\ $, for instance, $\ n_1(1)=1\ $ and $\ n_j(1)=0\ $ for $\ j=2,3,\dots,6\ $. – lonza leggiera Jul 10 '20 at 00:09
  • @lonzaleggiera why $n_2(1)=0$? –  Jul 10 '20 at 05:54
  • Starting from a total of $0$, the only way you can get to a total of exactly $1$ is from a single throw of a $1$. No other face can have occurred. – lonza leggiera Jul 10 '20 at 06:10
  • Why? You stop when the sum of digits reaches $\ge k$ (see the question). –  Jul 10 '20 at 06:35
  • But I don't believe your recurrence works unless you take $\ n_j(k)\ $ to the expected number of times face $\ j\ $ has been thrown given that the total is exactly $\ k\ $. The only way I can see how to use this recurrence to get get the expected number given that the total has reached $\ k\ $ or more, and was less than $\ k\ $ on the previous throw is the way I have done in my answer. – lonza leggiera Jul 10 '20 at 07:30
  • You may check that for small values of $k$, e.g., clearly $n_j(1)=1/6$ and $n_j(2)=7/36$. –  Jul 10 '20 at 08:27
  • On looking the problem for a $2$-sided dice, I find that the expected number of times each side has occurred after first reaching or exceeding a given total does appear to be the same for both faces. I presume your definition of $\ n_j(k)\ $ is the expected number of times the face $\ j\ $ has appeared when the total first exceeds $\ k-1\ $. If that's so, then I'm no longer convinced that it doesn't satisfy the recursion you've written down, although I still don't follow its derivation. I also now strongly suspect that my own answer, now deleted, must have contained an error. – lonza leggiera Jul 10 '20 at 13:25
  • That is fishy... to get 100, you need 100 ones but only 17 sixes. So ones should be much more frequent. – vonbrand Jul 10 '20 at 22:21
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    @vonbrand: It is true that you would need $100$ ones (if those were all you got) and only $17$ sixes (if those were all you got). And if we were to list all possible sequences that reached or exceeded $100$ only on the last roll, then ones would indeed be more common than sixes. But not all sequences are equally likely. The sequence consisting of $17$ sixes is $6^{83}$ times more likely than the sequence consisting of $100$ ones. In general, the preponderance of ones in the longer sequences is exactly offset by the fact that those longer sequences occur less frequently. – Brian Tung Jul 10 '20 at 23:46
  • @vonbrand Here is a quick simulation: http://tpcg.io/fSyWaEFl –  Jul 11 '20 at 08:24