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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.

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    $\begingroup$ How about working it out for a much smaller number than 100, such as 5? Then 6... $\endgroup$ – Steve Kass Jul 8 '20 at 18:54
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    $\begingroup$ ...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? $\endgroup$ – David G. Stork Jul 8 '20 at 18:55
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    $\begingroup$ You may find some relevant discussion at this rather older question, which @SteveKass may remember (he made a comment on it). $\endgroup$ – Brian Tung Jul 8 '20 at 19:02
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    $\begingroup$ 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? $\endgroup$ – Steve Kass Jul 8 '20 at 20:26
  • $\begingroup$ @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. $\endgroup$ – user293794 Jul 8 '20 at 21:21
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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?

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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 $j$.

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  • $\begingroup$ 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\ $. $\endgroup$ – lonza leggiera Jul 10 '20 at 0:09
  • $\begingroup$ @lonzaleggiera why $n_2(1)=0$? $\endgroup$ – d.k.o. Jul 10 '20 at 5:54
  • $\begingroup$ 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. $\endgroup$ – lonza leggiera Jul 10 '20 at 6:10
  • $\begingroup$ Why? You stop when the sum of digits reaches $\ge k$ (see the question). $\endgroup$ – d.k.o. Jul 10 '20 at 6:35
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    $\begingroup$ @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. $\endgroup$ – Brian Tung Jul 10 '20 at 23:46

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