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A friend of mine plays a game where they roll $N$ 10-sided dice (denoted as rolling $Nd10$) and group the dice together in partitions of 1 or more die such that they have as many possible groups whose sum is greater than or equal to 10. For example, consider a roll of 4d10 resulting in 3, 6, 6, and 10. They would be grouped into (10) and (6,6) with a 3 left over, so 2 is the maximum number of groups. It's permissible to throw the 3 in with either of the other groups, but it doesn't effect the final count of 2 groups whose sum is greater than or equal to 10. Let's call the maximum possible number of groups satisfying this constraint for a roll $M$.

I'd like to find the distribution of $M$ with respect to $N$. I'd also like to know things like the expected value of $M$, the modal value, and the variance for a given $N$.

For example, if $N=1$, we have $p(M = 0) = \frac{9}{10}$ and $p(M=1) = \frac{1}{10}$. This is from direct calculation by looking at all possible partitions of the dice rolls, which is trivial, since we're only rolling 1 die. I can also do $N=2$ since there are few enough partitions for direct computation, but it quickly gets out of hand as $N$ increases.

One related problem is given a particular $Nd10$ roll, how do I figure out $M$ algorithmically? I feel like if I understood how to do this, I might have the proper way of thinking to approach the original problem, but I'm not sure how to express what I want mathematically.

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    I like this problem. Off hand, it seems that if you order N throws $a_1\le a_2\le\dots\le a_N$, then the best you can do is take the largest rolls and group them with the smallest, as needed to reach $10$. So the rolls $1,1,2,3,3,5,7,7,7,9,10$ would be grouped $\{10\},\{9,1\},\{7,2,1\},\{7,3\},\{7,3\}$ for $5$ groups (with the $5$ roll left over). We can tell this is best possible, since $1+1+2+3+3+5+7+7+7+9=45$, so at most $4$ groups can be made from these rolls. I'll have to think some more on this and try to prove something. Cheers! – 2017-01-28
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    Okay, that idea is no good. If the rolls are $2,2,3,6,7$, the best thing to do is put the 7 with the 3, and the 6 with the 2s. So a better strategy must be to match up pairs that sum to exactly $10$ (so no "points" are wasted); then something must be done with whatever remains. – 2017-02-01

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