For N 6-side dice, what are the odds that more than 1/3 will come up 5s or 6s?

Question

If I roll a single die and win on a 5 or a 6, but lose on 1-4, I will lose 2/3 of the time.

I I roll n 6-sided dice, on average, 1/3 of them will come up 5's or 6's.

For 3d6, if my math is correct there are:

• 2x in 216 of 3 5's or 6's
• 3x in 216 of 2 6's and a 5
• 3x in 216 of 2 5's and a 6

For a total of 8/216 rolls where the result is 2 greater than (1/3 * 3 = 1) 5 or 6.

And another - 2*2*4 =16 rolls with 2 5's or 6's and a 1-4

For a total of 12 rolls where the result is more than 1 5 or 6 on 3d6.

Use case: in the Tabletop RPG Shadowrun when you roll your Nd6, you forfeit any successes (5's or 6's) greater than to your "limit", which you increase with XP or gold.

I know that if the limit were not resource-bound I could raise it to N and never forfeit successes. And I know that on average a limit of N/3 would not forfeit any successes. Now I'm trying to determine either how often I will be forfeiting successes with a limit of N/3 OR if there's a standard deviation I could add to N/3 to optimize for not forfeiting successes.

Answer 1

You don't need to distinguish between $5$s and $6$s. You have a biased coin with chance $\frac 13$ to get a (5 or 6) and chance $\frac 23$ to get (1 to 4). You have a binomial distribution of results. It is easier to compute the chance of getting strictly less than $\frac 13$ (5 or 6)s and subtract from $1$ as there are less cases. If you roll $6d6$ the chance of no (5 or 6)s is $(\frac 23)^6$. The chance of one is ${6 \choose 1}\frac 13(\frac 56)^5$. Alpha gives the chance of getting at least $2$ as $\frac {473}{729}$