Kenji and Manbu want to play a different game of craps with three dice. Kenji's dice are all 6-sided die while Manbu has one each of a 4-, 6-, and 8-sided die. Manbu reasons with Kenji that his dice will have an equal advantage as (4+6+8)/3 = 18/3=6. So, Kenji agrees to play.

Whoever rolls a 7 or 11 with their three dice on the first roll wins. Whoever rolls a 3, 4, or 18 on the first roll craps out. Does Kenji or Manbu have a greater probability of winning the first roll?

Guest Feb 15, 2020

#1**+1 **

\(\text{In order to win a player either has to roll 7 or 11, or the other player must roll 3, 4, or 18}\\ \text{Basically we are finding 3 element integer partitions of these numbers}\\ \text{In Kenji's case we have the element values 1-6 available 3 times each}\\ \text{In Manbu's case we have 1-4, 1-6, 1-8 available, once each}\\ P[\text{Kenji win}]=\dfrac{42}{216} + \dfrac{5}{192}-\dfrac{42}{216}\dfrac{5}{192} = \dfrac{1489}{6912} \approx 0.215422 \\~\\ P\text{Manbu win}] = \dfrac{37}{192}+\dfrac{5}{216}-\dfrac{37}{192}\dfrac{5}{216}=\dfrac{8767}{41472} \approx 0.211396\\ \text{Kenji has a slight advantage of winning on the first roll}\)

.Rom Feb 17, 2020