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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?

 Feb 15, 2020

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