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BGonline.org Forums
OT: Probability of getting specific dobule in two rolls of the dice
Posted By: leobueno
Date: Monday, 1 August 2011, at 5:09 p.m.
We play a folk game akin to Yahtzee where we try to achieve poker-type hands in 3 rolls of 5 dice. So, if in the first roll you get three of a kind, then you have two rolls to try to make five of a kind.
What is the probability of getting those two dice in the two subsequent rolls? Note that you roll, as needed, 2, 1 or 0 dice in either roll (obviously, you would be silly to roll anything but 2 dice in the second roll).
Below is my probability tree; I just don't know how to compute the ultimate number. For clarity's sake, let's say we are trying to get a 66. A, B, and C will indicate the outcome of the first roll. XX will indicate the desired result, two sixes, which can occur in 3 possible combination, represented as 1, 2, and 3.
First roll (2 dice):
1/36 of getting 2 sixes [A] [XX1]
10/36 of getting 1 six [B]
25/36 of getting 0 sixes [C]
XX1 = 1/36 = 0.028
Second roll of A:
(Roll no dice; you have a double six)
Second roll of B:
(roll only one die since you already have one six)
1/6 of getting 1 six [XX2]
5/6 of getting 0 sixes
XX2 = 10/36 * 1/6 = 0.046
Second roll of C:
(roll both dice since you do not yet have a six)
1/36 of getting 2 sixes [XX3]
10/36 of getting 1 six
25/36 of getting 0 sixes
XX3 = 25/36 * 1/36 = 0.019
I am inclined to say that the probability of getting two sixes is thus
XX1 + XX2 + XX3 = 0.028 + 0.046 + 0.019 = 0.093
Advise please.
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