What syntax do I use to compute this in Mathematica?

Game of Yahtzee:

1. Suppose you get three 5â€™s on the first roll. What is the probability of rolling the other two dice and

completing the Yahtzee on the second roll?

2. Suppose you get three 5â€™s on the first roll. What is the probability of completing the Yahtzee on the

second roll or the third roll?

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1 Answer

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1. Yahtzee Problem

For throwing a single die we have x, the number shown on the thrown die, follow a discrete uniform distribution:

dist = DiscreteUniformDistribution[ {1, 6} ]

Now the probability for getting two fives in the next throw is:

Probability[ x1 == 5 && x2 == 5, { x1 \[Distributed] dist, x2 \[Distributed] dist } ]

136\frac{1}{36}

2. Yahtzee Problem

Now completing the Yahtzee in the second or third throw can be seen split into three disjunct possibilities.

p(Yahtzee)=p(2,_)∨p(1,1)∨p(0,2)\begin{align}

p(Yahtzee) = p(2,\_) \vee p(1,1)\vee p(0,2)

\end{align}

where p(2,_)p(2,\_) denotes the case of getting 2 fives in in the first throw while the third does not matter and p(1,1),p(0,2)p(1,1), p(0,2) to be understood similarily. Since the events are mutually exclusive we can add them up:

p[2,_] = Probability[

x1 == 5 && x2 == 5,

{x1 \[Distributed] dist, x2 \[Distributed] dist }

];

p[1,1] = 2 \[Times] Probability[

x1 == 5 \[Xor] x2 == 5,

{x1 \[Distributed] dist, x2 \[Distributed] dist }

];

p[0,2] = p[2,_] \[Times] Probability[

x1 != 5 && x2 != 5,

{x1 \[Distributed] dist, x2 \[Distributed] dist }

];

p[yahtzee] = p[2,_] + p[1,1] + p[0,2]

7811296\frac{781}{1296}