Question

Suppose that two players A and B take turns rolling a pair of balanced dice and that the winner is the first player who obtains the sum of 7 on a given roll of the two dice. If A rolls first, what is the probability that B will win?

Short answer

The probability that B will win is 0.454545

Step-by-step solution

Given information

There have A and B players, they have two balanced dice.

Step 2:Calculation of the probability that B will win

The sample space S is given by,

\(S = \left\{ {\begin{aligned}{{}{}}{\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right)}\\{\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right)}\\{\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right)}\\{\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)}\\{\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right)}\\{\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)}\end{aligned}} \right\}\)

The probability of sum 7 i.e., the win is,

\(\frac{6}{{36}} = \frac{1}{6}\)

The probability of not sum 7 i.e., the loose is,

\(\frac{{30}}{{36}} = \frac{5}{6}\)

For 1^sttrial,

Player A rolled first and lost i.e.,

The probability of player A losing is\(\frac{5}{6}\)

Player B rolled and do 7 and win.

The probability of player B win is\(\frac{1}{6}\)

The probability of player B winning in 1^sttrial is,

\(\frac{5}{6} \times \frac{1}{6} = \frac{5}{{36}}\)

The probability of player B winning in 2^ndtrial is,

\(\frac{5}{{36}} \times \left( {\frac{5}{6} \times \frac{5}{6}} \right)\)

The probability of player B winning in 3^rdtrial is,

\(\frac{5}{{36}} \times {\left( {\frac{5}{6} \times \frac{5}{6}} \right)^2}\)

The probability of no player winning is,

\(\frac{5}{6} \times \frac{5}{6} = \frac{{25}}{{36}}\)

So, the probability of player B winning at the n^th round is,

\(\begin{aligned}{}\frac{5}{{36}}{\sum\limits_{n = 0}^\infty {\left( {\frac{{25}}{{36}}} \right)} ^{n - 1}} = \frac{5}{{36}}\left[ {1 + \frac{{25}}{{36}} + {{\left( {\frac{{25}}{{36}}} \right)}^2} + ...} \right]\\ = \frac{5}{{36}} \times \frac{1}{{\left( {1 - \frac{{25}}{{36}}} \right)}}\\ = \frac{5}{{36}} \times \frac{1}{{\left( {\frac{{36 - 25}}{{36}}} \right)}}\\ = \frac{5}{{36}} \times \frac{{36}}{{11}}\\ = \frac{5}{{11}}\\ = 0.454545\end{aligned}\)

Therefore, the probability of player B win is 0.454545