Question

A and B alternatively throw a pair of dice. A wins if he throws 6 and B wins if he throws 7. Whoever gets their number first wins the game. Find their respective chances of winning if A starts the game.

Answer 1

We first find out the chances of them getting their respective numbers. Then we get to their respective chances of winning. We find out the rolls where their respective numbers roll. Then we get an infinite G.P. series. We solve it to get their winning chances.

Complete step-by-step solution:
A dice contains six numbers which are 1 to 6. If a pair of dice is rolled then the combined outcome can be from 2 to 12.
Every dice has 6 outcomes when it’s rolled. So, for a pair of dice, the possible combination of outcome is $6\times 6=36$. We denote this event as event S. so, $n(S)=36$.
Now the given condition is that in a game A wins if he throws 6 and B wins if he throws 7, keeping in mind that whoever gets their number first.
Now we find out their individual chances of winning in a game.
A wins when he gets 6 in rolling of a pair of dice. We denote this event as event A.
The outcome of 6 for a pair of dice is only possible when the dice numbers are $\left( 1,5 \right),\left( 2,4 \right),\left( 3,3 \right),\left( 4,2 \right),\left( 5,1 \right)$. So, the number of chances or outcomes are $n(A)=5$
Here the total possible outcome of rolling a pair of dice is $n(S)=36$.
The probabilities of A getting his required number is denoted by P(A).
So, $P(A)=\dfrac{n(A)}{n(S)}=\dfrac{5}{36}$.
The probabilities of A not getting his required number is denoted by $P({{A}^{c}})$.
So, $P({{A}^{c}})=1-P(A)=1-\dfrac{5}{36}=\dfrac{31}{36}$.
B wins when he gets 7 in the rolling of a pair of dice. We denote this event as event B.
The outcome of 7 for a pair of dice is only possible when the dice numbers are $\left( 1,6 \right),\left( 2,5 \right),\left( 3,4 \right),\left( 4,3 \right),\left( 5,2 \right),\left( 6,1 \right)$. So, the number of chances or outcomes are $n(B)=6$
Here the total possible outcome of rolling a pair of dice is $n(S)=36$.
The probabilities of B getting his required number is denoted by P(B).
So, $P(B)=\dfrac{n(B)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6}$.
The probabilities of B not getting his required number is denoted by $P({{B}^{c}})$.
So, $P({{B}^{c}})=1-P(B)=1-\dfrac{1}{6}=\dfrac{5}{6}$.
Now, we find the probability of them winning.
A wins when he gets 6 before B gets his 7. So, A only wins only when B never gets his 7.
A starts the game it can be A gets 6 on the first roll.
If that’s not the case then the next case will be- A doesn’t get 6 in first roll and B doesn’t get 7 in his first roll, but then A gets 6 in his second roll.
So, this continues where A will win only when A gets his number after B never gets his number. This means the winning roll is in the odd number roll.
All the chances of P(A) and P(B) are independent.
So, A’s chances of winning are $P(A)+P({{A}^{c}})P({{B}^{c}})P(A)+P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P({{B}^{c}})P(A)+....$
This continues to the infinite number of rolls.
So, the probability is

& P(A)+P({{A}^{c}})P({{B}^{c}})P(A)+P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P({{B}^{c}})P(A)+.... \\\ & =\dfrac{5}{36}+\dfrac{31}{36}\times \dfrac{5}{6}\times \dfrac{5}{36}+\dfrac{31}{36}\times \dfrac{5}{6}\times \dfrac{31}{36}\times \dfrac{5}{6}\times \dfrac{5}{36}+.... \\\ \end{aligned}$$ This becomes an infinite series of G.P. where the common ratio is $$\dfrac{31}{36}\times \dfrac{5}{6}$$. The solution is $\dfrac{\dfrac{5}{36}}{1-\dfrac{31}{36}\times \dfrac{5}{6}}=\dfrac{5\times 6}{36\times 6-31\times 5}=\dfrac{30}{216-155}=\dfrac{30}{61}$. So, the chances of A’s winning are $\dfrac{30}{61}$. Now, B’s chances of winning are $$P({{A}^{c}})P(B)+P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P(B)+P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P(B)+....$$ This continues to the infinite number of rolls. So, the probability is $$\begin{aligned} & P({{A}^{c}})P(B)+P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P(B)+P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P({{B}^{c}})P({{A}^{c}})P(B)+.... \\\ & =\dfrac{31}{36}\times \dfrac{1}{6}+\dfrac{31}{36}\times \dfrac{5}{6}\times \dfrac{31}{36}\times \dfrac{1}{6}+\dfrac{31}{36}\times \dfrac{5}{6}\times \dfrac{31}{36}\times \dfrac{5}{6}\times \dfrac{31}{36}\times \dfrac{1}{6}+.... \\\ \end{aligned}$$ This becomes an infinite series of G.P. where the common ratio is $$\dfrac{5}{6}\times \dfrac{31}{36}$$. The solution is $\dfrac{\dfrac{31}{36}\times \dfrac{1}{6}}{1-\dfrac{5}{6}\times \dfrac{31}{36}}=\dfrac{1\times 31}{36\times 6-31\times 5}=\dfrac{31}{216-155}=\dfrac{31}{61}$. **So, the chances of B’s winning are $\dfrac{31}{61}$.** **Note:** We need to consider that the process of them getting their number can go on to the infinite times. We need to remember that the chances of them getting their numbers is not similar to them winning their match.