Question

Alice and Bob take turns in rolling a fair dice. Whoever gets "6" first wins the game. Alice starts the game. What are the chances that Alice wins?

Answer 1

Answer:

Probability of Alice win the game is $\frac{6}{11}$

Step-by-step explanation:

When a die is rolled then probability of getting 6 is $=\frac{1}{6}$

So probability of winning the game $P(W)=\frac{1}{6}$

Now probability of losing the game $P(L)=1-\frac{1}{6}=\frac{5}{6}$

It is given that Alice starts the game

So either Alice win the game in 1st attempt or 3rd or 5th .....

So probability that Alice win the game $=P(W)+P(L)P(L)P(W)+P(L)P(L)P(L)P(L)P(W)+.....$

So $P(w)=\frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times\frac{5}{6} \times \frac{5}{6}\times \frac{1}{6}......$

Now taking $\frac{1}{6}$ common and applying sum of infinite GP $P(W)=\frac{1}{6}\times \frac{1}{1-\frac{25}{36}}=\frac{1}{6}\times \frac{36}{11}=\frac{6}{11}$

Probability of Alice win the game is $\frac{6}{11}$