Person X and Y throw a Fair die one after another. Whoever throws 6 first wins. What is the probability that X wins?

Question

Two people take turns rolling a fair die. Person $X$ rolls first, then person $Y$ , then $X$, and so on. The winner is the first to roll a $6$. What is the probability that person $X$ wins?

Our teacher in class solved this question as follows:

Probability of winning in 1st round = Probability that 6 occurs = 1/6

So Probability of $Y$ losing in first round becomes $= 5/6$

Where it is assumed that X starts the game.

suppose if X throws and if 6 comes then the game will stop and X will win. but if X loses then Y will throw the die (here Y will have to lose because according to question we want X to win the game) and then X will throw . and so on.

it will be like $$X+ \neg X\neg YX+ \neg X\neg Y\neg X\neg YX+\dots$$

HERE $X$ denotes X wins and $\neg X$ means X loses i.e. 6 does not come . similarly $\neg Y$ means Y loses.

This will form an infinite geometric progression series. Converting into probabilities we have $$(1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(1/6)+ \dots$$ The sum of this geometric series is the answer.

MY question is: how the probability of $X$ in $2nd$ toss is $(5/6)(5/6)(1/6)$ and similarly in the following tosses? Why these terms are in multiplication not addition?

Answer 1

Let $p$ be the probability that player $1$ (X) wins.

Then, the probability that player $2$ wins is $$\frac{5}{6}\cdot p$$

Both events are mutually excluded and one of them must happen, so we have

$$p+\frac{5}{6}p=1$$

This gives $$p=\frac{6}{11}$$

Answer 2

For X to win on their second toss, three things need to happen:

1. X doesn't get a six on their first toss (else X wins earlier)
2. Y doesn't get a six on their first toss (else Y wins)
3. X does get a six on their second toss.

Now these things have probability $\frac56,\frac56,\frac16$ respectively. Also, they are independent events - what happens on one toss of the die doesn't affect what will happen on the next. (Here we can think of the players continuing to roll until X's second toss, whether or not the game is already over.) So the probability of all three happening is the product of the individual probabilities $\frac56\times\frac56\times\frac16$.