Simulation for An Interview Question

I was asked this question last year, and saw someone post it in misbbs yesterday.

The question is: A and B toss a dice by turns, the person who first gets 6 will win. Now A toss it first. What is the probability of A winning?

The answer should be: P(A)=6/11 and P(B)=5/11.　We need to use conditional probability to solve it.

Solution:　for first run, if A get 6, prob is 1/6. If A fails, prob is 5/6; then B get 6 is 1/6. So the prob of B win is A fails and B gets 6, which is 5/6*1/6=5/36.

For the second run, both A B fials in the first run, then A get 6, it is 5/6*5/6*1/6. For B, it is 5/6*5/6*5/6*1/6.

And so on.

So final A wins is: 1/6 + 5/6*5/6*1/6 * (5/6*5/6)^2*1/6 +... = (1/6)/(1-5/6*5/6) = 6/11.

Or we can see every time P(A) = 5/6 * P(B) and P(A) + P(B)=1. So P(A)=6/11.

It can be verified by this small program:

### veiry the dice toss question

rm(list=ls())
cat("\014")

rec=rep(-1,1000000)

for (jj in 1:1000000){
s=sample(6,1)
i=1
while (s != 6){
  s=sample(6,1)
  i=i+1
}
  rec[jj]=i
}

sum(rec %% 2 ==1)/1000000

The result is 0.545172 which is close to 6/11=0.5454545.