Thread: a dice game

A and B are playing a dice game with one dice. A starts the game first. 6 will win. What is
the probability that A will win the game?

I found the answer by the help of 1/1-n,using series and limit. The answer is 6/11. But there is another solution and I can't understand the logic of it.The solution is as follows:

A starts first.The probability that A will win is "x".
The probability that B will win is 5
-----.x
6
x + (5/6)x= 1 x=6/11
There is also another solution:
x= (1/6)(1) + (5/6)(5/6)u

x[1 - 25/36] = 1/6

x(11/36) = 1/6

x = 6/11

Originally Posted by serhanbener

A and B are playing a dice game with one dice. A starts the game first. 6 will win. What is
the probability that A will win the game?

Here is another way.

Originally Posted by Plato

Here is another way.

Yes. I had also used this method.

P(A) = 1/6 + (5/6)(5/6)(1/6) + (5/6)^4.(1/6) + (5/6)^6.(1/6) + ...

= (1/6).[1 + (5/6)^2 + (5/6)^4 + (5/6)^6 + .... to infinity]

1 1 36
= (1/6)----------- = (1/6)--------- = (1/6).------
1 - (5/6)^2 1 - 25/36 36-25

P(A) = 6/11

I want to learn the logic of the solutions that I have mentioned in #".

In the second alternative solution I think they mean this. A either wins on first throw or wins but not on first throw. For first throw p=1/6. For not on first throw
p=(5/6)(5/6)x (This is because after A and B have failed on first throw its like starting a new match so probability of A winning overall is still x)
So x=(1/6)+(5/6)(5/6)x

I think you are right.Many Thanks.

Originally Posted by serhanbener

A and B are playing a dice game with one dice.
A starts the game first. 6 will win. What is the probability that A will win the game?

Try this version (A starts, then B 2 in a row, then A 2 in a row....) :
A B B A A B B .....

Originally Posted by Wilmer

Try this version (A starts, then B 2 in a row, then A 2 in a row....) :
A B B A A B B .....

I am sorry. I couldn't understand.