Probability :Two gamblers

[Fahim Shahriar]

Two gamblers A and B play a game throwing two ordinary dice. A wins if he obtains the sum 6 before B obtains 7, and B wins if he obtains the sum 7 before A obtains 6. Which of the players has a better chance of winning if player A starts the game?

Show your full solution so that it can be understood easily.

[Fahim Shahriar]

A few months ago I couldn't solve that one. But now I can.
Learning...
Prob of $A = \dfrac {5}{36} = \dfrac {180}{36^2}$

Prob of $B = \dfrac {6}{36} \times \dfrac {31}{36} = \dfrac {186}{36^2}$

So $B$ has better chance.

[SMMamun]

Does 'Prob of $A$' in your calculation mean the probability of A's winning or the probability of just getting the sum 6 in any throw of the dices? Your conclusion is right, but not the reasoning you provided in your comment, especially because the dice throwing may go on for round after round before one is finally the winner, but you have considered only two rounds of throwing in total.

[Fahim Shahriar]

You're right. I thought it to be a two round game. Even I missed the first round. Then I have to calculate the probability for every possible rounds. It's full of calculations but not difficult. Isn't there any one liner solution ? I saw someone giving the solution in just one line in fb but can't remember.