Probability - Gambler A and B

The first one is easy. There are four states (0,3), (1,2), (2,1), and (3,0). The first and last are terminal states, and the game starts in state (1,2). The probability of going from (1,2) to (0,3) is 1/3. The probability of going from (1,2) to (2,1) to (1,2) is 2/3 * 1/3 = 2/9. Therefore, the probability of A losing is 1/3 + 1/3*2/9 + 1/3*2/9*2/9 + ...

This is 1/3 * (1 + 2/9 + (2/9)^2 + (2/9)^3 + ...), which is 1/3 times the sum of a geometric series with r=2/9.
Therefore the probability is 1/3 * 1/(1-2/9) = 1/3 * 1/(7/9) = 1/3 * 9/7 = 3/7.

The second one would seem to be 1 since B has an infinite amount of money and can't get to 0. No matter how much money A has, there is a non-zero probability that consecutive losses will wipe him out.

The previous post is incorrect. When starting with (1,N), the probability of A losing is (2^N - 1) / (2^(N+1) - 1) which approaches 1/2 as N approaches infinity. Therefore, the second answer is 1/2.

For the third one, starting with (m,n), the probability of A losing is:
(2^n - 1) / (2^(m+n) - 1)

I don't have proofs for 2 and 3, but after modeling with Markov chains, it's pretty clear that the answers are correct.

     I am not sure that it is 1/2. The probability of A to lose on the first round is 1/2. The probability to win is also 1/2, thus becoming with 2 USD. The probability to loose them both in the next two rounds is 1/2*1/2=1/4 and the probability to have that outcome is 1/2(win)*1/2(lose)*1/2(lose)=1/8, from where follows that the probability to lose up to the 3rd round is 1/2 + 1/8. As you are so knowledgeable what is the probability of A to lose everything after two successive wins in the beginning (1/2*1/2)? IMV it must be 1/4*1/8, and the probability to A to become absolute loser at any time is 1/2 + 1/2.4 + 1/4.8 + 1/8.16 + 1/16.32 + ...

You're overlooking the fact that the probability of A winning any bet is 2/3 - not 1/2.

     We are talking about case ii), aren't we. You don't have 'any bet' - A is putting 1 USD on the table and B is putting 1 USD on the table - gambling (toss a coin or throw dice or s.th.) - and the outcome is either A wins 1 USD, or B wins 1USD - where do you see three outcomes here?

I can think of one obvious way to interpret this sentence:
"We assume the probability of A winning any betting is 2/3."

I don't know how you're coming up with something other than that. There are two outcomes: A wins or B wins. The probability that A wins is 2/3.

If my interpretation of that sentence is incorrect, what interpretation do you propose?

     O.K., it is in the assignment. Then let's see what we have after Awins on the 1st round, and then follows the worst case scenario ending up with losing everything: 1rst round - Awin = 2/3; Alose = 1/3; 2nd - Alose = 2/3.1/3; 3rd round: Alose=2/3.1/3.1/3. It must be something of the kind: 1/3 + 2/27 + 4/243 ... and further the polynomial starts acquiring negligibly small members of the series.