Let W stand for the winning of a game and L for losing it. Then there are 4 mutually exclusive possibilities

(i) W, W, W

(ii) W, W, L, W

(iii) W, L, W, W

(iv) L, W, W, W.

[Note that case (i) includes both the cases whether he losses or wins the fourth game.]

By the given conditions of the question, the probabilities for (i), (ii), (iii) and (iv) respectively are

Hence the required probability

=

[Since the probability of winning the game if previous game was also won is 2/(1+2) = 2/3 and the probability of winning the game if previous game was a loss is 1/(1+2) = 1/3].