Probability of A winning the series

[thecat]

Two opponents A and B play a series of matches. The one who gets 12 first wins the series. At the moment the result is 6x4 in favor of A. What is the probability of A winning the series knowing that in each match the odds of A and B winning are 0.4 and 0.6 respectively?

I had solved this problem by negative binomial distribution, but another solution says the distribution below is correct. I couldn't understand why, since the game should end as soon as A won 6 games. Thank you in advance.

 

[Romsek]

Yeah I don't agree with that solution. The correct solution I believe would be

\(\displaystyle P[\text{A wins in $k+1$ games}] = \dbinom{k}{5}(0.4)^5(0.6)^{k-5}\cdot (0.4),~k \in [5,12]\)

\(\displaystyle P[\text{A wins}] = \sum \limits_{k=5}^{12} \dbinom{k}{5} (0.4)^6 (0.6)^{k-5} \approx 0.425604\)

I.e. the various ways A can win 5 games out of k, and then win the k+1 game.

Oddly enough their solution provides the same numerical answer.