Count conditional probability of winning a game

Question

In a certain game of tennis, Alice has a 60% probability to win any given point against Bob. The player who gets to 4 points first wins the game, and points cannot end in a tie. What is Alice's probability to win the game?

When solving in terms of a random walk it gives a probability around 83%, but the actual solution for this particular problem is around 71%.

What's the difference between the methods used and how correctly solve such kind of problems?

Answer 1

There are 4 possible outcomes:

1. Alice wins $4-0$. There is only one way this happens, i.e. Alice wins 4 points. The probability of that is $0.6^4$.
2. Alice wins $4-1$. There are $4$ ways this can happen, Bob can win the first, second, third or fourth point. So the probability of this event is $4\cdot 0.6^4\cdot0.4$
3. Alice wins $4-2$. Bob needs to win $2$ of $6$ points for this, and he cannot win the last point, so he can win $2$ of the first $5$ points, the probability here is ${5\choose 2}\cdot 0.6^4\cdot0.4^2$
4. Alice wins $4-3$, the probability here is ${6\choose 3}\cdot 0.6^4\cdot 0.4^3$

Summing those up, you get $\approx0.7$. I cannot comment on the random walk method because you didn't give enough details for me to tell you where you went wrong. But the linked answer considers a completely different game in which a player needs an advantage of 2 points to win. Your game has no such condition and will always end in at most 7 points. The linked answer's game can be arbitrarily long (and if you watch tennis, they sometimes get very long)

Answer 2

Just as a generalization for the answer above, formula for getting N points in a game first

$\sum_{i=0}^{N-1}{N-1+i\choose i}\cdot p^N\cdot (1-p)^{i}$

So in our case we have N=4, p=0.6, q=1-p=0.4

$\sum_{i=0}^{3}{2+i\choose i}\cdot 0.6^4\cdot 0.4^{i}$

which gives us 0.71 as a result.