i'm trying to figure out how to solve this.
Here is the problem:
Two players face each other off. At each iteration, a player has a probability $p = 1/3$ to get a victory. There are no draws. The first player to reach 4 victories wins the entire tournament. What's the probability that the player wins the tournament?
I am trying to solve this using conditioning.
I define $Xi$ a as a Bernoulli random variable of parameter $p = 1/3$ $$ X = \sum_{i=0}^n X_i$$ and N the number of games in the tournament.
$N$ can be either $4,5,6 or 7$ and $X$ can be either 0,1,2,3(loses the tournament) or 4 (wins the tournament)
I am trying to calculate $P(X=4)$ i.e he wins the tournament
I conditioned $X$ on $N$ so $P(X=4) = P(X=4 | N=4)*P(N=4) + .... + P(X=4 | N=7)*P(N=7)$
I am not sure i calculated $P(N=n)$ and the conditional $X$ probabilities but here is what i tried
$$P(X=4 | N=4) = p^4$$
$$P(X=4 | N=5) = \binom{5}{4}p^4(1-p)$$
$$P(X=4 | N=6) = \binom{6}{4}p^4(1-p)^2$$
$$P(X=4 | N=7) = \binom{7}{4}p^4(1-p)^3$$
and for $P(N=n)$, in order to have n games, either the player won 4 times or lost 4 times for n between 4 and 7
hence
$$P(N=4) = p^4 + (1-p)^4$$
$$P(N=5) = \binom{5}{4}p^4(1-p) + \binom{5}{1}p(1-p)^4$$
$$P(N=6) = \binom{6}{4}p^4(1-p)^2 + \binom{6}{2}p^2(1-p)^4$$
$$P(N=7) = \binom{7}{4}p^4(1-p)^3 + \binom{7}{3}p^3(1-p)^4$$
And then i just use $P(X=4) = P(X=4 | N=4)*P(N=4) + .... + P(X=4 | N=7)*P(N=7)$ and find $P(X=4) = 0.1009$
It doesn't seem right because $P(4\le N\le7)$ isn't equal to 1 and neither is $P(X=4)+P(X\le3)$
I might have calculated my conditional probabilities wrong somewhere over here, please help me out, thank you in advance
When you do the arithmetic, $$ \sum_{w=4}^7 \left( \frac13 \right)^w \left( \frac23 \right)^{7-w} \binom7w= \frac{379}{2187} \approx 0.1733 $$