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Probability of winning a game
If $X$ is the number of wins a player needs to win the game, what is the probability that the player wins the game in the first $10$ rounds, i.e. does he win with the minimum number of wins.
I tried to solve it by $Y = X + W$, where $W$ is a uniform random variable for the number of losses, and so $Y \sim B(2, \frac12)$. But that would imply that $X \sim B(10, \frac{1}{2})$ but not $X \sim B(2, \frac12)$, and that’s the difference I don’t see.
A:
$Y$ is a binomial random variable with parameters $2$ and $1/2$ hence $X$ is a binomial random variable with parameters $2$ and $1/2$.
Let $A$ be the event that $X$ is a number of wins that is smaller than $10$.
$$P(X>10)=1-P(X\le 10)\\=1-P(A^c)\tag1$$
Using the definition of the binomial distribution, we have:
$$P(A^c)=P(Y\le 9)=P(\max(Y,0)\le 9)\\=P(\text{win}\oplus \text{loss}\le9)\tag2$$
using the fact that $Y$ and $X$ are independent.
Notice that the only possible events that change the probability for
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