Probability that Alice Wins n Games Before Bob Wins m Games
Problem 746
Alice and Bob play some game against each other. The probability that Alice wins one game is $p$. Assume that each game is independent.
If Alice wins $n$ games before Bob wins $m$ games, then Alice becomes the champion of the game. What is the probability that Alice becomes the champion.
If Alice won at least $n$ games in the first $n+m-1$ games, then Bob won at most $m-1$. On the other hand, if Alice won at most $n-1$ games in the first $n+m-1$, then Bob won at least $m$ games. Hence, it is necessary and sufficient for Alice to be the champion that Alice wins at least $n$ games in the first $n+m-1$ games.
(Remark that even if Alice won $n$ games before finishing $n+m-1$ games, we may assume that they continue playing until $n+m-1$ games.)
The probability that Alice wins exactly $k$ games in the first $n+m-1$ games is given by
\[{n+m-1 \choose k} p^k(1-p)^{n+m-k-1}\]
since each game is independent.
Therefore, the probability that Alice becomes the champion is obtained by summing this probability over $k=n, n+1, \dots, n+m-1$.
Hence, we get
\[P(\text{Alice becomes the champion}) = \sum_{k=n}^{n+m-1}{n+m-1 \choose k} p^k(1-p)^{n+m-k-1}.\]
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(1) At least one die lands on the face 5 in the first $n$ rolls.
(2) Exactly $k$ dice land on the face 5 […]
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Let $E$ and $F$ be independent events. Let $F^c$ be the complement of $F$.
Prove that $E$ and $F^c$ are independent as well.
Solution.
Note that $E\cap F$ and $E \cap F^c$ are disjoint and $E = (E \cap F) \cup (E \cap F^c)$. It follows that
\[P(E) = P(E \cap F) + P(E […]
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For each pair of these events, determine whether […]
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Let $E$ be the event that the selected card is a king and let $F$ be the event that it is a heart.
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Definition of Independence
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Assuming that each toss is independent of each other, what is the probability that Alice tossed the coin exactly three […]
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Let $n$ be an integer greater than $1$. Let $k$ be a natural number with $k\leq n$. Then prove the formula
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\[E(X+Y) = E(X) + E(Y).\]
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Note that the […]