# Calvin and Hobbes

Calvin and Hobbes play a match consisting of a series of games, where Calvin has probability $p$ of winning each game (independently). They play with a "win by two" rule: the first player to win two games more than his opponent wins the match. Find the probability that Calvin wins the match (in terms of $p$) in two different ways:

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Find the probability that Calvin wins the match (in terms of $p$) by conditioning, using the Law of Total Probability.

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Let $C$ be the event that Calvin wins the match. Consider the first two games. There are 4 possible outcomes, 3 of which result in Calvin not losing immediately. Therefore, we condition on each of these possibilities: Let $WW$ represent Calvin winning both games, and $WL$, $LW$ represent a win then a loss or a loss then a win respectively. Using LOTP, we get:

\begin{aligned} P(C) &= P(C | WW) P(WW) + P(C | WL) P (WL) + P(C | LW) P(LW) \\ &= p^2 + 2pq \cdot P(C) \\ &\Downarrow \\ P(C) &= \frac{p^2}{1 - 2pq} \\ &= \boxed{\frac{p^2}{p^2 + q^2}}\end{aligned}
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Find the probability that Calvin wins the match (in terms of $p$) by interpreting the problem as a gambler's ruin problem.

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We can interpret this problem as a Gambler's Ruin where each player starts out with $2. Calvin wins if he gets to$4. Therefore, we have $N = 4$ and $i = 2$, so the probability of Calvin winning is:

$\frac{1 - \frac{q}{p}^2}{1 - \frac{q}{p}^4} = \frac{p^2(p^2 - q^2)}{p^4 - q^4} =\boxed{ \frac{p^2}{p^2 + q^2}}$

which is indeed the same answer as before.

# Symmetry

For the following 2 questions, think about how symmetry may be used to avoid unnecessary calculations.

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Suppose $X$ and $Y$ are i.i.d. $Bin(n, p)$. What is $P(X < Y)$?

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First, we note that $P(X < Y) = P(Y < X)$ by symmetry, and that $P(X < Y) + P(X = Y) + P(Y < X) = 1$. Hence,

\begin{aligned} P(X < Y) &= \frac{1}{2}\left({1 - P(X = Y)}\right) \\ &= \frac{1}{2}\left({1 - \sum_{k=0}^n P(X = Y | X = k)P(X = k)}\right) \\ &= \frac{1}{2}\left({1 - \sum_{k=0}^n P(Y = k)P(X = k)}\right) \\ &= \frac{1}{2}\left({1 - \sum_{k=0}^n\left({\binom{n}{k}p^k(1 - p)^{n - k}}\right)^2}\right) \end{aligned}
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Can you construct two random variables X and Y both distributed ($3, \frac{1}{2}$) such that $P(X=Y)=0$?

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Yes, let $Y = 3 - X$. Then, there is no way that $X$ and $Y$ take on the same value because their sum would have to be 3.

# Counting Cards

In the game Texas Hold'em, players combine two of their cards that are hidden to everyone else with five community cards to make the best possible five-card hand. The game is played with a standard deck of 52 cards. A flush is where all 5 cards belong to the same suit.

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Suppose you are holding 2 spades in your hand, and there are 2 spades showing among the three community cards. What is the probability that you hit the flush?

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Since we currently see 4 spades, there are 9 other spades that can be used. Of the 52 cards in the deck, we know what 5 of them are, so we have 47 card values that remain to be seen. To find this probability, we can use the Hyper-Geometric distribution, specifically of $HGeom(9, 38, 2)$, since there are 9 desirable cards, 38 undesirable cards, and 2 cards that are being drawn. Now, we are interested in the probability that we observe either one or two spades in the two cards we draw. Or $P(X = 1) + P(X=2)$, where $X$ is distributed as above.

$\frac{{9 \choose 1}{38 \choose 1}}{{47 \choose 2}} + \frac{{9 \choose 2}}{{47 \choose 2}} = 0.3497$