Rejoining the San Diego Padres

A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem . Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way. Standard notation: probability of A given B is written $ \Pr(A \mid B) $ probability of B is written $ \Pr(B) $ Bayes' Theorem: Using the notation above, Bayes' Theorem can be written: \[ \Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \] Let's apply Bayes' Theorem to the Monty Hall problem . If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random. Assume we pick Door 1 an...

Simplified Multinomial Kelly

Here's a simplified version for optimal Kelly bets when you have multiple outcomes (e.g. horse races). The Smoczynski & Tomkins algorithm, which is explained here (or in the original paper): https://en.wikipedia.org/wiki/Kelly_criterion#Multiple_horses Let's say there's a wager that, for every $1 you bet, will return a profit of $b if you win. Let the probability of winning be $p$, and losing be $q=1-p$. The original Kelly criterion says to wager only if $b\cdot p-q > 0$ (the expected value is positive), and in this case to wager a fraction $ \frac{b\cdot p-q}{b} $ of your bankroll. But in a horse race, how do you decide which set of outcomes are favorable to bet on? It's tricky, because these wagers are mutually exclusive i.e. you can win at most one. It turns out there's a simple and intuitive method to find which bets are favorable: 1) Look at $ b\cdot p-q$ for every horse. 2) Pick any horse for which $ b\cdot p-q > 0$ and mar...

Probability and Cumulative Dice Sums

Let a die be labeled with increasing positive integers $a_1,\ldots , a_n$, and let the probability of getting $a_i$ be $p_i>0$. We start at 0 and roll the die, adding whatever number we get to the current total. If ${\rm Pr}(N)$ is the probability that at some point we achieve the sum $N$, then $\lim_{N \to \infty} {\rm Pr}(N)$ exists and equals $1/\rm{E}(X)$ iff $(a_1, \ldots, a_n) = 1$. The direction $\implies$ is obvious. Now, if the limit exists it must equal $1/{\rm E}(X)$ by Chebyshev's inequality, so we only need to show that the limit exists assuming that $(a_1, \ldots, a_n) = 1$. We have the recursive relationship \[{\rm Pr}(N) = p_1 {\rm Pr}(N-a_1) + \ldots + p_n {\rm Pr}(N-a_n);\] the characteristic polynomial is therefore \[f(x) = x^{a_n} - \left(p_1 x^{(a_n-a_1)} + \ldots + p_n\right).\] This clearly has the root $x=1$. Next note \[ f'(1) = a_n - \sum_{i=1}^{n} p_i a_n + \sum_{i=1}^{n} p_i a_i = \rm{E}(X) > 0 ,\] hence this root is als...

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Probability and Cumulative Dice Sums