Imagine we have a standard 52 card deck, with cards going up in value from $A=1, 2=2, …, K=13$. Imagine we randomly and uniformly draw three cards without replacement from the deck. We are tasked with finding the expected value of the sum of the three cards. By linearity of expectation, this is clearly 21.
However, imagine now that someone comes up to you and offers to buy the sum of the three cards for \$25 (where the expected value is still \$21, in dollar terms). You have, clearly, a \$4 theoretical edge. Let's say you have $100 in your bankroll/bank account. What fraction of that bankroll should you wager?
There's a theoretically correct answer involving the generalized Kelly criterion, but it's long-winded and complex and near-impossible to do by hand. If we wanted to quickly approximate this value, what would we do?
One thought I had was to use the traditional Kelly Criterion: $f^* = \frac{p}{a}-\frac{q}{b}$, where $f^*$ is the fraction of our bankroll to bet, $p$ is the probability of a win, $q=1-p$ is the probability of a loss, and $1+b$ and $1-a$ are the average values of an investment of \$1 if we win or lose, respectively.
But how would we quickly approximate those values without the use of a computer?
Best Answer
If I understand you correctly, you've already decided to approximate the generalized Kelly criterion with the basic ($2$-outcome) Kelly criterion, and you're just asking how to calculate $p, q, a, b$ so you can use the formula for $2$-outcome Kelly criterion. Is that right?
If so, and since you're already approximating, you might as well approximate $p, q, a, b$. One easy way that can be done without computers is to realize that, if you draw $3$ cards from a deck of $52$, the dependence among those $3$ cards are very low. So just treat it as $3$ separate bets of $1$ card each, where the opponent offers to pay you $\$25/3 = \$8.33$ for that one card (whose average is $7$). Then trivially we have $p = 8/13, q = 5/13, b = 8.33 - 4.5, a = 11 - 8.33$. This gives the $f^*$ value for the $1$-card game, but it's a fractional value (of your bankroll), so I imagine (with another approximation) you can just use the same fraction for the $3$-card game.