## HCOL 195 11/4/09

Today we discussed “gambler’s ruin.” This is the idea that if we bet over and over small amounts, we will under some circumstances eventually go bust. I mentioned it in connection with the following scenario: Someone has borrowed money from a loan shark. The loan shark wants his money and interest back, and threatens to break the legs of the borrower if he doesn’t come up with the \$200 owed tomorrow. The man has only \$100, and his only plan is to go to the casino and attempt to win another \$100. What should he do?

A student suggested that he should bet the whole \$100 on one spin of the roulette wheel (odds of winning are 18/38 at an American casino). That was felt to have the greatest probability of success.

So we considered the idea of betting against a casino that has an infinite amount of resources. I drew a picture on the board: Gambler's Ruin

In the diagram, we imagine that we start with \$1 and bet it. The probability that we win is p=18/38, and the probability that we lose is 20/38. If we lose, we bust. If we win, we will go from \$1 to \$2. But we still might go bust, for example, we might lose the next two tries and end up with nothing, or we might bounce around a bit but eventually end up with nothing. If P1 is the probability that we eventually go bust, starting with \$1, and P2 is the probability that we eventually go bust, starting with \$2, then

P1=q+p*P2,

that is, we either bust immediately with probability q, or we obtain a second unit but eventually bust with probability P2.

But P2=P12. The reason is that if we have \$2, then (since the bets don’t know how much money we have), the probability that we will eventually have a situation of having \$1 is the same as the probability that we will bust if we start with \$1, since it’s just the probability that eventually we will have \$1 less than we have now. Therefore,

P1=q+p*P12

This is a quadratic equation for P1. There are two roots: 1 and q/p. If q>p then that root is not a possible root (since probabilities have to be no bigger than 1), and the only solution would be 1. But that means that in a real casino, where q/p>1, the only usable root is 1 and if we start with \$1 and keep playing indefinitely, we will eventually bust. But that means that no matter how much money we start with, we will eventually bust, because if we start with \$100, say, the probability is 1 that we will at some point have \$99 (\$1 less than we have now). But then the probability is 1 that we will at some point have \$98, then \$97, and so forth until we bust. We cannot beat the casino because the odds are in its favor.

The calculation of the roots is shown below and then on the right side of the board in the picture above. Gambler's Ruin 2

Next, we consider the situation where the casino has a finite amount of money (or equivalently where we have m dollars and want to gamble until we get (m+n) dollars, then quit with enough money to pay off the loan shark). So, we would like to calculate the probability that if we start with m dollars, and gamble \$1 at a time, that at some point in the future we will have (m+n) dollars. We can calculate this, using what we have done with the infinite casino. That’s because if Q is the probability of never getting to (m+n) dollars, that’s the same as the probability that we eventually bust without ever getting to (m+n) dollars, and (1-Q) would be the probability that we do eventually get to (m+n) dollars. But that means that

Pm=Q+(1-Q)Pm+n, since there are two ways to bust against an infinite casino, starting with m dollars: Either bust without ever getting to (m+n) dollars (that’s Q), or get to (m+n) dollars (probability 1-Q) and then go bust against the infinite casino (probability Pm+n). From what we’ve already learned, the probability of busting against an infinite casino, starting with m dollars, is rm where r is the smaller of 1 or (q/p). That yields the equations on the last chart. But for us, (q/p)>1 so that doesn’t help us to find out the probability of starting with m dollars, and gambling \$1 at a time eventually getting to (m+n) dollars. But if we reverse the roles of the gambler and the casino, the probability that we will get to (m+n) dollars is the same as the probability of a casino that has n dollars busting against an infinite casino. That exchanges the roles of m and n, and of (q/p) and (p/q), and a similar formula works. That’s shown on the last chart. Gambler's Ruin 4

If we evaluate this for just one bet, the probabililty that we win is just 18/38=0.474. If we divide our bet in two so that m=n=2, the probability of our winning is less than this, which confirms the student’s intuition that BOLD PLAY, that is, betting the maximum amount, is the best way to keep our legs from getting broken. So bet the \$100 on one spin of the wheel, and hope for the best.

I remarked that this is why the lottery isn’t a good way to plan for your retirement. If you were to bet \$40 a week for your entire life, your expected net worth from gambling at the end would be \$80. (In class I said \$40, but I checked it out; still not a good retirement plan!)