HCOL 196, April 1, 2011

Here is a copy of the study sheet for the second quiz. We will start discussing it next week, probably on Wednesday.

On Monday I plan to discuss personal finances and planning for your economic future.

Today we discussed the question: Suppose you have to raise a large amount of cash immediately, and have only half the amount needed. If you do not raise the cash, dire consequences will result. The only way to raise the cash is to go to the casino in New York and gamble the amount you have. Should you bet it all at once, or gamble a little at a time?

I intended this to be in the context of a roulette wheel, with a fixed payoff of 18/38. One student noted that the payoff in Blackjack or Poker is variable. Poker is a game of skill, and I do not think that the casino in New York has poker. They must have blackjack, but the times that the advantage is to the player are occasional and you can’t just go into a blackjack game an expect the odds to be in your favor at that time. And, when the odds are in your favor, they are generally only a bit in your favor. So the point of the exercise is to look at a fixed payoff situation.

In the picture below, if you have one unit ($1 in the picture, but it could be any fixed amount), then the probability of doubling your money is p, and the probablility of busting immediately is q, where p+q=1.

But if you get to $2, you could still bust, eventually, even if you went to $10, $15, or even more. Call the probability of eventually busting, starting at $1 (whatever happens in the interim) to be P1. Similarly, the probability of eventually busting, starting at $2, is P2. There are only two ways to bust, starting at $1. One is to bust immediately (probability q), and the other is to bust by winning the first bet (probability p), getting you to $2, and then eventually busting starting at $2 (probability P2). These are independent events, so the total probability is just the sum. Thus, P1=q+p*P2.

But, P2=P12. To understand this, suppose you are at $2. The probability of eventually busting is P2. But if you are at $2, then the probability of eventually getting to $1 is P1, since that is just the probability of ending up with one dollar less than you have now. (The casino doesn’t know how many dollars you have in your pocket; that is illustrated by the red dashed line in the above picture). But if you get to $1, the probability of your busting is P1, so the probability of starting at $2, eventually getting to $1, and then eventually busting, is the product of those two probabilities, that is, P12.

Similar reasoning shows that Pn=P1n for any integer n.

Replacing P2 with P12 gives us a quadratic equation for P1, shown on the chart below.

There are two roots, 1 and q/p, as shown below.

But q/p can’t be a probability for us because (for the roulette wheel) it is greater than 1, and probabilities cannot be greater than 1. So, for us, since we aren’t the casino, our probability of busting if we just bet and bet and bet is 1.

We then asked, if you start with $M, and want to end up with $(M+N), and then stop, what is the probability of doing this? Suppose Q is the probability of eventually busting without getting to $(M+N), and (1-Q) is the probability of getting to $(M+N). Then the probability of your eventually busting (if you don’t stop gambling when you get to $(M+N)) is the sum of your probability of busting without getting to $(M+N), which is Q, plus the probability of getting to $(M+N), which is (1-Q), times the probability of eventually busting from $(M+N), which is PM+N.

But that doesn’t work, since for us the only root of the quadratic equation that we can use is 1. But if we look at it from the point of view of the casino, things change. The casino’s probability of winning a bet is 20/38 so that q/p for the casino is 18/20, which is less than 1. So that is the figure to use. The casino’s goal is to make you lose $M, that is, to start out with $N and end up with $(M+N) without first losing $N (which, if you are sticking to the plan to quit the game once you get to $(M+N), would end the game).

This leads to a similar equation that expresses PN in terms of the probability of the casino “busting” (losing $N), which is \tilde{Q} in the equation and the probability of getting to $(M+N) at which point the game ends (because you have run out of money). But the probability that the casino loses $N is the same as the probability that you win $N. So that is the probability we need.

This leads to a simple equation for \tilde{Q}, shown below, where r=p/q. When we plugged numbers in, we found that the probability of eventually going from $M to $(M+N) is greater if M=N=1 than if M=N=2. The larger M and N are, the lower the probability that you’ll get to $(M+N).

Here’s the calculation for M=N=2.


So, in this situation, BOLD PLAY is the best strategy.

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