One evening with nothing else much to do, we caught a re-run of CSI. One of the characters learned a lesson about probability the hard way. He took a large pot of money and proceeded to start with a small bet. The idea was that if he lost, he would double his bet on the next hand, and so on. If we assume (generously) that the odds are essentially even, then sooner or later even after a long losing streak, he would win a bundle. The show didn’t say what he would do after a winning hand because the poor guy ran into a losing streak that was long enough to break his bank. What he did next was the basis of the rest of the show.

Does this strategy actually work? That is, if you have an infinite amount of money to bet, and follow the strategy of doubling after a losing bet, regardless of what you do on a winning bet, can you come out ahead? Well, things get sticky whenever infinities (yes, there are more than one infinity) are involved. The answer for a finite amount of money to start is that you will, on the average, lose it all in a normal casino where the odds are always against you. The interesting thing is that you will lose it faster than by continuously making the minimum bet. So if you enjoy gambling, this strategy is actually bad because you not only lose your money, you don’t have as much fun doing it!

That result should be obvious. After all, someone has to pay for all those lights in Las Vegas. However, the proper analysis of cumulative betting games are not always so obvious. Since I am as lazy as the next person, instead of trying to invent a new betting system as an example, I did the obvious and looked on Google for something relevant. This site had what I was looking for, but it was not cast in the form of a casino game. So I will re-word it a bit.

Assume you are playing 21 at a table with rules such that you have an even chance of winning. The table is unusual in that it has a $1 minimum but no maximum. Also the betting in not standard. If you win on the first hand, you get $2. If you don’t win on the first hand, but win on the second hand, you get $4. If you don’t win until the third hand, you get $8 and so on. The sequence starts over every time you win.

The question is: What do you have to bet to play this game fairly? Consider the payoff at each hand and the probability of getting that payoff. Interestingly enough, the payoff times the probability for each hand is the same! In each case the probability times the payoff is equal to $1. ($2 X 0.5; $4 X 0.25 etc.) That looks like a fair game would have you betting a dollar for the first hand, and simply throwing in another dollar for each additional hand until you win. Obviously if you stay in long enough, you will always be guaranteed a huge payoff for a relatively small investment because the payoff grows so much faster than the accumulated bet. Is that right? In fact, you probably would benefit the most from a long losing streak.

That doesn’t seem right. Is it?

This is posted while I’m on vacation, so I might not respond to communications for a while.

In response to the interest my original tutorial generated, I have completely rewritten and expanded it. Check out the tutorial availability through Lockergnome. The new version is over 100 pages long with chapters that alternate between discussion of the theoretical aspects and puzzles just for the fun of it. Puzzle lovers will be glad to know that I included an answers section that includes discussions as to why the answer is correct and how it was obtained. Most of the material has appeared in these columns, but some is new. Most of the discussions are expanded compared to what they were in the original column format.

Tags: decision theory, losing streak, big win, puzzle, casino, betting strategy