January 17, 2007

My Roulette System

Well now, Dan comes back from vacation and steals my spot at the top of the blog! With real posts about minimum wages and such things no less! I'll put a stop to this!

In my last post about gambling I failed to really make clear what my system entails. This really didn't help anybody offer an evaluation of the system. Without further ado:

The game is Roulette*. I play the outside, which means I don't put chips directly on any numbers but play groups. In particular, the roulette board is divided into three collumns. Betting successfully on a collumn pays 3-to-1. You can also bet on half the field(the two zeros are excluded from both the collumns and the evens/odds and red/black) with a pay rate of 2-to-1.

When I play, I bet two equal stacks of chips, one on each of two collumns. This means if I win I receive three stacks of chips back, covering my bet and giving me the same payout as if I had bet half the board. On the one hand this gives me better odds than betting red/black/odd/even as I'm covering slightly less than 2/3 of the board. It also means if I lose I lose twice as much.

Now, when I do lose I double down, placing double the original bet on the same collumns. If I lose again the bet is redoubled and again placed on the same collumns. This process is repeated until one of my two collumns wins, bringing me all the money that I'd lost in that stretch and one stack the size of the original bet. If I win on a particular spin, I move my bets to the other two collumns. I admit that this doesn't effect my odds of winning, but at the very least it gives me something to do and assuages that nagging feeling that it *has* to be more likely to hit on the other two collumns after a string of hits on one of them. I know, not true, but it sure makes sense and if it doesn't hurt my odds than it really doesn't matter.

There are obvious problems with this which were entailed in the Gambler's Falacy and Martingale System pages linked in the previous post. A very long string of losses is not impossible, even if it may be unlikely. Given the doubling of bets, a sufficiently long string of losses would bankrupt the player eventually.

The difference I think between my system and the Martingale system it that the latter seems to assume, based on the Wikipedia page, that the odds of success on a given bet are slightly less than 50%. That being the case it makes sense that over time you would eventually lose all your money. But my system has odds in my favor in execess of 60%. This in particular is where I need help from math inclined folks. Do the higher odds of success in my model make it viable where the Martingale system is not? It is granted that a player using my system needs a bank to draw from far in excess of the minimum bet they are using (I figure that a bank of $400-$500 for a table with a $15 minimum bet is sufficiently "safe"). It is also granted that given limited funds this system could be beated by a bad streak of luck, say hitting the same collumn and/or the zeros four or five times in a row. But I do think that the odds of such a streak are pretty poor and represent probably the best odds you can get in a casino.

Any thoughts?

*It should be noted at the outset that I always play the table with the lowest minimum bet and I *always* bet the minimum, except as specified in the explanation.


Noumena said...

First, in order to end up ahead by the original bet when you win, you'll need to triple down instead of doubling down. (NB I calculated all this under the assumption that your bet is not returned if you win. In other words, if you bet $1 and the pay is 2-to-1, you're only ahead $1 on a win. After finishing all this, I realised that your bet may be returned if you win, so a $1 bet that pays 2-to-1 might mean you're ahead $2. If this is the case, then I'd have to rework the numbers to be sure of my conclusions, but I don't expect them to be really different.)

Suppose you just double down, and m is the minimum bet. The first round you bet 2m. Say you lose, so you're now down by 2m. Next round you double your bet, and bet 4m. You've now invested 6m. If you win, you only win (4m/2)*3 = 6m, so you've broken even. Say you lose round two, so you're now down 6m. On round three you bet 8m, and you've invested 6m+8m = 14m. If you win, you win (8m/2) = 12m, so now you're down by 2m. Things will only get worse as the rounds progress.

On the other hand, suppose you triple down. The first round you bet 2m. On round two you bet 6m, with a total investment of 8m. A win gets you (6m/2)*3 = 9m, for a net profit of m. Similarly, if you lose on round two and win with a bet of 18m on round three, you net (18m/2)*3 - 26m = m.

Now, with a minimum bet of $15 and a pot of $390, you have 26m to work with. Since you're tripling your bet on a lose, you can afford to follow your strategy for three rounds (betting 2m, 6m, and 18m, for a total investment of 26m, or $390) before going broke. So, to evaluate your strategy, we need to calculate the expectation value for playing it over three rounds.

The expectation value is the sum of the products of probabilities times the payoff. For example, suppose we are playing a game where we bet $1 and a fair coin is flipped. If the coin lands heads, we lose our bet, for a net outcome of -1. If the coin lands tails, we win our bet, and are paid $2, for a net outcome of 1. Hence the expectation value is (1/2)*(-1) + (1/2)*(1) = 0. (Incidentally, a fair game is defined to be one whose expectation value is 0.)

Things are slightly more complicated with the roulette strategy, because we have several statistically independent events (the different rounds). Suppose, on any given round, the probability of winning is p. (Keeping p variable will make it easier to compare fair roulette and actual roulette later.) The first round, then, the probability is p of winning, with a net profit of m. There's also a (1-p) probability of losing, so that we proceed to round two. On round two, there's again a probability of p of winning back the total investment with a net profit of m, and a (1-p) probability of losing. And similarly for round three -- though if we lose there, the net `profit' is -26m.

Putting all three rounds together, we end up with the following formula:

p*m + (1-p)*(p*m + (1-p)*(p*m + (1-p)*(-26m)).

Now, to evaluate this for real roulette, we plug in m = 15 and p = 24/38 (the bet covers 24 spaces on the wheel, out of the total of 38). This is gives us an expectation value of about -$5.25. $0 would be breaking even, so, for every initial $30 bet, on average, you will end up down by about $5.25. Another way to put this is, you can expect to be able to place that initial $30 bet about 71 times before you're too broke to afford the buffet dinner. Incidentally, the odds of getting the evening-ending three-in-a-row loss is 1/27, or about 3.7%.

By way of comparison, if you bet m on a single number and the payoff is 35-to-1, the expectation value is

(1/38)*(35m) + (37/38)*(-m),

which, for m=$15, is about -$0.79. If you bet m on each of two numbers under the same conditions, the expectation value is

(2/38)*(34m) + (36/38)*(-2m)

which is about -$1.58. That makes for about 171 turns of the wheel.

So, Ben, I'm afraid that your strategy is actually quite a bit worse than just betting on single numbers. There seems to be two fundamental problems with your strategy. First, playing it out more than a few rounds requires an enormous amount of working capital (round four requires a bet of 54m, or $810, and that's on top of the $390 you've already invested). Second, the expectation value actually gets worse the more rounds you can play, because the cost of staying in rises so rapidly.

Finally, for ideal or fair roulette, we take away spaces 0 and 00, which give the house its advantage. Then p = 24/36 = 2/3, and we have an expectation value of $0, whatever value m happens to take.

MosBen said...

You're right, I'm not ahead after a successful win following losses. I'm merely winning back all the losses. If I bet $2 in the first round, $1 per collumn, then bet $4 in the second because I lost, I make back $6 if I win, which covers the first and second round of betting.

I'll address the rest of your analysis when I've got a spare moment at work.

Anonymous said...

Noumena said: "Things are slightly more complicated with the roulette strategy, because we have several statistically independent events (the different rounds)." Emphasis mine, for no particular reason.

It seems I have further misunderstood your system, Ben. I thought that when you lost on rnd 1, you increased your bet such that a win on rnd 2 would put you where you would have been if you had lost rnd 1. I see now that that isn't the case.

With that in mind, and with the benefit of Noumena's calculations, I conclude that I think your system is for sucks.

Anonymous said...

Bugger. My comment doesn't make sense. I meant that I thought that Ben's system was to bet enough on rnd 2 that winning would give him the result he would have had if he had won on rnd 1.