A Strategic Fallacy

Brian Alspach

Poker Digest Vol. 4, No. 23, November 2 - 15, 2001

Recently, I came across Bib Ladder and Willie the Hummer in an animated conversation. When Bib saw me, he signalled for me to come over to their table.

``Professor, Willie here has an interesting strategy he says is giving him lots of winning sessions in roulette.''

``That's right, Professor, I think I've found a nice little way to beat roulette,'' interjected Willie.

``Tell me about it,'' I replied.

``There's gotta be a catch,'' murmered Bib, earning him a glare from Willie.

Willie began, ``I start by betting $15 on both red and black. If a red number comes, I bet $14 on red and $16 on black the next time. If a black number comes, I bet $14 on black and $16 on red. I keep doing the same shifting. Whenever a red number comes, I reduce the amount I bet on red by one, and increase the amount I bet on black by one; if a black number comes, I reduce the amount I bet on black by one, and increase the amount I bet on red by one.''

He completed his discourse by saying, ``Whenever you get back to betting $15 on each color, you've got a tidy profit, and since the number of red and black outcomes is the same in the long run, this is a good strategy.''

``Willie, are you playing on a wheel with a single 0, or a wheel with a 0 and 00?''

``I've found a wheel with a single 0,'' he replied.

``What do you do when a 0 comes up?'' I asked.

``I just swallow the loss of all my chips and make the same bet I just made.''

``Willie,'' I told him, ``I've seen this strategy before. Let's first look at a simpler game than roulette. Consider flipping an unbiased coin. The probability of heads is 1/2 as is the probability of tails. Suppose you are offered even money for betting either heads or tails. If you follow your betting strategy, it's easy to verify that you will be ahead $m after m rounds of betting if you are back to betting $15 on both H and T. This means that heads and tails have occurred the same number of times. In fact, even reaching a stage where the number of heads and tails is almost the same is sufficient to give you a profit.''

I continued by working out what the profit would be for the different possibilities after 20 plays. The following table contains that information. The first column shows the split between the number of occurrences of H and T. The second column shows the profit, where a minus sign indicates a loss. The third column shows the number of ways the number of occurrences can split as shown in the first column.

Split Profit Number
10,10 20 184,756
11,9 16 335,920
12,8 4 251,940
13,7 -16 155,040
14,6 -44 77,520
15,5 -80 31,008
16,4 -124 9,690
17,3 -176 2,280
18,2 -236 380
19,1 -304 40
20,0 -380 2

``Willie, for each possible split, if you multiply the profit by the number of ways the split may occur, and then add all of them together, you obtain precisely zero. This is no surprise since the expectation for betting a single chip on a 50-50 even-money bet is zero. Betting more chips in different patterns does not allow you to escape the fact the expectation is zero. What your strategy does do is give you about a 74 percent chance of being ahead at the end of 20 plays because there are 772,616 ways of finishing ahead out of the total of 1,048,576 ways for the outcomes to occur.

``What we just did was for a game with zero expectation. Using this strategy for roulette with a single zero is going to be even worse. For example, if you play 37 times and get the best possible split of 18 reds, 18 blacks and one 0, your profit will be only 16. So you can see things are going downhill because that single 0 is going to come up now and then.

``You'd be better off giving up this strategy while you are ahead, and listen to Bib when he tries to convince you to concentrate on poker.''

Willie left muttering to himself while Bib was quietly saying to me, ``I knew there was a catch.''

``Bib, it's even worse than what I indicated because the entire scheme is based on a false premise.'' I then went on to explain the following to my old friend.

Let T(n), H(n) denote the number of tails and heads, respectively, after n flips of the coin. Succinctly, Willie's strategy is based on the fact of being nicely ahead whenever T(n) - H(n) is zero or close to zero.

He undoubtedly is confused about the true meaning of the phrase ``in the long run H(n) and T(n) are the same.'' It is true that as n grows larger, the probability of having H(m) = T(m) at some point m, for m smaller than or equal to n, approaches one. In other words, you are essentially certain of reaching a point where the same number of heads and tails has occurred.

The problem is that before you reach that point, you may have such a big discrepancy between the number of heads and tails that either you do not have enough chips to cover the betting pattern, or the number of chips you need to bet exceeds the table limit.

What is really meant by saying ``they are the same in the long run'' is that as n gets very large, the ratio H(n)/T (n) becomes very close to one with probability close to 1. This says nothing about the behavior of H(n)- T(n). The strategy requires some control over H(n) - T(n) and one does not get it because the ratio H(n)/T(n) behaves nicely in the long run. I think many people suffer the same confusion as Willie.

When you combine the fundamental flaw in Willie's premise together with the destructive power of a 0 or 00, you have a poor betting strategy for roulette.

Above, I mentioned how a wheel with a single zero diminishes his strategy considerably. Let's look at a wheel with both a 0 and 00.

The probability of a 0 or 00 coming up is 1/19. So over 19 plays of Willie's scheme with one appearance of a 0 or 00, and a perfect split of nine reds and nine blacks, he loses $2.

It certainly isn't pretty.

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