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Best of Donald Catlin

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Taking Advantage of an Advantage: Part 2 - How Not To

6 March 2011

Last month I wrote an introduction that postulated a simple game with a probability p of winning and a probability q of losing where p was larger than q. Think of a biased coin. The expected return e for this game is e = p - q. If our stake after k bets is Sk, what should the size of our next bet be in order to best take advantage of our situation?

We let f be the fraction of our current stake that we are willing to risk; this is known as proportional betting. The number f is obviously a number between 0 and 1. After k + 1 bets our stake Sk+1 will either be Sk + fSk with probability p or Sk - fSk with probability q. Calculating the expected stake for our k + 1st step we have

exp(Sk + 1) = pexp(Sk + fSk) + qexp(Sk - fSk)

which is equivalent to

exp(Sk+1) = (p + q)exp(Sk) +(p - q)exp(Sk)

Recalling that p + q = 1 and p - q = e we then have

exp(Sk+1) = (1 + ef)exp(Sk) (1)

If our starting stake is S0 then applying (1) repeatedly, after n steps we would have

exp(Sn) = (1 + fe)nS0 (2)

In analyzing most gambling games, it is typical to maximize our expected return. Since e is positive it would seem that we can maximize our expected return here by taking f equal to 1. This means betting everything each time.

Whoa, not so fast. Using our example from last month, p = 0.51 and S0 = 100, if we made 10 bets we would either win all ten bets with probability (0.51)10 or we would go broke. This means our probability of winning would be approximately 0.0012 and our probability of losing would be 0.9988. Of course if we succeed we will win $204,600 but the overwhelming likelihood is that we will lose our $100. Not very good is it?

Let's back off a bit. How about setting f = 0.9? Well, in principle one cannot go broke since one always has a bit of money left, but in practical terms as soon as the stake falls below one unit, the player can no longer bet and is in essence broke. Using our same example, let's see how this would work in ten bets.

Let me define a path as a string of the characters W and L standing for win and loss, respectively. Thus LWLWWLWWLW would represent 10 bets with 6 wins and 4 losses. I will call a path a ruin path if at any point the stake falls below 1 unit; otherwise it is called a non-ruin path. I should like to count the number of non-ruin paths, take the probability of each, add them up, and thereby obtain the probability of not falling below a unit bet at any point.

If one starts with S0 units, then for w wins and l losses with w + l = n, the stake would be

Sn = (1 + f)w(1 – f)lS0 (3)

So for 6 wins and 4 losses in our example we would have

S10 =(1.9)6(0.1)4100 = 0.47

Yikes! All such paths are ruin paths. In some of them the ruin point is reached before the 10th step. I am not going to bother you with the details of the counting but the following table describes the situation.

# Wins

All Paths

Non-Ruin Paths

Prob./Path

Probability

1

1

1

(.51)10

0.001190

9

10

10

(.51)9(.49)

0.011437

8

45

45

(.51)8(.49)2

0.049450

7

120

100

(.51)7(.49)3

0.105579

6 or less

848

0

irrelevant

0.000000

Totals

1024

156

---

0.167656

Figure 1
Non-ruin probability with f = 0.9

Subtracting the figure in the lower right corner from 1 we see that the probability of our stake falling below 1 at some point is 0.832344. The wins are big but not very likely, much like a state lottery. In 10 plays your expected win is $19.53 and your most likely outcome is that you're broke.

Very well, let's not be so greedy. How about f = 0.3? Here the situation looks better. It turns out that for 10 plays there are no ruin paths at all; the probability of ruin is zero. Now that's more like it.

Or is it? Let's calculate our expected stake after 10 plays using formula (2). It turns out to be approximately $106.16. Think about that. Here you have a nice positive game and you are going to play to win $6.16.

If you have a positive game you want to play it for a long time and make as much as you can. Suppose we play 200 times with f = 0.3.

Well, there are too many paths to count them as we did in Figure 1. Let's look at the collection of most likely paths. Since the probabilities of winning and losing are 0.51 and 0.49, respectively, the most likely scenario is 102 wins and 98 losses. (Please don't write to me about the Arc Sine Law. I am aware of it.)

S200 = (1.3)102(0.7)98100 = 0.02766

Oh no! All of these are ruin paths. As n is raised higher and higher the proportion of ruin paths rise and the probability of ruin as we have defined it rises as well. Playing a positive game isn't as easy as it seemed, is it? Is there any value of f that makes sense? I'll address that next month. See you then.


Don Catlin can be reached at 711cat@comcast.net

Donald Catlin

Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers