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General Gambling >> Probability

BruceZ
Pooh-Bah


Reged: 09/03/02
Posts: 1636
Re: Normal distribution of hourly rate? I, for one, don't think so.....
      05/13/04 02:47 PM

That is the correct risk of ruin formula. It does not assume that your hourly results are normally distributed. It is based on the assumption that the risk of ruin can be determined accurately based solely on your win rate and standard deviation. That is, the higher moments of the distribution can be ignored (the mean or win rate is related to the first moment, and the standard deviation is related to the second moment). This has been found to be a robust assumption for gambling games such as blackjack and poker. If you would really like to see a thorough analysis of this assumption, you could refer to this article, but this is for advanced and highly determined readers only. You can also refer to the thread that Robk linked to above for more of a hand waving discussion.

The central idea is that we can construct a coin flip game in which a single coin flip has the same win rate and standard deviation as a single hand of poker or blackjack. Certainly the payoff distributions of the coin flip and the poker hand are not the same; however, if they start out with the same mean and standard deviation, they will always have the same mean and standard deviation after any number of flips and hands N. Moreover, the central limit theorem guarantees us that for N sufficiently large, both games will be well approximated by the same distribution, namely a normal distribution with the same mean and standard deviation. We can then simply determine the risk of ruin for this coin flip game, and this will be taken as the risk of ruin for our game of poker or blackjack. This will be true so long as the number of hands N is relatively short so that the "fine structure" of the game is negligible as far as risk of ruin is concerned. This fine structure is related to the higher order moments of the probability distribution function which are found to not come into play over time periods of length N over which the central limit theorem applies. This is the only assumption that is necessary, and this is the assumption which the above mentioned lengthy paper addresses.

The derivation which I linked to earlier assumes some prior knowledge of the risk of ruin formula for coin flip games. It also contains a slight error in the construction of the coin flip game as pointed out by Ralle, where sigma is used instead of sqrt(sigma^2 + E^2). The end result is still correct since this approximation is made later. This latter approximation need not be made at all, it just serves to greatly simplify the final formula while still being highly precise for most practical cases of interest. For these reasons, I have written down my own derivation which assumes no prior knowledge of risk of ruin problems, and which does not make the above approximation.


Derivation of bankroll and risk of ruin formula

Let’s start with something really simple.
Coin flip, P(win) = p, P(lose) = q = 1-p
$1/flip, bankroll = $1, opponent is infinitely wealthy.

What is risk of ruin r?

r = q + (1 – q)*r^2.

This is the first neat trick. It says that we can go bust one of two different ways.. We can lose on the first flip with probability q and be finished, or we can win the first flip with probability 1-q, and then proceed to lose a $2 bankroll with probability r^2 since this is the probability of losing $1 twice.

Solving for r:

r = q/(1-q) = q/p.

What if our bankroll were B? Then to bust we would have to lose $1 B times, so

r = (q/p)^B

Now what if the bet size were b? Then our bankroll would be B/b bets, and

r = (q/p)^(B/b) (eq. 1)

Now here is the main idea. Let’s see if we can set p and the bet size b so that the EV and variance of the coin flip game is the same as the win rate WR and sigma^2 of our poker game.

Let: (eq. 2)

bet size b = sqrt(WR^2 + sigma^2)
p = ˝ + WR/2b
q = ˝ - WR/2b.

Note that this is the same form as the equations in Sileo's paper , but sigma has been (correctly) replaced by b. We can verify that

EV = p* b + q*(-b) = WR

and

Variance = p*b^2 + q*(-b)^2 – EV^2 = b^2 –WR^2 = sigma^2

as advertised. Substituting p and q into (eq. 1)

r = [ (1 – WR/b) / (1 + WR/b) ]^(B/b)

This is our formula for the risk of ruin for any game with win rate WR and standard deviation sigma. Remember that b is function of both WR and sigma from (eq. 2). This can be put into a simpler form. Take the log of both sides.

ln(r) = (B/b)[ ln(1-WR/b) – ln(1 + WR/b) ]

Since WR is much smaller than b, we can use the approximation ln(1 + epsilon) is approximately equal to 1 + epsilon, when epsilon is a small number. Then we have

ln(r) = (B/b)(1 – WR/b -1 –WR/b) = (B/b)(-2WR/b)

r = exp(-2WR*B/b^2).

Rearrangement gives the bankroll B needed for a risk of ruin r.

B = -b^2/2WR*ln(r).

Replacement of b by sigma for sigma >> E gives the formula in its familiar form.

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Entire topic
Subject Posted by Posted on
* Normal distribution of hourly rate? I, for one, don't think so..... Johan 05/13/04 12:11 PM
. * * Re: Normal distribution of hourly rate? I, for one, don't think so..... BruceZ   05/13/04 02:47 PM
. * * Re: Normal distribution of hourly rate? I, for one, don't think so..... Johan   05/14/04 08:50 AM
. * * Re: Normal distribution of hourly rate? I, for one, don't think so..... Robk   05/13/04 01:10 PM
. * * Re: Normal distribution of hourly rate? I, for one, don't think so..... Johan   05/13/04 03:03 PM
. * * Re: Normal distribution of hourly rate? I, for one, don't think so..... Jimbo   05/13/04 11:54 PM
. * * Re: Normal distribution of hourly rate? I, for one, don't think so..... Johan   05/14/04 08:19 AM
. * * Re: Normal distribution of hourly rate? I, for one, don't think so..... Johan   05/13/04 12:45 PM

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