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Johan
stranger

Reged: 11/06/03
Posts: 10
Loc: The Netherlands Normal distribution of hourly rate? I, for one, don't think so.....
#682745 - 05/13/04 12:11 PM Edit Reply Quote Quick Reply

T.W.I.M.C.

I was writing an article about how much bankroll was needed at a certain standard deviation with a certain win rate...
The formula i came across was the following.

Bankroll needed = -(standard deviation^2/2*avg, win per hour)ln(the risk you are willing to accept).

On first sight, this formula looks ok. I found it several times on the internet, so I assumed the general consensus was that this formule did what i wanted to do.

However, someone pointed out to me that this formula accually assumes a normal distribution of the hourly outcomes around the avg per hour! Since my pokertracker doesn't have enough hands in it to be able to say something meaningfull about this i have no way of saying if this is accually the case.

The longer I think about it, the more i think this formula is wrong, the distribution should at least be different for different types of players (loose agressive has bigger wins, bigger losses than tight aggrissive, though problaby larger wins than losses, even when the hourly rate is set to 0, also i could imagine a player with lots of small losses and few Big wins, the formula wouldn't be (entirely)
correct.).

Would someone with either a lot of math experience or someone who does have the 200000 or so hands in some kind of database be so kind to shed some light on this? This is quite an important concept I think.

Johan Rensink

Edited by Johan (05/13/04 12:12 PM)

 Post Extras:   Johan
stranger

Reged: 11/06/03
Posts: 10
Loc: The Netherlands Re: Normal distribution of hourly rate? I, for one, don't think so..... [Re: Johan]
#682830 - 05/13/04 12:45 PM Edit Reply Quote Quick Reply

Or, doesn't the distribution matter and is the formula always correct?.... now I'm confused.

 Post Extras:   Robk
veteran Reged: 10/28/02
Posts: 1242
Loc: Chicago Re: Normal distribution of hourly rate? I, for one, don't think so..... [Re: Johan]
#682893 - 05/13/04 01:10 PM Edit Reply Quote Quick Reply

The formula is indeed correct. I think your misunderstanding lies in the role of the Central Limit Theorem, which roughly states that the sum of n independent, identically distributed random variables is approximately normal (the distribution convereges to normal as n increases). Thus we don't need to assume anything too special about the distribution of poker results. This thread should prove helpful (it includes a link to a derivation).

Edited by Robk (05/13/04 01:12 PM)

 Post Extras:   BruceZ
Pooh-Bah

Reged: 09/03/02
Posts: 1636 Re: Normal distribution of hourly rate? I, for one, don't think so..... [Re: Johan]
#683150 - 05/13/04 02:47 PM Edit Reply Quote Quick Reply

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

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 = ½ + WR/2b
q = ½ - 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.

 Post Extras:   Johan
stranger

Reged: 11/06/03
Posts: 10
Loc: The Netherlands Re: Normal distribution of hourly rate? I, for one, don't think so..... [Re: Robk]
#683178 - 05/13/04 03:03 PM Edit Reply Quote Quick Reply

Robk,

I think you are correct for blackjack, since blackjack is played "perfectly". Poker however doesn't have that, since the same conditions never occur twice, nor does it have a calculable avg. win rate given number of hands an hour and a certain strategy.

So I think this formula was too easily copied from the blackjack world. I'm still not convinced that <i>pokerresults</i> are normally distributed. This is, in my opinion one of the game's caracteristics, then again who am I? I'm no math pro or something like that, maybe we should ask mason.... Johan

 Post Extras:   Jimbo
Pooh-Bah Reged: 09/02/02
Posts: 2193
Loc: Planet Earth but relocating Re: Normal distribution of hourly rate? I, for one, don't think so..... [Re: Johan]
#684219 - 05/13/04 11:54 PM Edit Reply Quote Quick Reply

Quote:

I'm no math pro or something like that, maybe we should ask mason....

I would normally never laugh at someone who wrote this but when you are questioning BruceZ it becomes absolutly hilarious!

Jimbo

 Post Extras:   Johan
stranger

Reged: 11/06/03
Posts: 10
Loc: The Netherlands Re: Normal distribution of hourly rate? I, for one, don't think so..... [Re: Jimbo]
#684606 - 05/14/04 08:19 AM Edit Reply Quote Quick Reply

try working out the order the messages were posted in, i didn't react to BruceZ 's post yet, i will momentarily.

bye,

Johan

 Post Extras:   Johan
stranger

Reged: 11/06/03
Posts: 10
Loc: The Netherlands Re: Normal distribution of hourly rate? I, for one, don't think so..... [Re: BruceZ]
#684639 - 05/14/04 08:50 AM Edit Reply Quote Quick Reply

Quote:

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.

Thanks, it really had me thinking, especially because I could imagine two distributions with the same stddev and averages. But one with a lot of small losses, and one normally distributed. I felt (no mathametical proof though)that the small losses one had a bigger change of losing it all.

I haven't had the time (and i will need time ) to read the utah article, but i will as soon as possible. I've read the introduction and it indeed should tell me everything i want to know.

Thanks for taking the time to explain this to me, great post!

Johan

Edited by Johan (05/14/04 08:53 AM)

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