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 09-03-2010, 11:11 PM #1 bubbba82 journeyman   Join Date: Jan 2009 Location: newport pembrokeshire Posts: 337 How does standard deviation relate to poker? Is there a thread or can someone help me work out how standard deviation relates to poker? ( a guide for dummies)
09-04-2010, 01:17 AM   #2
Donk_Vader
journeyman

Join Date: Jun 2010
Posts: 272
Re: How does standard deviation relate to poker?

Quote:
 Originally Posted by bubbba82 Is there a thread or can someone help me work out how standard deviation relates to poker? ( a guide for dummies)
Your equity in a hand = X% which standard deviation can be figured for?

 09-04-2010, 02:14 AM #3 halftilt veteran   Join Date: Jul 2010 Location: no FDs, no OESDs, FINAL DESTINATION Posts: 2,399 Re: How does standard deviation relate to poker? The only place in poker where standard deviation is really relevant is when you're trying to figure out how good/bad you're running.
 09-04-2010, 04:56 AM #4 Huggy veteran     Join Date: Nov 2006 Location: Tag on the streets Lag under sheets Posts: 2,919 Re: How does standard deviation relate to poker? wtf, does anyone here know what SD is?
 09-04-2010, 06:34 AM #5 jewbinson Pooh-Bah     Join Date: Jul 2009 Posts: 4,247 Re: How does standard deviation relate to poker? SD is always a number. Variance = (SD) squared, so variance is also a number SD can be a different number depending on what you want it to represent. The SD of your "actual winrate" is a number, say, X, which measures how good or bad you have run (i.e. lots of bad beats = bad run here) over whatever sample you take. Your "expected winrate" is what your winrate is expected to be if poker were variance-free (i.e. no upswings or downswings, but you always run at EV). Experiencing or giving lots of bad beats from (resp. to) villains will yield a large X relative to your winrate. "Winning the pots you are supposed to" roughly speaking will yield a small X relative to your winrate. But this is only one type of SD in measuring spread in poker. There are others such as "SD of your actual winnings" or "SD of how card dead you are (i.e. are you card dead for like 400 hands, then run good for 400 hundred hands, or are you card dead the amount you are "supposed to be")". The most important point to make is that you cannot control your variance in any of these cases, apart from the variance of your winrate, which can be reduced by playing much tighter. But if you're too tight, then your winrate is not likely to be very positive because you're easily exploitable by LAGs or loose-passives. You do need to worry about variance BEFORE YOU PLAY and this is taken into account when people advise you to stick to a 20+BI bankroll if you play 100BB deep poker. Once you agree to this principle, there is nothing you can do. You cannot control variance. If you are "scared of variance", then you don't understand the concept of risk-ruin, which is what poker is all about. But now I'm going into BR management so I think I'll stop
09-04-2010, 09:32 AM   #6
bubbba82
journeyman

Join Date: Jan 2009
Location: newport pembrokeshire
Posts: 337
Re: How does standard deviation relate to poker?

Quote:
 Originally Posted by jewbinson SD is always a number. Variance = (SD) squared, so variance is also a number SD can be a different number depending on what you want it to represent. The SD of your "actual winrate" is a number, say, X, which measures how good or bad you have run (i.e. lots of bad beats = bad run here) over whatever sample you take. Your "expected winrate" is what your winrate is expected to be if poker were variance-free (i.e. no upswings or downswings, but you always run at EV). Experiencing or giving lots of bad beats from (resp. to) villains will yield a large X relative to your winrate. "Winning the pots you are supposed to" roughly speaking will yield a small X relative to your winrate. But this is only one type of SD in measuring spread in poker. There are others such as "SD of your actual winnings" or "SD of how card dead you are (i.e. are you card dead for like 400 hands, then run good for 400 hundred hands, or are you card dead the amount you are "supposed to be")". The most important point to make is that you cannot control your variance in any of these cases, apart from the variance of your winrate, which can be reduced by playing much tighter. But if you're too tight, then your winrate is not likely to be very positive because you're easily exploitable by LAGs or loose-passives. You do need to worry about variance BEFORE YOU PLAY and this is taken into account when people advise you to stick to a 20+BI bankroll if you play 100BB deep poker. Once you agree to this principle, there is nothing you can do. You cannot control variance. If you are "scared of variance", then you don't understand the concept of risk-ruin, which is what poker is all about. But now I'm going into BR management so I think I'll stop
perfect ty. So does a bell curve have anything to do with this at all?

 09-04-2010, 10:42 AM #7 Calhoun137 enthusiast   Join Date: Jan 2009 Posts: 63 Re: How does standard deviation relate to poker? Ok here's the best way to think about it. Let's say we are flipping coins, and each time it lands heads you give me a dollar, and each time it's tails I give you a dollar. Then let's keep track of how much money I have made or lost. On average I expect to make nothing, but in practice I will most likely lose or win a finite amount. In fact if we flip the coin N times, then I am most likely to win or lose somewhere around the square root of N dollars. Why is this true? It's actually easy to show. Let's call my profit/loss M. Then M starts at zero, and the average change in M after one or more flips is zero. Here is the clever trick: let's consider the average change in M squared. Well after one flip it's quite clear that this will be 1, since +1 and -1 squared are both equal to 1. Notice that after one flip the average profit/loss is zero, but in reality I either gained or lost a dollar. M squared however is 1, so you can see it sort of measures the fluctuations in my profit/loss. Call M(N) my profit/loss after N flips. Then what is M(N+1) squared? Since I can either gain or lose 1 dollar, we have either M(N+1) = M(N) + 1 or M(N+1) = M(N) - 1 lets square these two expressions and take the average, which I write as ^2. If you do this, which is quite easy, you will find that: ^2 = ^2 + 1 If you think about it, this implies that after N flips, the average of M squared will be N; therefore the square root of the average of M squared is root N, which is just a fancy way of saying the SD for flipping a coin N times is root N. (to answer your question about the "bell curve" the answer is that the SD of a bell curve is half the width of the curve at half maximum height, the bell curve is called a Gaussian and it's really special, see the law of large numbers. For coin flips, as N becomes super large, the probability looks like a bell curve) Last edited by Calhoun137; 09-04-2010 at 10:53 AM.
 09-04-2010, 11:51 AM #8 Calhoun137 enthusiast   Join Date: Jan 2009 Posts: 63 Re: How does standard deviation relate to poker? for any graph of EV vs # of hands played, u can compute the SD. It just tells u how your profits/losses fluctuate. If your sample size is N hands, where N is a very large number; then your SD will be about root N if you mostly get into coin flips, and will be better or worse than root N depending on how good/bad you get your money in. So while SD tells you how good/bad you are running (by comparing your actual profit/loss fluctuations to the SD), it's absolute value can also tell you how good/bad you are getting it in. (If you want to play with the equations, try setting the probability in my previous to post to like 2:1 odds and see how the SD changes) Last edited by Calhoun137; 09-04-2010 at 12:04 PM.
09-04-2010, 01:08 PM   #9
bubbba82
journeyman

Join Date: Jan 2009
Location: newport pembrokeshire
Posts: 337
Re: How does standard deviation relate to poker?

Quote:
 Originally Posted by Calhoun137 Ok here's the best way to think about it. Let's say we are flipping coins, and each time it lands heads you give me a dollar, and each time it's tails I give you a dollar. Then let's keep track of how much money I have made or lost. On average I expect to make nothing, but in practice I will most likely lose or win a finite amount. In fact if we flip the coin N times, then I am most likely to win or lose somewhere around the square root of N dollars. Why is this true? It's actually easy to show. Let's call my profit/loss M. Then M starts at zero, and the average change in M after one or more flips is zero. Here is the clever trick: let's consider the average change in M squared. Well after one flip it's quite clear that this will be 1, since +1 and -1 squared are both equal to 1. Notice that after one flip the average profit/loss is zero, but in reality I either gained or lost a dollar. M squared however is 1, so you can see it sort of measures the fluctuations in my profit/loss. Call M(N) my profit/loss after N flips. Then what is M(N+1) squared? Since I can either gain or lose 1 dollar, we have either M(N+1) = M(N) + 1 or M(N+1) = M(N) - 1 lets square these two expressions and take the average, which I write as ^2. If you do this, which is quite easy, you will find that: ^2 = ^2 + 1 If you think about it, this implies that after N flips, the average of M squared will be N; therefore the square root of the average of M squared is root N, which is just a fancy way of saying the SD for flipping a coin N times is root N. (to answer your question about the "bell curve" the answer is that the SD of a bell curve is half the width of the curve at half maximum height, the bell curve is called a Gaussian and it's really special, see the law of large numbers. For coin flips, as N becomes super large, the probability looks like a bell curve)
This is pretty awsome ty .

So am i wrong to think of SD like this -

good luck = 0
bad luck = 1

over infinite number of hands there would be equal 0's and equal 1's
and it would run out like - 01010101010101010101 (long term)

So if im running bad in a cash game the numbers would run like -
111111101111011110111 (short term)
If im running good the numbers would like -
0000000101010000000101000000

But over a lifetime of hands played the good luck and bad luck equals out?

Im not very academic so imn trying to translate to what i think in a as easy terms as possible.

This is great too because i think the people who understand standard deviation
Come to terms with variance alot easier.

 09-04-2010, 02:21 PM #10 Calhoun137 enthusiast   Join Date: Jan 2009 Posts: 63 Re: How does standard deviation relate to poker? Just because I love this stuff, let me also mention that we can consider things like the average of M^3 and M^4 and so on for different numbers of flips N. Each of these powers is called a "moment" of the probability distribution, and they are fundamental in statistics. The first moment is the average, the second moment is the variance, and so on... (the square root of the variance is the SD)

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