The enigma that is variance / std dev and risk
June 6th, 2014 at 5:00:33 AM
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First off, I'm not super educated on statistics and whatnot, but think I have at least a basic understanding of it.
If I did the math correctly, it looks like 9/6 JoB, $0.50 denom, 10 lines, and playing $100,000 coin in (or 4,000 pulls), the EV is -446 and the standard deviation is $3,050. We know 68.2% of the time our actual results will land between -1 SD and +1 SD. -1 SD is EV-SD and 2 SD's is EV+SD (-3,496 to 2,604).
So here is where I'm confused (if what I wrote above is correct)...Is it actually true your results will land between -1 SD and +1 SD 68.2% of the time? Are we equally likely to be at -1 SD as we are to be at +1 SD? First, I think yes, but after thinking over it, I say no. A majority of the variance comes from the RF. When playing VP, for the most part, you are losing more than expected, and once you hit your RF, your actual is much closer to the EV that it should.
I think most sessions will be closer to -1 SD than +1 SD....but the few sessions that are closer to +1 SD, will outweigh the negative sessions (so the actual tends to EV).
Let's say I wanted to calculate the same thing, but figure out the std dev for playing 2 pulls (instead of 4000) as an extreme case. There are $50 of coin-in, SD is $68.xx and EV is -$0.233. So basically landing between -1 SD and +1 SD is going to be -$68 and +$68. Of course, you can't end up at -$68 if you only played $50 worth of action. I *believe* we could say 15.9% of the time we'll end up lower than -1 SD (loss greater than $68). But that, again, doesn't make sense, since it's impossible to ever be outside of -1 SD for this 2 pull scenario. This begs the question -- how many pulls are required in a sample to say "this will happen X% of the time"? In other words, do I need 10 pulls, 1K pulls, 100K pulls in a sample to say "15.9% of the time my results will be outside of -1 SD"? Because clearly, it doesn't work for 2 pulls.
On a similar note, what would you say the risk is in a play? For the first example (10 lines, $0.50 denom, $100,000 coin in, 4000 pulls), how much money would you say is at risk? Risk meaning how much money I'm risking to lose. At first glance, I think of risk being associated with EV, and for this one, it'd be -$446. After thinking about it for a while, that doesn't quite make sense (I don't think?), since there would be risk involved even if the game was +EV (say, full pay deuces wild, where EV is +$762). And the risk is not $100,000, since you pretty much have 0% chance of losing all $100K coin in. 1] The risk cannot be 0 or less, because you are clearly risking money. 2] It cannot realistically be $100,000. 3] And risk cannot be EV, since if it was a +EV play, that'd mean the risk is negative, which also cannot be true because statement 1. So...what would be the associated risk with such a play? PS: This is not a play that I'm considering, just an example.
Again, I'm not the best with statistics and all that kinda stuff, so really appreciate any help and guidance on the subject. Thanks for reading!
If I did the math correctly, it looks like 9/6 JoB, $0.50 denom, 10 lines, and playing $100,000 coin in (or 4,000 pulls), the EV is -446 and the standard deviation is $3,050. We know 68.2% of the time our actual results will land between -1 SD and +1 SD. -1 SD is EV-SD and 2 SD's is EV+SD (-3,496 to 2,604).
So here is where I'm confused (if what I wrote above is correct)...Is it actually true your results will land between -1 SD and +1 SD 68.2% of the time? Are we equally likely to be at -1 SD as we are to be at +1 SD? First, I think yes, but after thinking over it, I say no. A majority of the variance comes from the RF. When playing VP, for the most part, you are losing more than expected, and once you hit your RF, your actual is much closer to the EV that it should.
I think most sessions will be closer to -1 SD than +1 SD....but the few sessions that are closer to +1 SD, will outweigh the negative sessions (so the actual tends to EV).
Let's say I wanted to calculate the same thing, but figure out the std dev for playing 2 pulls (instead of 4000) as an extreme case. There are $50 of coin-in, SD is $68.xx and EV is -$0.233. So basically landing between -1 SD and +1 SD is going to be -$68 and +$68. Of course, you can't end up at -$68 if you only played $50 worth of action. I *believe* we could say 15.9% of the time we'll end up lower than -1 SD (loss greater than $68). But that, again, doesn't make sense, since it's impossible to ever be outside of -1 SD for this 2 pull scenario. This begs the question -- how many pulls are required in a sample to say "this will happen X% of the time"? In other words, do I need 10 pulls, 1K pulls, 100K pulls in a sample to say "15.9% of the time my results will be outside of -1 SD"? Because clearly, it doesn't work for 2 pulls.
On a similar note, what would you say the risk is in a play? For the first example (10 lines, $0.50 denom, $100,000 coin in, 4000 pulls), how much money would you say is at risk? Risk meaning how much money I'm risking to lose. At first glance, I think of risk being associated with EV, and for this one, it'd be -$446. After thinking about it for a while, that doesn't quite make sense (I don't think?), since there would be risk involved even if the game was +EV (say, full pay deuces wild, where EV is +$762). And the risk is not $100,000, since you pretty much have 0% chance of losing all $100K coin in. 1] The risk cannot be 0 or less, because you are clearly risking money. 2] It cannot realistically be $100,000. 3] And risk cannot be EV, since if it was a +EV play, that'd mean the risk is negative, which also cannot be true because statement 1. So...what would be the associated risk with such a play? PS: This is not a play that I'm considering, just an example.
Again, I'm not the best with statistics and all that kinda stuff, so really appreciate any help and guidance on the subject. Thanks for reading!
June 6th, 2014 at 5:07:23 AM
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I have absolutely nothing constructive to add to this but I am also curious to the answer. I was thinking about going for the high level card this year because of a few benefits if would give me and this is a question I have had for a little while now.
Expect the worst and you will never be disappointed.
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June 6th, 2014 at 8:27:37 AM
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The statement above about 34% of the distribution lying on both sides of the mean applies to a normal distribution. A normal distribution is symmetric, you are as likely to be on one side of the mean as the other. Vp has a positively skewed distribution. However, as you play more hands, the distribution of your results looks more and more like a normal. This is the central limit theorem.
