A tough variance problem (at least for me)

[TNixon]

[ QUOTE ]
Having a downswing of a particular size is very different from losing that amount in a fixed number of hands.

[/ QUOTE ]
Did you read my example? I *think* I'm calculating a close approximation of what I'm trying to get at there. If the example doesn't work, can you explain why, and how to get a real approximation?

[ QUOTE ]
Losing $x starting at hand 1 is far from independent from losing $x starting at hand 2.

[/ QUOTE ]
Again, I'm not sure I understand what you're saying here. I mean, I understand what you're saying, I'm just not seeing how it's relevant.

Is it not reasonable to say that, at any given point, the probability of your current progress being within one standard deviation of the mean is 68%? And if it is reasonable to say that, then it also seems reasonable to carry that probability across multiple hands, so that over a 2 hand span, the probability of being within one standard deviation for the entire span would be 46.24%?

[pzhon]

[ QUOTE ]

Did you read my example? I *think* I'm calculating a close approximation of what I'm trying to get at there. If the example doesn't work, can you explain why, and how to get a real approximation?

[/ QUOTE ]
I read your example. It was a confused mess, and the method does not work. Sorry, I don't have time to explain why if my previous comments, and the comments of others, are not enough despite their relevance. Maybe you should read those more carefully and take some time to think about it, and the explanations I have given for why it isn't easy to get an approximation.

[jason1990]

This thread might be of interest to you.

[ncray]

I just ran some simulations, but they were a 5 minute hack job, so let me know if you spot any mistakes. Still, I hope they shed some light on the problem.
#Let's just say you win a $3.22 with probability .1
#and the times you don't win the pot, you lose $.18 with probability .9.
#Then for a given hand your expectation is .1*3.22-.9*.18 = .16
#Your standard deviation is (.1*(3.22-.16)^2+.9*(-.18-.16)^2)^(1/2)=1.02

There are an infinite number of choices of p, the amount you win when u win a pot, and the amount you lose when you lose a pot, but I just took this one particular vector to run simulations on. My question(s) were "what is the prob. of a downswing of $q at any point during a 1000 hand run?" I let q vary in increments of $2 to $2...$30. My code in R is at the bottom.

[1] 2 (q = $2)
[1] 1 (the prob. of a downswing of $2 over 1k hands is 1 (not exactly, but rounded internally in R to I think around 15 sig fig accuracy))
[1] 4 (for q = $4)
[1] 1 (the prob. is still 1)
[1] 6 (for q = $6)
[1] 0.877 (the prob is .877...)
[1] 8
[1] 0.475
[1] 10
[1] 0.196
[1] 12
[1] 0.05
[1] 14
[1] 0.02
[1] 16
[1] 0.005
[1] 18
[1] 0.003
[1] 20
[1] 0
[1] 22
[1] 0
[1] 24
[1] 0
[1] 26
[1] 0
[1] 28
[1] 0
[1] 30
[1] 0

What to take away from this? It is extremely unlikely with the above choices of variables to encounter a downswing of more than $20 over a 1000 hand run.


for(q in 2*1:15)
{
data = matrix(ncol=1,nrow=1000)
for(j in 1:1000)
{
max = 0
cur = 0
flag = 0
for (i in 1:1000)
{
cur = cur + sample(c(3.22,-.16),1, prob = c(.1,.9))
if (cur > max) max = cur
if (cur < max - q) flag = 1
}
data[j]=flag
}
print(q)
print(mean(data))
}

[TNixon]

[ QUOTE ]
It was a confused mess, and the method does not work. Sorry, I don't have time to explain why if my previous comments, and the comments of others, are not enough despite their relevance.

[/ QUOTE ]
Nobody else has commented on it at all, and your comments were not exactly self-explanatory.

My example was based on this very simple chain of logic:

[ QUOTE ]
Is it not reasonable to say that, at any given point, the probability of your current progress being within one standard deviation of the mean is 68%? And if it is reasonable to say that, then it also seems reasonable to carry that probability across multiple hands, so that over a 2 hand span, the probability of being within one standard deviation for the entire span would be 46.24%?

[/ QUOTE ]

There are only two links in the chain there, but if they hold true, then the method I used would be a valid approximation (given that I'm willing to adjust the question slightly). If one of the links is bad, then it shouldn't be a great exercise in thought or time to point out which one and why.

I'm more than happy to admit I'm in over my head here, and would be very happy to be given some real direction, rather than just being told "that doesn't work, but I don't really feel like telling you why".

[TNixon]

[ QUOTE ]
Having a downswing of a particular size is very different from losing that amount in a fixed number of hands.

[/ QUOTE ]
Ok, I think I understand why you pointed this out, and either my thinking is severely off, but I still don't understand why (likely, like I said, I'm way over my head here, just trying to figure things out), or you didn't take much more time examining what I was trying to do in that example than you did explaining why it was wrong and missed out on the fact that I knew it would only be an approximation that actually answers a different question. (not quite so likely, but it's entirely possible that I didn't sufficiently explain what I was trying to do).

So let me try to be as thorough as possible in explaining my thought process, to make it as easy and as painless as possible to poke the relevant holes:

First of all, in the interest of what appears to be a fairly simple approximation, I'm willing to modify the problem from chance of an absolute downswing to the chance of being a certain amount below expectation.

For a string of 2,-4,-2,3,-6,2,4,2,4,-3, the average expectation is 0.2/hand. At hand 1, we would be 1.8 above expectation, at hand 2, 2.4 below expectation, at hand 3, 4.66 below expectation, and at hand 5, 6.2 below expectation.

This is obviously only a very loose approximation to what most people think of as a "downswing", since at hand 2 we're at a "downswing" of -4, at hand 3, a swing of -6, and hand 5, a swing of -9, but I'm willing to find 6.2 instead of 9, and 4.6 instead of 6.

It's a very rough approximation, but the question itself isn't exactly a precise one to begin with, and this is better than nothing at all.

My thought was that by changing the question in this way, it could be simplified to a level where I was actually capable of figuring out an answer, by trying to determine the probability of being a certain distance away from expectation instead of the actual size of the swing, and I *believe* the example I laid out did just that. It obviously didn't solve the original problem, but I think it did solve this modified version of the problem. If you disagree, feel free to do so with a little more substance than "that doesn't work."

Of course, people far smarter than me seem to have already solved the original problem. Thanks for that link, jason1990. [img]/images/graemlins/smile.gif[/img]

[TNixon]

Ok, after using jason's formula on my example (which is a little bit difficult to begin with, since my "approximation" attempt didn't require a winrate), it seems my attempt at approximation is so far off as to be worse than useless.

I don't understand why, but obviously there's some massive hole in my logic or my math. Ah well. It sounded good. I thought. [img]/images/graemlins/smile.gif[/img]

[TNixon]

[ QUOTE ]
I just ran some simulations, but they were a 5 minute hack job, so let me know if you spot any mistakes. Still, I hope they shed some light on the problem.

[/ QUOTE ]
Is your simulator measuring the same thing as jason's formula in the other thread?

If so, I should be able to get the numbers to match at least somewhat closely (they'll be different, of course, but should be at least in the same neighborhood), but I can't, which means there's a bug in the simulator, a problem in the formula, or that they're both really measuring different things. I'm not sure which of those is true, but I think one of them has to be.

It's also entirely possible I mangled the formula getting it into excel, but I was able to match jason's results from the other thread putting in his numbers, so that seems somewhat unlikely.

[TNixon]

Since I'm being harassed by a regular from this forum who I am convinced is a complete idiot, I would very much appreciate any comments on the validity of this post from the heads-up forum.

The thread containing this post was asking about the likelihood and frequency of downswings of various sizes while playing NLHE heads-up cash, with 20 buy-ins being a common example.

Obviously I gave up on my own (very misguided) attempts to approximate this problem, so all I've done here is plugged numbers into a formula provided by jason1990. Comments on both the validity of the formula itself, as well as comments on my application of the resulting math, would be greatly appreciated.

Thank you

-----------------------

According to a formula generated in this thread:

http://forumserver.twoplustwo.com/sh...age=0&vc=1

If I managed to get the formula typed in right into excel (I'm fairly confident, since my output matched his numbers for the few examples he did run, but anything's possible, and I've screwed up far simpler things than transitioning a formula into excel)...

With a winrate of 8PTBB/100, and a std dev of 50PTBB/100, the chance of having a 20 buyin downswing in 50k hands is about 4.2%. Over 100k hands, it's 8.25%. Which (assuming the math is even close) basically means that one in twelve people will have a downswing that big in their first 100K hands. So yeah, that seems fairly common-ish. Of course, 100k hands is a *lot* of hands, so this probably shouldn't be a monthly occurence, but if you play enough, you could very easily see a swing that big multiple times.

And btw, this does depend HEAVILY on your actual variance. If the std. dev is increased to 75PTBB/100, the odds shoot through the roof, with an 57.3% chance of having a 20BI downswing over just 50k hands, and a 15.6% chance over just 10k hands. 75 does seem pretty high though, just based on the very few real std. devs I have heard.

[TNixon]

btw pzhon, I clearly owe an apology for this:

[ QUOTE ]
Nobody else has commented on it at all, and your comments were not exactly self-explanatory.

[/ QUOTE ]

You were obviously assuming I had read the plethora of other threads on this topic, which were I not a lazy-ass bastard, would have been a perfectly valid assumption.

I'm reading through some of them now, but I'd still greatly appreciate a comment on the validity of jason1990's formula as a gross approximation to the problem, which is likely more than sufficient for the question that is typically asked here: "does this downswing even make sense"? Being able to spout off exact percentages would be ideal (but is pretty obviously not possible), but being able to spout off approximate probabilities is better than nothing at all.