How likely is this to happen?

[Apanage]

I´m having a horrible downswing and I am down 360 BB over 13293 hands played 5-handed.
Before this downswing I was 1.95 BB/100 over 150K hands with a standard deviation of 16.02 BB/100.
Can someone calculate the probability of such a downswing to occur given that 1.95 BB/100 is my true winrate?
And do anyone know if there is a calculator somewhere that can calculate what your true winrate is within certain confidence intervals?

[Orlando Salazar]

This is like a 10 buying downswing in NL, I think everyone has this.

[boccage]

Assuming thats your true winrate and standard dev, there's about a 0.0004 chance of run like that. Pretty unlikely. Oh, and that's also assuming my actuary skills aren't failing me. No promises; It's kinda late.

[DWarrior]

if this is NL, this is so standard.

if this is Limit, you were way overdue.

PS, the odds of it happening are low, but it will probably happen after you play many hands.

[Voltaire]

5-handed?

NL this could happen, but if it's limit I have to say, you are getting beaten.

[jason1990]

[ QUOTE ]
I´m having a horrible downswing and I am down 360 BB over 13293 hands played 5-handed.
Before this downswing I was 1.95 BB/100 over 150K hands with a standard deviation of 16.02 BB/100.
Can someone calculate the probability of such a downswing to occur given that 1.95 BB/100 is my true winrate?

[/ QUOTE ]
Let X(t) be your bankroll at time t, where t is measured in 100's of hands. Let Z(t) be the maximum of X(s), where s ranges over the whole interval [0,t]. Then Z(t) - X(t) measures the size of the current downswing you are on. Let T be the first time Z(t) - X(t) = b, where b = 360 in your case.

In other words, T is the number of hands you will play before your first 360BB downswing. We want to know the distribution of T.

This actually appears to be a very difficult problem. By modeling X as a Brownian motion with drift, using something called the "Girsanov transform," and applying some known results about something called "local time," I have come up with some information. Let m be your winrate and s your standard deviation. Define r = -m/s and x = b/s. It appears that, for all y > 0,

E[e^{-yT}] = g(a),

where a = y + r^2/2, and

g(a) = e^{rx} sqrt{2a}/(sqrt{2a} cosh(x sqrt{2a}) + r sinh(x sqrt{2a})).

Here, "sinh" and "cosh" are the hyperbolic sine and cosine. It follows that

E[T^n] = (-1)^n g^(n) (r^2/2),

where g^(n) is the n-th derivative of g.

Here are some numbers for you (courtesy of Mathematica). If m = 1.95 and s = 16, then

E[T] = 7890.86 and SD(T) = 7800.23.

So the expected number of hands you need to play to get a 360BB downswing is about 789,000. But the standard deviation is about 780,000, so you are well within one standard deviation.

Now, if you used Poker Tracker to compute your standard deviation, then you may have a biased estimate, as we have discussed in other threads. Your true standard deviation may be higher than 16. Suppose m = 1.95 and s = 18. Then

E[T] = 3019.08 and SD(T) = 2944.42.

Or maybe your true winrate is not 1.95. Suppose m = 1.5 and s = 18. Then

E[T] = 1706.28 and SD(T) = 1629.93.

Notice that in all cases, the mean and standard deviation of T are about equal. If T were exponentially distributed, then they would be exactly equal. So let us guess that T is approximately exponentially distributed. In that case, we have (approximately)

P(T < t) = 1 - e^{-t/M},

where M = E[T]. You have played 163,293 hands, so you would like to know p = P(T < 1632.93), the probability that you would have a 360BB downswing in your first 163,293 hands. Using the above estimate for this probability, here are some numbers.

m = 1.95 and s = 16 ==> p = 0.187
m = 1.95 and s = 18 ==> p = 0.418
m = 1.50 and s = 18 ==> p = 0.616

If this estimate is accurate, then you should not be surprised by this downswing, even if your true winrate is 1.95BB/100 and your true standard deviation is 16BB/100. About 1 out of 5 people with those stats will have a 360BB downswing in their first 163,000 hands.

[jason1990]

After some more work, I have a formula to share with the forum. First I will give the formula. Then I will share some technical remarks about it. Here is the formula.

m = winrate (in BB/100)
s = standard deviation (in BB/100)
b = downswing you are worried about (in BB)
(b = 360 in the OP)
T = time until first b downswing (in 100's of hands)

P(T < t) = 1 - e^{-2tm^2/(s^2(e^{2bm/s^2} - 1) - 2bm)}.

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

Now for the technical remarks. After some additional calculations, it seems pretty clear to me that the normalized variable T/E[T] converges in law, as b -> infinity, to an exponential random variable with mean 1. This means that

lim_{b->infinty} P(T/E[T] < t) = 1 - e^{-t}

The expected value of T is

E[T] = (s^2(e^{2bm/s^2} - 1) - 2bm)/2m^2.

Combining these formulas gives the above expression for P(T < t). It should be noted that the formula for P(T < t) is only an estimate, which is valid for "large" b. I do not yet have error estimates to indicate how "large" b needs to be.

[TNixon]

Can you give me a quick double-check on some math?

After plugging your formula into excel, and throwing some numbers through it, I've come up with a probability of 8.25% of having a 20 buyin downswing at some point in 100k hands, with a winrate of 8PTBB/100, and a std dev of 50PTBB/100.

Does that seem about right?

[jason1990]

If 20 buy-ins is 1000 PTBB, then it appears you have used the formula correctly.

Here are some comments on the validity of the formula. It is based on a Brownian motion model. Brownian motion is the scaling limit of a random walk. This model is appropriate in situations where you are considering a large number of hands/sessions which are independent and over which game conditions do not change. It is an idealization. However, it is the same model from which one can derive the standard risk of ruin formulas that everyone relies on. I think it is a reasonable model, though not perfect.

Within this model, the formula itself is valid only for "large" b. As I said, I have not computed any margins of error. However, I did post the exact Laplace transform of the time T until the downswing. This is valid for any b (within the Brownian motion model, of course). In principle, one could use a computer to numerically invert the transform and calculate the exact probabilities.

I realize that these comments may be over your head, but I wanted to post them anyway. Perhaps someone else out there might be interested in this formula, and may even want to try the numerical inversion I mentioned.