[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.