Average longest winning streak

[obloni]

Say you have pair over pair before the flop and you get it all in with your opponent. Your chance is winning is roughly 82%, what is the average longest losing streak you can have over a sample of like 1000 hands? Probably helps to know the math to console yourself after the your opponents hit 2-outers on you one after another.

I believe Barry Greenstein wrote about something of that to the effect about flipping a coin and then the average longest winning streak given that you flip it a 1000 times is like 10 or something like that.

Anyone got the proper formula to calculate this? A link would be sufficient. Thanks.

[jay_shark]

You want E(x) where x is a random variable that takes on the values 1,2,3,...1000 .

So we wish to solve :

E(x)= 1*P(1) + 2*P(2) + 3*P(3) + ...+ 1000*P(1000)

where P(i) is the probability that i is your longest losing streak . If P(i) is your longest losing streak then P(k) for k not equal to i equals 0 .

We do know that P(0) + P(1) + P(2) +...+ P(1000)=1 but there is no simple relationship between the P(i's) that allows for a simple formula .

[nlwolf]

I think (but I'm not 100% sure), that following is correct:

n - number of 'experiments' - here its 1000 pair/overpair situations.
p - probability of winning

avg longest winning streak is log_(1/p){n}
sorry dont know how to write it correctly - logarithm base is 1/p

So with 1024 coinflips that would be log2(1024) = 10
with pair/overpair we have p = 0.82, 1/p = 1.22

avg longest winning streak: log_(1.22){1000} ~= 34.7
avg longest losing streak : log_(5.56){1000} ~= 4.03

[pzhon]

Here are two ways to do this problem.

[img]/images/graemlins/diamond.gif[/img] The probability that the lonest streak is at least k can be computed. This thread is related. Some of the 1/2 values need to be replaced by 0.18, and some need to be replaced by 0.82.

The average value of a random variable which takes nonnegative integer values is the sum of the probability that the value is at least k for k=1,2,3,...

[img]/images/graemlins/diamond.gif[/img] Inductively, we can build up the probability distributions for the longest streak in n hands by conditioning on the last streak.

Given that the last streak has length k, and the first 1000-k-1 has longest streak m, the longest streak of the union is max(k,m).

For n<1000,

Probability(longest streak has length n)
=
Probability(last streak has length n) * Probability(longest streak out of 1000-n-1 is at most n)
+
Sum from k=0 to k=n-1 (Probability(last streak has length k) * Probability(longest streak out of 1000-k-1 is exactly n))

The probability that the last streak has length n<1000 is p^n (1-p), where p is the probability of losing each hand.

The boundary case is that the probability that all hands out of h were lost is p^n.

[pzhon]

I used the second method. It took a few hours of computation, which might have been saved if I had done the calculations numerically instead of exact arithmetic.

Here is part of the distribution of the lengths of the longest losing streak, assuming 18% losses and no ties.

0: 7x10^-67
1: 4x10^-13
2: 0.00719
3: 0.41464
4: 0.43443
5: 0.11577
6: 0.02240
7: 0.00408
8: 0.00074
9: 0.00013

The average length of the longest losing streak is 3.7463922433208756643 .

If this were important, it would make sense to check by a Monte Carlo test, or by using the first method as well.

Mathematica code:
<ul type="square">
Clear[prob];
prob[streak_, length_, p_] := prob[streak, length, p] =
If[length &lt; 0, 0,
If[streak &gt; length, 0,
If[length == streak, p^length,
p^streak (1 - p) Sum[prob[i, length - streak - 1, p], {i, 0, streak}] +
Sum[p^k(1 - p)prob[streak, length - k - 1, p], {k, 0, Min[length - 1, streak - 1]}]]
]
][/list]

[obloni]

hmmm thanks a lot pzhon. the computations are pretty complex, and i lost you tbh.. guess it's obvious i don't hold a degree in mathematics, heh.

Seems like nlwolf's simplified approximation works pretty ok too, comparing the results.

Thanks a lot guys!

[nlwolf]

pzhon,

Are you sure there is no simple formula for expected longest winning streak?

I found this link http://www.flatstats.co.uk/horseraci...cle&amp;id=160

They state: expected winning streak = log(number of runs)/-log(prob of winning)

[pzhon]

[ QUOTE ]
pzhon,

Are you sure there is no simple formula for expected longest winning streak?

[/ QUOTE ]
No, but I doubt there is one, and that is a lot more than a guess. For any formula to work, it needs to fit the extreme cases, and it is very hard for a simple formula to do that.

[ QUOTE ]
I found this link http://www.flatstats.co.uk/horseraci...cle&amp;id=160

They state: expected winning streak = log(number of runs)/-log(prob of winning)

[/ QUOTE ]
That is not what the article says. First, it says that that formula is an estimate, not the exact answer. Second, it doesn't say that it was estimating the average value of the longest streak. It said that it was estimating the mode, "what the longest winning run of favs is likely to be."

My method was exact, as you can check using the code I posted.