Question for BruceZ (or others) regarding streaks

[alThor]

[ QUOTE ]
Look at the 1000 hands as 900 overlapping windows of width 100.

[/ QUOTE ]

I found no mistake in your formulas, but wouldn't use want to use 901 windows?

OTOH, I am confused by some entries in your sim post. Cell E1 should be =D1; the other 900 (or 899?) cells Ei should max cells Di and E(i-1). Is that what you actually did? Perhaps this explains the discrepancy.

alThor

[irchans]

I agree, there are 901 windows. That changes the theoretical answer to 11.7447%.

For the Excel simulation, I agree that cell E1 should be =D1. In the other E cells I had max of cells D(i-1) and E(i-1) which I think works as well as your formula.

I reran the Excel simulation with your corrections. Among the 200,000 virtual deals there were 20,789 deals that had 14 pairs or more in a window. That simulation gives a 10.4% estimate of the probability. I wrote a simulator in Mathematica and got 22,545 deals with 14 pairs or more. That's 11.3%. Given the size of the simulations, either there is a bug one of the simulation programs and the theoretical derivation, or these random generators are not good enough. Hmm.

[DougOzzzz]

I haven't tried to understand your post yet explaining your method for solving the problem

I can speak for Excel's RNG being crappy though. There are some very obvious patterns if you analyze a large sequence of randomly generated numbers. I almost always get incorrect answers when I write simulations in Excel even with millions of trials.

Still, it seems a bit strange that BruceZ's method could be so far off. I understood his method, and it makes sense. The flaw in the method also makes sense, and logically I concluded that the error would cause the odds to increase.

However the difference still seems large. I'll try to understand your method at a later time. I don't have as much mathematical training as most of you here so sometimes it's difficult for me to translate these formulas into something that makes sense to me.

I do appreciate the effort that everyone put into answering my question, though.

[BruceZ]

I have PMed irchans about this, and he agrees with my objections.

This method has a problem similar to the one in my solution.


[ QUOTE ]

The Probability p[n] that the nth window will have exactly 14 hands is 0.0014279740790007608. Let g[n] be the probability that the nth window has exactly 14 hands and that no previous window had 14 hands. Let r[n,i] be the probability that the nth window will have exactly 14 hands given that the ith window had exactly 14 hands. Note that r[n,i] = r[n-i] because it only depends on the difference between n and i. Then

p[n] = Sum[ g[j] * r[n-j] , {j, 1, n}].

[/ QUOTE ]

For this formula to be correct, r[n,i] must be the probability that the nth window will have exactly 14 hands given that the ith window had exactly 14 hands, and given that no window previous to i had 14 hands. This latter condition will change the probability.

Further, it is not true that r[n,i] only depends only on n-i, because when the windows overlap, small values for the starting position of the first window i place different constraints on the window starting at n than larger values of i, since there would be more potential windows with 14 pairs that could start before a large value of i, and the hands in the window starting at n would be constrained to preclude this.


[ QUOTE ]
Now we just need the sequence r[n] to compute g[n]. If n<100, r[n] is the sum over k=0,...,14 of the probability that the first n hands of a window has k pairs times the probability that the next n hands after the window have k pairs. (If n>=100, r[n] = 0.0014279740790007608.)

Code:


r[n] = Sum[ C[n, i]
*(1/17^i)*(16/17)^(n - i) * prfirst[n, i],
{i, 0, Min[n, 14]}]



The probability that the first n hands of a 100 hand window have i pairs is

Code:


prfirst[i, j] = If[ j < 14 - (100 - i),
0,
C[i, j]*C[100 - i, 14 - j] / C[100, 14]]

[/ QUOTE ]

Because of the conditional issue discussed above, all C[100,14] do not have the same probability when the window starting at n overlaps with the first window. In this case, arrangements which put more pairs at lower values of n would be less likely, since these arrangements are constrained to preclude a window of 14 starting before the first window. The C[i, j]*C[100 - i, 14 - j] would be similarly constrained.

[irchans]

I wrote two quick simulations. Here are the results for the probability that at some time in a sequence of 1000 hands, there exists a 100 hand window with 14 or more pocket pairs.

In the first simulation of 10000 1000 hand sequences, I got 14 or more pairs 1091 times. The second simulation I got 1040. So, I think the correct answer is around 10% or 11%.


For the second simulation I copied the following formulas into excel:


cells c1 to c1000 >>>> +IF(RAND()<1/17,1,0)
cells d1 to d1000 >>>> +SUM(C1:C100)
cell e1 >>>> 0
cells e2 to e1000 >>>> +MAX(E1,D1)

Each time you hit the F9 button, the worksheet shows the maximum number of pairs in a 100 hand window in cell e1000.