Calculating the % Change of Winning X SNG's Consecutively?

[dabluebery]

Ok, I need some math help. In a 9 person SNG for an average player, your chances of winning X number of tournaments in a row is simple math, (1/9)^x. Pretty easy, and pretty indicative that winning 4 or 5 in a row is pretty unlikely.

However, I'm interested in the % chance of a person winning 4 or 5 sng's in a row over a long period, say 100 or 1000 sng's. I don't know how to do math like this, and the only thing I can think would be to Montecarlo it using Excel, which is kinda crappy.

Any help is much appreciated.

[pzhon]

The probability of winning at least 4 SNGs in a row in 100 SNGs is 0.8705%. In 1000 tournaments, it is 8.586%. Of course, these are extremely sensitive to the assumption that you win exactly 1/10 of the time.

The way I did this was by considering 5 states.
0-3: Have not yet won 4 in a row, and currently have a streak of 0-3 wins.
4: At some point, you have won at least 4 tournaments in a row.

By playing a tournament in state n=0,1,2, or 3, you move to state 0 with probability 9/10 and state n+1 with probability 1/10. In state 4, you stay in state 4 with probability 1. This can be encoded in a transfer matrix M whose ijth entry is the probability of moving from state i to state j. In M^n, the ijth entry tells you the probability of moving from state i to state j by playing n tournaments. The 04 entry in M^100 is 0.008705. The 04 entry in M^1000 is 0.08586.

There are many techniques for approximating the values. The probability can be estimated from above by counting the expected number of streaks. This is (the probability of winning tournaments 1-4) + (the probability of winning tournaments 2-5) + ... + (the probability of winning tournaments 97-100) = 97/10^4 for example. This estimate overcounts the sequences with more than one streak of length 4, but it would allow you to say that the probability is less than 0.97% that you have a streak of 4 wins in a sequence of 100 tournaments.

A much more accurate estimate comes from saying that the average wait before the next defeat is 10/9, so there are about 100/(10/9)=90 chances to win 4 times in a row, and the probability of not having a 1/10^4 occurence in a row in 90 attempts is (9999/10000)^90. 1-(9999/10000)^90 = 0.8960%. 1-(9999/10000)^900 = 8.607%. There are more accurate estimates, but they are more complicated.