[jcm4ccc]

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OK, here's the math on playing 6 accounts in the same tournament, and the chances that two of them will end up on the same table.

Let's say that the tournament has 2500 people and that ZeeJustin plays against, on average, 50 people for one tournament entry. In other words, for one account and one entry, he plays against 2% of the entire field. 98% he doesn't play against. Does that sound about right?

Let's name his accounts A, B, C, D, E, F. The chances that A and B do NOT play on the same table are 98%. The chances that A and C do not play on the same table are 98%. The chances that both of the events are true is 96% (98% * 98%).

so how many different possibilities are there?
A vs. B
A vs. C
A vs. D
A vs. E
A vs. F
B vs. C
B vs. D
B vs. E
B vs. F
C vs. D
C vs. E
C vs. F
D vs. E
D vs. F
E vs. F

15 different possibilities. Each possibility has a 2% chance of happening, and a 98% chance of not happening. .98 ^ 15 = .74. So there is a 74% chance that this will not happen, and a 26% chance that two of his accounts will end up playing against each other at some point.

This is all assuming that ZeeJustin would play against 2% of the field, on average. He may be better than that.

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Because I suck, what would then be the odds of him seeing himself over the course of 10 MTTs? How likely?

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There's a .74 * .74 * .74 chance that he will NOT play against himself in 3 tourneys. That's a 40% chance that he will NOT play against himself, and a 60% chance that he will play against himself. Over 10 tournaments? There's a 5% chance that he will NOT play against himself, and a 95% chance that he will play against himself.