[AaronBrown]

Although LuckyLucky got his answer, I think the original question is interesting for the reason pzhon mentions. I think the actual probability of having no money finishers from table 1 is less than 2% rather than 6.5%.

Think about it this way. The tables play for some period of time until there are 20 or fewer players left and the organizers break one table up. At that time, there could be as few as one or as many as ten survivors from table 1, but they will share 1/3 of the chips between them.

At this point, the chance that the winner comes from original table 2 or 3 is 2/3. The chance that the second place finisher comes from original table 2 or 3, given that the winner did, is roughly 2/3 minus the expected proportion of chips held at time of breakup by the eventual winner. In effect WhiteWolf's calculation assumes the amount to subtract is 1/30, but it has to be greater than that. WhiteWolf is computing the probability as of the beginning of the tournament, which ignores the fact that money must stay within table 1 until at least ten players are eliminated from the tournament.

The chance that the third place finisher comes from the original tables 2 or 3, given that the first two finishers did, is roughly 2/3 minus the expected combined proportion of the chips held by the first two finishers. And so on through the first six.

Even this calculation overestimates the chance because 2 times in 3 table 1 is preserved after the breakup (assuming the tournament organizers choose to break up the smallest table).

Making some reasonable guesses for the chip stack distribution at time of break up, I get about a 1.3% chance that no original table 1 player will finish among the top six.