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I believe the point at which blinds could be considered "infinitely small" would be where they are small enough that even if the short stack was stealing them every time, enough hands would be dealt that the leader would expect to have several chances with monster hands.
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No.
Even if your subsequent analysis were valid, it would require more than a few hundred times the blinds. You required AA versus AA matchups, which happen 1/270,725 of the time, but further you require that this not split the pot, but a flush happens only about 1/23 of the time, so you need the stacks to be tens of million of times the blinds.
"Infinitesimal" already means something in non-Archimedean ordered fields such as the hyperreals. However, real numbers are Archimedean. That means for every positive real number x, there is some positive integer n so that x is greater than 1/n. If you keep giving up a real blind x, you will give away your whole stack in n steps.
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In each of those hands, he could expect to get back at least the blind-steal bet from the SS, and eventually, the SS would have a hand he thought was worth playing, even though he knew he was up against a monster. At that point, he would either double up or lose out. So let's say the Leader will only play if he gets AA, and shorty, knowing that, will only come along if he has AA as well.
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If the leader adopts a strategy that bad, then the short stack should throw away AA when the leader bets. If the leader adopts this brain-dead strategy, the short stack will steal the blinds 220/221 of the time, and will give up a minimum raise 1/221 of the time. The result is a random walk that increases 1 chip 220/221 of the time, and decreases perhaps 2 chips 1/221 of the time. If the trailer had 1/6 of the chips at the beginning, then the trailer's chance of winning will be very close to 1, and will be closer to 1 for larger numbers of chips/smaller blinds.
<ul type="square">That is the exact opposite of the blinds becoming negligible. [/list] A crucial statistic to watch is the variance:disadvantage ratio, properly normalized in terms of the total stacks, not chips. The average total variance of a (uniformly bounded) martingale that starts at x and ends at 0 or 1 is fixed, x(1-x). If your variance:disadvantage ratio is high, then you are close to a martingale, so your equity is close to x. If your variance:disadvantage ratio is low, then the drift dominates, and you may win with probability very far from x.
Suppose you construct simple strategies that have arbitrarily large variance:disadvantage ratios against all opposing strategies as the number of chips each side has increases. Then, as the blinds decrease, either player could fall back on a high-variance strategy to win with probability close to the fraction of the chips. That would be an argument for saying that the skill would be negligible as the blinds shrink. However, I don't believe such a family of strategies exists. If you disagree, the burden is on you to construct such a strategy.
Your suggested strategy has a variance:disadvantage ratio against the above counterstrategy that decreases to 0 as the blinds shrink. The counterstrategy wins with probability approaching 1 as the blinds decrease even if the counterstrategy starts with 1/1000 of the chips.