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I reread recenly this interesting thread by Aleo (
A bad way to play on the bubble ), and had some new thoughts.
I want to specifically adress this paragraph (and calculation):
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If I take a coinflip, I have a 50% chance of busting and a 50% chance of being the big stack with three left.
So I have 50% chance of $0
and a 50% chance getting into the final 3 with about 4000 to 2000 to 2000
this should mean 1st 50% of the time I survive- $25 equity (10+1)
2nd 25% of the time - $7.5 equity (10+1)
3rd 25% of the time - $5 equity (10+1)
so all together this means .5(0)+.25(50)+.125(30)+.125(20)
or, $18.75 equity
BUT...
if I avoid confrontation when I know it's gonna mean a showdown I have the same equity (slightly less if I'm in the blind) as before. This is
1st 25% of the time - $12.5 equity
2nd 25% of the time - $7.5 equity
3rd 25% of the time - $5 equity
4th 25% of the time - $0
so all together this means .25(0)+.25(50)+.25(30)+.25(20)
or, $25 equity
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The point of Aleo here is, that getting into coin-flip situations on the bubble, with equal stacks and equal ability, is -$EV, since by folding you remain in a +$25 EV position, and taking the coin-flip reduces your EV to +$18.75.
However, according to this reasoning and evaluation, taking a 7:3 showdown, is only marginally +$EV:
Taking it:
0.7*37.5 (your overall portion of the prize pool, according to the same calculation, when stacks are 2x,x,x) = $26.25
Avoiding it: $25.
And of course, any situation where it's 66:33, is neutral in terms of $EV (about 0 $EV), for instance: AQ vs. KJ.
I believe that part of the problem is in the assumption of having "only" $37.5 EV, once in the money, with stacks at 2x,x,x.
I would suggest it's in the vicinity of $40, for a strong player (I'd hope someone who's very familiar with these calculations, like Bozeman, will help here), and on the other hand - a player that is constantly avoinding confrontation once it's 4 handed with equal stacks (I assume pretty massive blinds, of course), as suggested in the original post, has probably less than 25% of the prize pool as an approximate EV, especially if he's dealing with loose-aggressive players.
Any thoughts?
Edit: all the $EV numbers are calculated for a $10 SNG, but that's only for the sake of convinience, of course.
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In TPFAP (Second Edition; P109) Sklansky discusses determining your chances of finishing in each place as part of the 'Making Deals' chapter. He explains that while, with equal skill levels, determining your chances of finishing in first place is easy, determining your chances of finishing in any other position is more difficult. However, he employs a method that he says gives you a reasonably good idea of the correct answer. With three people remaining, Sklansky suggests starting from the point of view of the last place player and working your way up. Using this method, here's the proof that AM's percentage calculations are correct:
After the hypothetical coin flip, Hero has 50% of the total chips, and Soandso and Whatshisface have 25% each. Since there isn't a last place player (Soandso and Whatshisface are tied for second), we'll compare them to one another. Each has a 25% chance of finishing in first, since that's their portion of the total chips. The ratio of last place player's chips to second place player's chips is 1:1, so they each have a 50% chance of finishing in second *if* they don't finish in first. Using this information, we know that Soandso and Whatshisface each have a 25% chance of finishing in first and a 37.5% chance of finishing in second, and therefore a 37.5% chance of finishing in third.
Combined, they have a 50% chance of finishing in first (which gives Hero a 50% chance), a 75% chance of finishing in second (which gives Hero a 25% chance), and a 75% chance of finishing in third (which gives Hero a 25% chance).
Using that information, AM's percentages and EV numbers are correct. (One must logically assume that with equal skill levels and equal stacks, there can be no difference between chances of finishing in different positions, so his percentages and EV numbers for the situation in which the coin flip was Not taken must also be obviously correct.) You mentioned in your post that you believe there are situations in which better players should take the odds (coin flip in the first example) because the difference in skill makes the chance at having many more chips a much better prospect. AM was assuming equal skill level, but let's look at examples with a major difference in skill between Hero and the other players. The most logical way to show this seems to be with a certain percentage boost for Hero, so that the more of the total chips he has, the greater his advantage.
Let's say that compared to his opponents, Hero is actually 10% better than his portion of the chips, and that the other players are equal to one another. After our coin flip, hero has 50% of the chips, but according to us has a 55% chance to win, which means that each of the other two players has a 22.5% chance to win, and an equal chance to come in second or third. This gives us 22.5%, 38.75% and 38.75% for first, second, and third respectively with regard to Soandso and Whatshisface, and 55%, 22.5% and 22.5% for Hero. Our equity is $27.50 + $6.75 + $4.50 = $38.75 after the coin flip, and $19.38(rounded) before.
If we can prove that there's a bigger increase in this number than in the number we get when not taking the coin flip, we'll know that there's a point at which skill outweighs the other factors. The hard part is figuring out Hero's percentages when he *doesn't* take the coin flip. There are still four people remaining, and I don't know that Sklansky meant to imply that his method will work with any number other than three. I'll give it a try, though.
When Hero doesn't take the coin flip, he's still 10% better than his stack would indicate, which means that he has a 27.5% chance for first. Soandso, Whatshisface, and Whatshisname (the player that wasn't in our calculations before) are still of equal skill level, so they each have a 24.17% (rounded) chance for first.
Here's where I hit a wall, and begin to think that I can't use the same method. I know that each of the three worse players has the same chances of finishing in each position, but how do I figure out what the chances are for second, third, and fourth? I would be willing to bet that because Hero is a certain *percentage* better than his stack, he can play hands closer to a coin flip in our hypothetical situation than one of the worse players. Does this make sense, or did I complicate something that should be simple? As always, take what I post with a grain of salt.
Brett