The old coin-flip debate

[AleoMagus]

Yes, we are in agreement.

99 and TT are definitely strong enough hands to push with and a coinflip confrontation if you are called is ok, because you have the greatly added value of the steal attempt.

It is calling all-in raises against other decently sized stacks that this and the other thread is addressing

Regards
Brad S

[PrayingMantis]

Reading my original post again, I've realized that the point that seemed clear to me, wasn't too clear from the way I've written it. Although the discussion that origniated was exactly what I was looking for.

My point was, that according to the original calculation by AM, calling all-in as a 7:3, with equal stacks on the bubble, is only marginally +$EV. However, I cannot believe this is correct. I'm pretty much positive that calling all-in as such a favorite, is significantly +$EV, and the same probably goes for calling it as a 2:1 favorite.

For my thinking here to be true, we have to assume that by folding in that situation we are not securing an $EV of 25$ (but less), and by calling, we are gaining more than merely the $37.5 * P (while P is our probability of winning the showdown). So basically, I'm arguing against these 2 numbers (25$ and $37.5, for a $10 SNG), and my argument is that calling all-in as, say, 2:1, favorite, is higher +$EV (against certain, but not few, opponents) than was suggested by AM.

Some of the replies here (from Pitcher, and also AM), are in the line of what I'm thinking.

I think this is a very important discussion (and not because I started, or reopened it). It's possible we're talking here about a small but meaningful increase in ROI, although it's difficult to say exactly how much.

[BrettK]

[ QUOTE ]
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):

[ QUOTE ]
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


[/ QUOTE ]

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.

[/ QUOTE ]

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

[Phil Van Sexton]

I just can't see how you can have an equal chance of finishing 2nd or 3rd. You would have to be more likely to finish 2nd.

Let's assume the one of the shortstacks is going to win. What are the chances that you finished 2nd? Well, you have 4000 and the other shortstack has 2000....so I'd say it's 67% (4000/6000) that you "win" 2nd place.

therefore...
33.33% * $30 = $10
16.67% * 20 = $3.33

So it's $38.33, not $37.50.

Did I just go through all that for 83 cents?

[BrettK]

[ QUOTE ]
I just can't see how you can have an equal chance of finishing 2nd or 3rd. You would have to be more likely to finish 2nd.

Let's assume the one of the shortstacks is going to win. What are the chances that you finished 2nd? Well, you have 4000 and the other shortstack has 2000....so I'd say it's 67% (4000/6000) that you "win" 2nd place.

[/ QUOTE ]

I'm not sure what you mean when you say that you're assuming that one of the short stacks is 'going to win'. Would you be more specific? It would help me respond. Using Sklansky's method for three people, the chances of the first player finishing in each position is determined by figuring out the chances of the second and third players finishing in each position and subtracting those from 100%.

Brett

[fnurt]

One thing that's missing here is the impact of the blinds. If you're somehow on the bubble with the blinds at 10/20 maybe it doesn't matter. But in the real world, the blinds are much higher, and you're either in the BB with money already in the pot, or in another position where the blinds provide you with an overlay. A model that basically assumes the blinds are 0/0 and you can keep this up forever doesn't reflect reality.

[Jsb]

in your calculations you came out with that Hero had a 55% chance of getting first, a 22.5% chance of second, and a 22.5% chance of third. i think Phil was talking about those percentages in particular. i don't know how to do most of this statistics stuff yet, but it seems against my intuition for hero to have an equal chance of catching second and third. In this case that one of the lesser players gets first, you say that Hero has an equal chance of getting either second or third, but my instinct is that he has a greater chance of getting second than third. my instinct could very well be wrong, it just strikes me as kind of odd. and i think that was what phil was saying as well.

[Phil Van Sexton]

We have already calculated that 50% of the time, one of the shortstacks will win.

In this case, you will be playing against the other shortstack for 2nd. Think of this as a 2 player tournament where the winner gets $30 and the loser gets $20. We know that your chance of winning a tournament is equal to your chip count, so the calculation is simple....4000/6000=66.67%.

Since we are only talking about the 50% of the time that you don't get first, you're chance of getting 2nd is 66.67% * 50% = 33.33%

The calculation is easy because the shortstacks both have the same amount. If they didn't, you'd have to do this calculation for each player and do a weighted average. This will get much messier for more than 3 players, but likely doable with a computer.

[PrayingMantis]

I completely understand the model AM has used in his calculation, and I'm familiar with Sklansky's statements in TPFAP. However, as other posters have stated here, and combined with other reasons, it seems to me as it's lacking in some respects. I don't have a better model (I know Bozeman has worked on simulating similar problems), and AM also admitted that his calculation is not perfect, but I can suggest a few different variables that can be added here:

1. (as was already said), It doesn't look reasonable that big-stack has the same probability for finishing 2nd as 3rd (with stacks at 2x,x,x). This has to do with the $EV of getting into the money as the big-stack.

2. In the title here I wrote "coin-flip", but I'm actually more interested in situation where hero is, say, 2:1, against someone who pushes against him (this has implications for coin-flips too, of course). According to the original model here, calling is only marginally +$EV. However, it seems that if hero is consistently avoiding 2:1 confrontations (over whatever long-run), he's consistently making -CEV moves. This is clear, especially if the blinds are significanly high, which means he's getting great pot-odds. By doing so (folding), he is *by definition*, increasing his opponent (aggressor) $EV, and by that reducing his own. Another point (that really complicates it, IMO), is that we can no longer assume all players have equal ability, if Hero is making a consistent CEV mistake against his opponents.

3. That leads to another, similar, complication (or a "paradox"): if all opponenets are equally skilled, Hero should take ANY +CEV opportunity he has, since he hasn't got any skill advantage. Not taking even the slightest +CEV opportunity is, according to our "equally skilled" assumption, a mistake. Therefore - our Hero should call all-in even if he's less than a coin-flip, if the pot-odds justify it. With high blinds, these spots are very common.

There are some other points to consider. For instance: how high and what is your position in relation to the blinds, when is the next level coming, etc.

Anyway, the main question remains: what are the criteria for judging whether a certain all-in call (equal stacks, on the bubble), is +$EV.

[fnurt]

What you didn't mention is that when two players go all-in, everyone who is not in the hand gains EV, because someone is going to get busted out.

So when you fold in a favorable situation, your opponent gains EV, but you don't take the corresponding negative EV hit all by yourself. The other players also lose significant EV that they would have realized if you had called.

So you lose EV by making these repeated decisions, but everyone else is in the same boat; either they lose the same EV when they're on the hot seat, in which case you're back to even, or they call and someone gets eliminated. You rate to gain EV when you force someone other than yourself to make the decision.