The old coin-flip debate

[PrayingMantis]

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.

[stupidsucker]

These equations baffel me a little, but I understand them enough to get the jist of what you are saying. Forgive me for asking a stupid question, but how do you know its a "coinflip" without seeing the cards.

My calling hands are of a higher standard then my pushing cards. If I push I dont know if I am getting called. If I call then I know its going to be a race, but I have no idea what the % chance is. I may be on either side of a domination.

Please understand that I am not trying to belittle the thread. I really want to understand what you mean.

[PrayingMantis]

You can never *know* of course, if it's a coin-flip, until you see the other-opponent's cards (whether you called, or he called your push). But against most opponents, you might have a strong enough "feel" of their pushing (raising) and calling standards, and have a pretty good guess of where you would "normally" stand.

Aleo is saying, for instance (he would say it better than me, and I hope he'll reply in this thread), that calling all-in on the bubble, with equal stacks, when you hold 99, is a wrong move, because most of the time you'll be in a coin flip situation (i.e, against over-cards), or worse: bigger pair (However, since it's short handed, and high blinds, you can not put your opponents on big pair every time they push, and it's much more probable they are on over-cards. Hence, coin-flip). Calling with 99, according to some assumptions, is -$EV. You're losing money, despite the fact you're probably getting the right, if not good, pot-odds.

Now comes the more subtle point of what is your read on the raiser "looseness". With what Ax will he push? Will he push with any A? any pair? etc. If he will, many times you'll be in a far better than coin flip - 7:3 or much better. Does it worth the call? Do you still consider it as a coin-flip in normal circumstances? is it +$EV? Marginally +$EV? Auto-call? That was the point of my post (or some of it). I hope this is more clear.

[AleoMagus]

Hi PM

For anyone who hasn't seen that thread, it is here

I was toying with those ideas when I first posted this and I think it is important to read Phil Van Sexton's criticism, which I think is valid.

Still, The idea of calling equal stack ALL-IN raises on the bubble with equal stacks still seems like a horrible move to me. As I said in that original post, I wouldn't even do it if I knew myself to be slightly ahead.

What troubles me about the reasoning though is the thought that in some way, we hurt our chances of making the money when we start to play too timidly and it is hard to know where to draw the line between what hurts us most, especially on an agressive bubble where your blinds are constantly being challenged.

The implications of my original argument, if correct, run both ways after all. We should be raising a lot on the bubble, and maybe even all-in against good players, because there is very little they can actually call with that makes the move correct for them, even if they are ahead. In a really strong game this might mean players going all-in almost any chance they can be the first in the pot until one player finally gets a big pair and can call down the raiser.

In that original thread I never did get any responses from math types about my reasoning and I too, would love to see some.

I'm gonna write more about this, I promise

Regards
Brad S

edit - In re-reading your post, I think you may be touching on at least Part of where this reasoning might be flawed. Having 2x,x,x stacks may give you greater $EV than my assumption, but I am also not an expert on those approximations. That combined with the potential -$EV implications of playing too timidly on the bubble almost certainly lessens my original argument. To what degree remains to be seen.

[PrayingMantis]

[ QUOTE ]
Still, The idea of calling equal stack ALL-IN raises on the bubble with equal stacks still seems like a horrible move to me. As I said in that original post, I wouldn't even do it if I knew myself to be slightly ahead.

[/ QUOTE ]

How much is "slightly ahead"? That is the main problem here. I think that in this forum there is a big tendency to see everything in terms of aggression and folding equity (I do it many times myself). However, It seems that by not calling (all-ins, that is), consistently, when there's a good possibility you're around, say, 2:1 favorite, is giving the aggressor in that spot a very big advantage (I'm talking high blinds of course), and in these last stages of a tourney, that can be significantly -$EV for you.

Of course the gap still applies: you can raise with a much wider range of hands, but against certain strong opponents, especially in the higher buy-ins, big calls are a dangarous weapon too. The higher the blinds, the narrower the gap. Part of what you achieve (if you manage to survive) is tightening your opponents, which is also +$EV for you. I know many in this forum won't accept it, but your opponents might actually *fear* your calls.

[AleoMagus]

For me, slighltly ahead usually means small to medium pocket pairs. These hands will really only be behind against a bigger pair and as you have stated, you can't always be scared of that.

I have softened somewhat on hands like TT and JJ because that I think so many opponents will push with Ax on the button or even K7+. The possible addition of a small card makes hands like TT (and maybe even 99) better than perhaps I have first suggested.

88 or smaller pairs are trouble though and I think still are better folded in the situations we are discussing. I'm actually still folding 99-TT to big raises on the bubble and I like my bubble stats of late. I'm sure this is relative to opposition and may be different at higher or lower limits. I am playing mostly Party 30+3 these days with a few 50+5 thrown in to the mix. If I think a player begins repeatedly going after my blinds, I will lessen these requirements. If he lets me keep my blind a lot of the time, I'll be happy to give him the occasional one even if I do have 99.

Slightly ahead may also mean hands like AT or smaller aces if you think an opponent will push with any two cards over 9 or even with Kx. I'd never call an all in raise from a big (equal) stack with even AJ in these situations unless I knew a player was continuously gunning for my blinds and I needed to make a stand. Most times, if I can leave myself with 3-4 BB after a loss, I will call with more. Another problem with medium aces is that so often you will be dominated if you guess wrong.

As for opponents fearing calls, the problem is that if you survive, that opponent is eliminated or crippled and his fear is no longer very important. I just don't give credit to players for actually paying attention to what happens to other people. I definitely think players pay attention when they are in a hand, but not much otherwise. If I want fear or need to slow a guy down, I think re-raises all-in are still ideal. That's pretty obvious I guess.

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