how often can you win (general theory ?)

[pfkaok]

So, mostly this question is for middle or late stages in relation to stack sizes, but i suppose it could apply to early stages as well.

anyways, with N players left (say 5-25% of the starting field), later in the tourney, with blinds large, but not stupidly large such that the average stack is red zone, how often can a good (say 100% ROI) player win when they have an average stack? also, with N players, how big of a stack does a very good player need in order to win 1/N times?

i know of course it will depend a lot on how many players are left, as a very good HU player probably can't win much more than 60% when the stacks are equal and the blinds are fairly large. so thats only 120% of if he won only his share. but say there's 50 players left out of 500. a very good player with an average stack will certainly win more than 1/50 * 120% = 1/42. there must be some equation that takes into account players left that will give a solution to how often you win. does anybody know of such an equation that will at least approximate this?

[Exitonly]

well if you're an average stack, just pretend it's a whole new tournament.

so if there 50 people left, and you're average, and you're a 100% ROI player, that means your 2x as good average, so you could probaly estimate that you'd win 1/25 times.

[AtticusFinch]

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so if there 50 people left, and you're average, and you're a 100% ROI player, that means your 2x as good average, so you could probaly estimate that you'd win 1/25 times.

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This only holds in a WTA scenario. In a normal structure, you can have a 100% ROI while being lousy at getting to the final table. (I seem to be a perfect example of this.)

I'm still working on my model for this. Will be posting a new version soon.

[pfkaok]

i don't think its that simple though. b/c later stages when the blinds are larger play slightly differently. also, common sense says this can't be the case. look at when its HU. obviously if you're a 100% ROI player, you won't win twice average, or 100% of the time heads up. it must depend on the number left to some extent.

[Exitonly]

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i don't think its that simple though. b/c later stages when the blinds are larger play slightly differently. also, common sense says this can't be the case. look at when its HU. obviously if you're a 100% ROI player, you won't win twice average, or 100% of the time heads up. it must depend on the number left to some extent.

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That's becuase you can't be a 100% ROI player heads up..

ok, a better way to phrase it, if you're twicee as good as average, then you'll win twice as much.

So if there are 50 players left, you'll win twice as much as everyone else (slightly less than 2/50). And if you were heads up you'd win twice as often as he would. (66%)

[pfkaok]

no. i was just using a 100% ROI player as an example of a very good player. to be that good you must be an all around good player. good full table, SH, and HU. or at least the assumptoin in my example is of an all around solid player in every area.

also, i don't think its even that simple as you state it for the beginning of a tourney. it would be if it were winner takes all, but in a normal tourney a 100% ROI player should almost certainly win more than 2x his share. you should finish 1st more than 2nd, 2nd more than 3rd, ect...

[limon]

a 100% roi player should have more than 100% roi...funny stuff...i know of one well known player and author who spends about half a million a year on tournament entries and hasnt broke even in 3 years i guarantee hes 100% roi... talking out my ass portion of post: i virtually guarantee the best FULL TIME TOURNAMENT player (52 5k+ events a year) in the world could go on a 5 year streak without ever placing 1st.

[ilya]

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well if you're an average stack, just pretend it's a whole new tournament.

so if there 50 people left, and you're average, and you're a 100% ROI player, that means your 2x as good average, so you could probaly estimate that you'd win 1/25 times.

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i don't think that works, because the remaining 10% of players figure to be a stronger field than the initial one you used to calculate your edge at the outset of the tournament.

[Dave D]

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So, mostly this question is for middle or late stages in relation to stack sizes, but i suppose it could apply to early stages as well.

anyways, with N players left (say 5-25% of the starting field), later in the tourney, with blinds large, but not stupidly large such that the average stack is red zone, how often can a good (say 100% ROI) player win when they have an average stack? also, with N players, how big of a stack does a very good player need in order to win 1/N times?

i know of course it will depend a lot on how many players are left, as a very good HU player probably can't win much more than 60% when the stacks are equal and the blinds are fairly large. so thats only 120% of if he won only his share. but say there's 50 players left out of 500. a very good player with an average stack will certainly win more than 1/50 * 120% = 1/42. there must be some equation that takes into account players left that will give a solution to how often you win. does anybody know of such an equation that will at least approximate this?

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OP referanced this thread in another thread, so I'll bump/respond to it cuz I think it's interesting.

I'm pretty sure OP means "win" as in place first , assuming a standard payout structure right?

I don't think there's any real clear answer to this question. I think a lot has to do with the player's individual style, ie some people are better at playing a small stack, others' chances go up basically exponentially as their stack is bigger. I realize that everyone would do better with a bigger stack obviously, but what I mean is some players have *more* of a chance than others with a bigger stack. This is distinct from some players are better than others because I'm trying to make the subtle distinction that some players rely more on the post flop game which they need a bigger stack for.

This is where this question relates to the other one Ed Miller posted. I think there are two things that are crucial to trying to answer this question:

1. What's the rest of the field like. Your expectation changes a lot depending on the type of tourney. You can be a lot more confident in a party 10+1 than a 100+9. You know there will be a good number of weak players towards this stage of a 10+1 than a 100+9. As such, stack size becomes more important as you go up in buy in.

2. How fast are blinds going up? What is the payout structure? Basically I think we all know that a slower structure favors better players. We know therefore that these factors are crucial in determining your chances of winning. Payout structure is relativly minor, but also affects the way other people play. Namely, if it's really steep you know that you're playing for first, but others might be trying to stall so you can take advantage of that.


Overall though, I'd say that if you have the avg, you're going to win way less than 1/N times (I know you said greater than avg, but I'm stating this as a baseline). I think you're going to need at least 3x the avg to even come close to making any sort of expectation as to your chances of winning. Less than that, and for me it's a toss up.

Maybe I'm just jaded, but the bottom line is that in order to win, you're going to still need to win several more coinflips. So you need a lot of chips to simply survive luck. I would view this situation as having so many variables that it's basically useless to try to figure it out. You don't know the skill level at your table, your table compared to the rest of the tourney, and all kinds of other variables. I think all these unknown variables makes it impossible to ever come up with an equation, so I just wouldn't worry about it.

[pfkaok]

well. i know its going to depend on the opponents a lot. but i'm just saying in this example, hero is a 100% ROI player against that field. so on average his edge at the final table should be pretty consistant.

and 100% ROI is a very good player. so i don't see how he could possible win less than 1/N, with N players left and his stack is average. if there's 16 players left, and he has 1/16th of the chips, that just means he has to double up 4 times to win. unless the blinds are VERY big, and you're super shortstacked to where you have to take -EV gambles to avoid getting blinded out, there's no reason you shouldn't be more than 50/50 to double up (if you're a good player)

for example. in the 16 person field, if the blinds were reasonable, and hero has an edge, of say 53% to double up, then his chance to win would be 53% ^ 4 = 7.9% = 1 / 12.7

now i'm not sure that you can have that much of an edge in most online tourneys with the blinds as high as they are, but its just an example, and i'm sure that in bigger buyin live events the top players have at least that much of an edge.

so overall, the solution to this would almost certainly have to include a factor of, say, M of the avg player.

i think the 3 things i'd like to know most are:

1) % wins for good player with average stack

2) % of average stack needed for good player to win his share

3) % of average stack needed for good player to win twice his share.

it'd be very nice to be able to find this for a few different common structures. stars' normal tourney, party normal tourney, stars rebuy, party rebuy, and big buyin live event.