Poker Theory SNG Question

[Kaz The Original]

A +EV method of colluding could occur on the bubble with the larger stack "losing" a big pot to his short stacked on the bubble friend.

(I do not of course condone collusion. It is despicable.)

[Alix]

1.84 x original equity, using the Sklansky-Peacock method (see 3 player case in TPFAP). Since this method slightly undervalues big stacks, you can probably treat this as a lower bound.

[ML4L]

Hey Alix,

Can you explain how you got that? Thanks.

ML4L

[Alix]

If you have 3 people with stacks 5000,2000,1000
big stack's chance of finishing second is
1/8 * 5/7 + 2/8 * 5/6

If the smallest stack gets 1st, we assume the chance of getting second for player1 is proportional with the fraction of remaining chips held.

I wrote an excel function that generalizes this for up to 10 players with SNG payouts.

[mikewvp]

I know this isn't really an answer to your question, but it would be extremely stupid to do this because you are eliminating the chance that both of you finish in the money. Thats how I see it.

[Louie Landale]

In a graduated pay-out tournament, the more chips you have the less each is worth, so doubling up at even money is clearly a bad thing to do. I don't think the other response suggesting the big stack has a big advantage eliminates that, but I suppose it mitigates it some.

The theroeticaly odds for doubling-up-or-busting out will depend on the payout schedule. I cannot do that math but suspect its in the 60% range early in the tournament. But that's only if everyone is evenly skilled. The higher skilled player needs a better reason to risk busting out since he's such a favorite even with a shorter stack.

- Louie

[ML4L]

Hey Alix,

Thanks.

[ QUOTE ]
I wrote an excel function that generalizes this for up to 10 players with SNG payouts.

[/ QUOTE ]

Badass.

ML4L

[Aisthesis]

Assuming an ROI at this kind of tournament of 0% for both buyer and seller, then you're just breaking even against everyone else. So, you've gained at least $11 as your share of the winnings from the opponent you bought out (1/9 of $100).

On the other hand, you've gained a disproportionate share, but I really don't think it's the full $100. You haven't doubled your expected winnings--although that might be the case if the whole prize pool went to first place.

Let's say the prize structure is $167 to 3rd, $333 to 2nd, and $500 to 1st, and that your own winnings are distributed equally at 10% over these three prizes. I would say that your chances go up equally for each place for an increase of only 3.3% to each place (if you've doubled your chances for 3rd with this deal, then you haven't increased your chances for 2nds at all, etc.).

So, with 3 prizes, I think your expected winnings have increased to $133 from $100. Hence, the opponent's stack is worth $33. Basically, for 0% ROI, I think the stack is worth 1/n times the prize pool, where n is the number of places paid.

But let's assume you have an ROI of 50% and your opponent has an ROI of 0% with the same prize structure. Now, you're expected to win each place 15% of the time. You just bought your opponent's 10% share in each place. But the other players had only a 9.375% share in each place.

If you yourself had been sitting on the sold stack, your strength relative to the remainder of the table could be viewed as 15/9.375 = 1.6

That would give you two 13.4% shots at the money vs. 8.37% on the part of the remaining opponents. So, you should now win each prize 26.8% of the time rather than 15%. So, the opponent's stack is now worth $268 - $150 = $118 to you. Basically, your chances of getting in the money have increased here (by this calculation) from 45% (at the initial stack size) to 80.4%.

I think the variables here are going to be "n" for the number of places paid, then your own strength and that of the opponent you're buying from. The stronger you are in the field, the more an opponent's stack should be worth (assuming you're equally skilled in playing big, small or medium stack). And the stronger the sellout opponent is, the more his stack will be worth. But your own skill in handling the extra money will be much more important than the skill of the opponent whom you bought out.

This is a pretty interesting question actually. Assuming the simple case of 0% ROI for all players (same prize structure), it translates nicely into a very practical consideration: If you double-up early by knocking out an opponent, then in a $100 tourney, you have only effectively won $33. You've only increased your EV by 33%... at least according to my way of figuring this. But if you're good at handling the big stack, it can be worth a lot more than that (more than the $100 he bought it for, actually).

[Aisthesis]

I really think the "collusion" direction that this question can take is much less interesting (for one thing, I just have no interest at all in that and would hope that others here don't either) than the question of what you've actually gained by increasing your stack-size early in a tourney: What have you gained by busting someone out early?

As LL has pointed out, I think the prize structure is a big factor, but I also don't think the question is solvable without bringing in both your own strength and that of the sellout opponent.

With a few modifications, I think one could also apply this to a final table in a big multi, where the buy-in has become essentially irrelevant but where the increments in the prize structure can make an enormous difference in your winnings.

I really think this is an issue worth examining pretty thoroughly and in a variety of contexts. It seems to me to provide a great method for making some kind of translation from tournament dollars to real dollars.

[Aisthesis]

In the 50% ROI case, I forgot to divide the extra winnings by 1/3, and this changes things dramatically. You only go up from a 15% chance at each prize to a 17.9% at each prize. Hence, you've only won $29 in this case--less than in the 0% ROI case.

Generally, if p(b) is the probability that the buyer will make a given place without the deal (p(b) <=33.3%) and p(s) is the respective probability for the seller, then the buyer's probability of reaching a given place increases to:
(4*p(b))/(3 + 3*p(b) - 3*p(s))

This can obviously be translated back and forth into ROI's and hence EV, but the equations get rather complicated due to the rake. It's not bad at all if you just go with $100 buy-ins and a $1,000 prize pool and at least bearable if you use a 10% rake rather than 9%.

If n places pay (e.g., final table situation), this works out to:
((n+1)*p(b))/(n + n*p(b) - n*p(s))