Poker Theory SNG Question

[Jason Strasser]

If you are in a sit and go, and the buy-in is $100+9, ten handed (like on party), what would you pay to buy someone elses stack before the game started? So that you would have 2k chips, and be nine handed, with 1k chips more than everyone else.

Thanks.

[Michael Davis]

My friends and I do this all the time. We split the money at the end.

Signed, PITTM.

[Jason Strasser]

Did I hear you correctly? You and your friends "combine stacks" at the opening of a sng? Isn't that collusion with questionable (or not, I'm waiting for someone to show me that its not profitable to pay a full $109 for a double stack--as my intuition seems to say) EV?

[Tosh]

It was a joke referring to a guy who admitted to cheating onm the internet forum.

[Iceman]

[ QUOTE ]
If you are in a sit and go, and the buy-in is $100+9, ten handed (like on party), what would you pay to buy someone elses stack before the game started? So that you would have 2k chips, and be nine handed, with 1k chips more than everyone else.

[/ QUOTE ]

Slightly less than $100 - maybe $90. The more chips you have, the less each one is worth, and vice versa. For example, a short stack that's in the money is worth a lot more than its chips would indicate, and no matter how close the big stack is to having all $1000 of the chips it's never worth $500 or more. This factor is small in the start of an SNG, and only becomes significant once you're down to 4-5 players, so I'd pay almost the full $100 for the stack but not quite.

[Jason Strasser]

AAAAAAAAAAAAAH.

I remember that guy! He casually referred to colluding with his friend over instant messenger or something. Excuse me for the harsh reply, I am not as sharp as I'd like to be.

[Louie Landale]

Perhaps you are talking about getting a friend to "lose" a big pot early to you.

If this is a winner-take-all tourney then your friends chips are worth their face value, $100 in this case. That is, you don't gain anything by cheating like this.

If its a graduated pay out tournament (pay top 3 places) then your friends chips are worth MORE to your partnership if he keeps them then if he gives them to you and gets busted out. I'd suppose these extra chips are worth about 75% of face value, but that's just a guess (and I don't know how to calculate it but know its a function of the payoff schedule). That is, you LOSE money by cheating like this.

- Louie

[tubbyspencer]

The actual cost of collusion would be an extra $109, not just $100; so I don’t see how collusion could be profitable. Not at the beginning any way.

But I don’t think this question is only relevant as regarding collusion. Think of the scenario where someone goes all in before you on the first round (a bad move by the all-inner usually, but not totally unheard of on the low buy in S&G’s I play!) Calling when you have a 50/50 chance is this situation is essentially the same as paying an extra entry fee for twice the chips if you win; because half the time you’ll be out – and the other half you’ll double up. It therefore would cost you $218 to double up once, and to be out once.

Unless you are a really bad player, this can’t be profitable. You’ll need to come in the $ almost every time to make this profitable. (If you are a bad player – this strategy won’t help you win, but it might help you lose less.)

The other question becomes then – at what percentage win rate is it profitable to double up at the beginning. I think clearly if you doubled up 70% and were out 30% you’d have a positive EV (over your normal EV on S&G’s). What about 60%? What about 55%?

I don’t know the answers – I’m interested to see if anyone has any ideas.

[Happy Appy]

I think that as was said it depends on your relative skill level. Risk is more advantageous to unskilled players because it overshadows the amount they lose because of errors. If you are even in skill to everyone at the table, having twice the chips should be twice as valuable because at that moment the marginal rate of exchange between your chips and theirs will cause you to be able to get more with less when against a short stack because chips are worth more to him than you. But I think that these theoretical considerations are trivial when compared to the fact that you will make so many mistakes when you play (as all players do). The decision of value should be based on how well you play as a big stack versus average stack. For example, Phil Helmuth plays a big stack quite well, according the McVoy, and so for him having a big stack could be more than twice as valuable, but this would be because other players are more likely to make more mistakes then he does when he has a big stack. BTW, I think Phil is kind of a douche, does anyone agree?

Theoretical argument:
assume a 3 player game in which one player can buy as many of the buy ins as he wants. And assume all players play perfectly. Then, if he buys in:

once his chances are .1 and EV = 300 / 3 - 109 = -9 (obviously)

three times his EV = 300 - 109*3 = -27 (once again, obviously)

two times he has to win the others players money with a 2 to 1 lead. Call the chance that the big stack will take out the short stack without becoming a short stack first (i.e. directly). That chance must be equal to the chance that the other player will double up because both players are essentially playing with the same amount of chips because of all-in protection(although the big stack can afford to lose once), call the probabilty that the short stack doubles up, P12. If the first player doubles up, then the roles are reversed (i.e. the big stack becomes the short stack and vise versa). Furthermore, we know these two quanties must be equal because they have all of the same odds associated with them. P21 = P12 = .5 then. And a player must win this coin flip as the big stack to win, otherwise the shortstack becomes the big stack. Therefore the player with a 2 to 1 chip lead has a 2 to 1 advantage (you can get this with algebra). Therefore the EV is -18 in this case. The algebra btw is P(big stack wins) = .5 + .5*P(short stack wins) and P(short stack wins) = .5 * P(big stack wins). One can sole this easily to find P(big stak wins) = 2/3.

This theoretical argument can be applied to any number of buy ins and for any number of entrances. And while you could make the argument that every hand isn't necessarily won with an all in, this doesn't really matter because these cases can be treated in a similar way and if you do the math you will see that it comes out the same. Therefore, I would conclude that double starting chips would be worth double in a fair game. In an unequal game, P21 and P12 are dependent on players and the more times chips change places, the more advantage will be given to the better player. In this case it would seem that having the double stack would be more than twice as valuable for the worse player and less than twice as valuable for the weak player. You could also say that this scenario only applies with 2 players, I am done thinking about this but my intuition tells me that it shouldn't matter.

Nick

[Sundevils21]

[ QUOTE ]
My friends and I do this all the time. We split the money at the end.

Signed, PITTM.

[/ QUOTE ]

Freaking hilarious [img]/images/graemlins/grin.gif[/img]. Keep up the good work, lol.