Bubbling

[pzhon]

[ QUOTE ]
Thanks for the analysis. Good stuff. I'm interested in the 'independent chip model'. Is this a program or is it a formula for calculating average values? If it's a formula, would you mind sharing?

[/ QUOTE ]

You are welcome.

I believe that what I've been calling the independent chip model is written up in TPFAP; it has been discovered independently by a few people. (I'll get around to ordering TPFAP soon.) It is among the models discussed on this thread.

One way of looking at it is to determine the winner of the tournament first, assuming that the probability of winning is proportional to the chip counts. Then assume that there is a separate tournament among the remaining players (whose chip counts are unchanged) to determine who gets second place, then another tournament among the losers to determine who gets third, etc. If the chip counts are (5,10,20), then the player with 5 has two ways of placing second: behind the player with 10 and behind the player with 20. The probabilities of each are supposed to be (10/35)(5/25) and (20/35)(5/15). 10/35 is the chance the player with 10 chips wins with a total chip pool of 35, and after those 10 chips are removed, 5/25 is the chance the player with 5 chips beats the player with 20.

Here is the Mathematica function I created for this thread. Unfortunately, I didn't make it with the idea it would be readable. places[chipcountvector,index] gives the vector of probabilities of places for the player in position index.

<font class="small">Code:</font><hr /><pre>
places[chipc_, i_] :=
If[Length[chipc] == 1, {1},
Table[If[j == Length[chipc],
chipc[[i]]/Sum[chipc[[k]], {k, Length[chipc]}],
0],
{j, Length[chipc]}]
+
Sum[If[t == i,
Table[0, {s, Length[chipc]}],
Append[places[Delete[chipc, t],If[t&lt;i,i-1,i]]
*chipc[[t]]/Sum[chipc[[k]],
{k,Length[chipc]}],
0]],
{t, Length[chipc]}]
]
</pre><hr />

Example: places[{3,2,1},1]
{3/20,7/20,1/2}

1/2 chance that the player with 3 chips wins.
7/20 chance that the player with 3 chips is 2nd.
3/20 chance that the player with 3 chips is 3rd.

This function computes the probability vectors recursively. It is ugly even to me, but it worked on the second try.

[PrayingMantis]

[ QUOTE ]
More accurate models might include position and the sizes of future blinds, but I think other effects are more important. In particular, how long will the BB keep bullying the table? The model assumes the BB will stop after this hand, but that's not right. Whether this increases or decreases the value of folding or raising is unclear to me.


[/ QUOTE ]

I was thinking about this aspect of the problem too. It seems to me that hero's position is favorable over the other medium stack's (OMS) position, since OMS is sitting to the immidiate right of big stack (i.e, playing before him). Playing "after" an aggressive big stack is favorable, IMO.


If this is correct, it will contribute to the $EV of folding - that's because hero's chips "worth more" than OMS's chips, and there's less reason to gamble.

[ZeeJustin]

I raise this and call an all-in. I think pushing is far better than folding. Limping seems to be the worst scenario, since he could check and bet any flop which is a disaster for you.

[ZeeJustin]

[ QUOTE ]
3rd place is a profit of $185 1st is a profit of 785 - that is is a 4.24x


[/ QUOTE ]
Net gain (4.24x) is clearly a more relevant number than gross gain (2.5x)

[eastbay]

[ QUOTE ]

Here is the Mathematica function I created for this thread. Unfortunately, I didn't make it with the idea it would be readable. places[chipcountvector,index] gives the vector of probabilities of places for the player in position index.

<font class="small">Code:</font><hr /><pre>
places[chipc_, i_] :=
If[Length[chipc] == 1, {1},
Table[If[j == Length[chipc],
chipc[[i]]/Sum[chipc[[k]], {k, Length[chipc]}],
0],
{j, Length[chipc]}]
+
Sum[If[t == i,
Table[0, {s, Length[chipc]}],
Append[places[Delete[chipc, t],If[t&lt;i,i-1,i]]
*chipc[[t]]/Sum[chipc[[k]],
{k,Length[chipc]}],
0]],
{t, Length[chipc]}]
]
</pre><hr />

Example: places[{3,2,1},1]
{3/20,7/20,1/2}

1/2 chance that the player with 3 chips wins.
7/20 chance that the player with 3 chips is 2nd.
3/20 chance that the player with 3 chips is 3rd.

This function computes the probability vectors recursively. It is ugly even to me, but it worked on the second try.

[/ QUOTE ]

Mathematica? Oi. Anyone have a C function for this? Or any sane language?

eastbay

[PrayingMantis]

[ QUOTE ]
[ QUOTE ]
3rd place is a profit of $185 1st is a profit of 785 - that is is a 4.24x


[/ QUOTE ] Net gain (4.24x) is clearly a more relevant number than gross gain (2.5x)

[/ QUOTE ]

This statement doesn't make sense.

You can very well see 3rd as a profit of $185, and 1st as profit of $785, but then you must see any out-of-money finish as -$215 profit. So still, the jump from out-of-money to 3rd is +$400, while the jump from out-of-money to 1st is +$1000. Therefore, the jump from out-of-money to 1st is 2.5 times the jump to 3rd.

I will be very glad if you can show any meaningful $EV perspective for which 1st is 4.24x 3rd.

BTW

Your "f**k 3rd" attitude (or more accurately: f**k 2nd), is clearly much more relevant once in the money, since the jump from 3rd to 1st (+$600) is 3 times the jump from 3rd to 2nd (+$200).

However, When you are still out of the money, taking risks for all your chips with a small-medium stack, when a mini-stack is close to busting, your hand is not a huge favorite AND the blinds are not high enough to justify the CEV - this can be very well a negative $EV move, as was shown by other posters here.

[Nepa]

[ QUOTE ]
There is no gaurantee that the short stack will go out - what do you do if he triples up? I have seen that happen over and over again.


[/ QUOTE ]

I think you deal with that when and if it happens. If you move all-in here and get called by the big stack and lose the short stack is jumping up and down. The most profitible play would be to fold here and wait for a better situation. You still have plenty of time to win the tournament. Folding isn't playing for 3rd it is playing smart.