Sklansky's Billion Dollar Freezeout Additional Question

[TomCollins]

Is there any heads-up matchup better than KzKy vs Kx2y? I'm talking purely EV. 7z7y vs 7x2y loses less, but ties more, so it has worse EV. Does it make sense to fold KK vs K2 because trading a win for a tie makes sense in terms of survival?

[slickpoppa]

Since you have basically unlimited time to wait for the right hand, you want to wait for the hand matchup that lowest chance of losing.

[jason_t]

http://twodimes.net/h/?z=64048
pokenum -h ks kh - kd 2c
Holdem Hi: 1712304 enumerated boards
cards win %win lose %lose tie %tie EV
Ks Kh 1597463 93.29 89530 5.23 25311 1.48 0.940
2c Kd 89530 5.23 1597463 93.29 25311 1.48 0.060

http://twodimes.net/h/?z=384033
pokenum -h 7s 7h - 7c 2s
Holdem Hi: 1712304 enumerated boards
cards win %win lose %lose tie %tie EV
7s 7h 1566600 91.49 66167 3.86 79537 4.65 0.938
2s 7c 66167 3.86 1566600 91.49 79537 4.65 0.062

[TomCollins]

You obviously misunderstood my question. It is obvious why you must wait for the best matchup. The question is, what IS the best matchup for this situation.

Suppose you have KK vs K2, and 77 vs 72. KK vs K2 will produce a higher EV for you, but will produce more losses.

77 vs 72 produces fewer wins, and more ties. So the strategy.

My question was, which is more important? Higher EV or Smallest loss %. Without doing the math, the intuitive answer seems to be the smallest loss %.

My other question is, what is the best EV matchup? What is the best no-loss matchup?

As far as I can tell, the answers are KxKy vs Kz2y and 7x7y vs 7z2y.

[TomCollins]

After doing some quick math on 77 vs 72o (sharing 1 suit) and KK vs K2 (sharing 1 suit), I came to the conclusion that if you wait until you have this matchup, and then go all in, and only play that one exact matchup, and you could coast to victory after 1 win (obviously this is simplifying the strategy tremendously), 77 vs 72 is better than KK vs K2.

My math is similar to how you figure odds in craps, the odds of making a 6 are (# of wins)/(# of losses + #losses+#wins), so there are 5 ways to make a 6, and 6 ways to make a 7, so you end up winning 5/11 of the time. So paying you 6-5 odds makes sense.

Similarly for this case, with KK, you win 94.16% of the time, and lose 4.31% of the time, and tie the rest. Since we keep playing on ties, this strategy will result in a win before a loss 95.6% of the time. With 77, you end up with a win before a loss 95.9% of the time. .3% may seem trivial, but with a billion dollars on the line, it may be worth it.

[TomCollins]

Additionally, the odds of waiting for one specific set of hands like KK vs K2 sharing 1 suit happens roughly 1/3248700 hands.

This means that you have a fairly good shot of waiting for this exact combination without losing a huge portion of your stack.

I think the odds are in the range of a couple billion to 1 that you will recieve one matchup of this type within 100 million hands. At that point, you will be down to 1.3Billion to .7Billion. Strategy can loosen up quite a bit once you have a lead, and it becomes quite simple to win with a near 100% success rate.

(note: I probably screwed up the math somewhere in here, hopefully not too bad).

The question remains, what combination of Wins/(Wins+losses) is the optimium. Anything better than 77 vs 72?

[slickpoppa]

[ QUOTE ]
You obviously misunderstood my question. It is obvious why you must wait for the best matchup. The question is, what IS the best matchup for this situation.

Suppose you have KK vs K2, and 77 vs 72. KK vs K2 will produce a higher EV for you, but will produce more losses.

77 vs 72 produces fewer wins, and more ties. So the strategy.

My question was, which is more important? Higher EV or Smallest loss %. Without doing the math, the intuitive answer seems to be the smallest loss %.

[/ QUOTE ]

I don't think I misunderstood your question. Aside from accidentally ommitting the word has, my answer was pretty clear:
[ QUOTE ]
Since you have basically unlimited time to wait for the right hand, you want to wait for the hand matchup that lowest chance of losing.

[/ QUOTE ]

Smallest loss %.

[TomCollins]

The more that I think about this, the more I think its more complex than that.

Suppose you lose .1%, win .0001%, and tie the rest of the time. This doesn't make sense to take a chance here.

Is the formula win/(win + loss) correct?

[soah]

You should look at the ratio of wins to losses while omitting the ties.

If you win 2%, lose 1%, and tie 97%, then you should view it the same as winning 67% and losing 33%.

[slickpoppa]

[ QUOTE ]
The more that I think about this, the more I think its more complex than that.

Suppose you lose .1%, win .0001%, and tie the rest of the time. This doesn't make sense to take a chance here.

Is the formula win/(win + loss) correct?

[/ QUOTE ]

Now that I think about it, I take back my perious answer. Pure EV is all that matters, which means KK v. K2o is the best matchup. The formula win/(win + loss) = EV.

The reason why tying more and losing less does not help is because tying more increases your chances of having to go all-in again, and thus having another chance to lose.

[TomCollins]

Your formula isn't pure EV.

Think about a situation where you tie 99.99% of the time, and win .01% of the time. Suppose you get $5000 for a tie, $10000 for a win, and $0 for a loss. Your EV is around $5000.01. However, this situation is great in this game. However, you may have a situation where you win 99% of the time, and lose 1% of the time and never tie. Your EV is $9,990, which is higher, but its better to do the former situation.

[slickpoppa]

[ QUOTE ]
Your formula isn't pure EV.

Think about a situation where you tie 99.99% of the time, and win .01% of the time. Suppose you get $5000 for a tie, $10000 for a win, and $0 for a loss. Your EV is around $5000.01. However, this situation is great in this game. However, you may have a situation where you win 99% of the time, and lose 1% of the time and never tie. Your EV is $9,990, which is higher, but its better to do the former situation.

[/ QUOTE ]

You are right that the formula is not EV. EV = W% + .5T%.

As for the formula that actually maximizes your chance of winning the freezout, I'm not sure, but I think I'm getting close.

[TomCollins]

I'm pretty sure its Win%/(Win% + Loss%). On that note, is there any better than KK vs K2 for that formula?

[soah]

The formula is correct but wouldn't 77 vs 72 be better than KK vs K2?

[DougOzzzz]

[ QUOTE ]
I'm pretty sure its Win%/(Win% + Loss%). On that note, is there any better than KK vs K2 for that formula?

[/ QUOTE ]

I am pretty sure that formula is correct ONLY if you have an infinite amount of chips. In the actual problem, you have a HUGE amount of chips relative to the blinds, so it is very close to the actual formula for the problem, however, EV is still a very small factor in the equation.

For example Win 90% Lose 9% Tie 1% is a little better than Win 80% Lose 8% Tie 12%... at least I think...

Look at it this way - a tie is not a good result for you, although you will still get another opportunity shortly. However, it costs you chips to wait for this opportunity.

Still 77 vs 72 is much better than KK vs K2 here.

[TomCollins]

It doesn't cost much chips to wait, as proved in a previous post. You do have the chips to wait for an exact hand matchup.

[TomCollins]

I think you are right. Is this the best matchup?

[DougOzzzz]

[ QUOTE ]
It doesn't cost much chips to wait, as proved in a previous post. You do have the chips to wait for an exact hand matchup.

[/ QUOTE ]

That's not the point. Yes, it doesn't cost many chips to wait. Yes you can wait for the best matchup... but what if you've waited 1 million hands, you get the "best" matchup, and you tie? Your situation is a little worse than it was before...

Therefore, EV is still a consideration. Wins/Wins+Losses is only correct if you have an INFINITE number of chips, not 1 billion or even 100 quadrillion. Maybe it's factor is comparably very tiny, but it is a factor.

[soah]

1 million chips in that game is virtually insignificant. 1/1000 of your stack. At 200 million you can start worrying.

[DougOzzzz]

[ QUOTE ]
1 million chips in that game is virtually insignificant. 1/1000 of your stack. At 200 million you can start worrying.

[/ QUOTE ]

keyword: virtually. Thus TomCollins' formula is virtually correct. But not 100% correct.

[TomCollins]

You simply wait for the matchup again. It is statistically impossible that you will have to wait more than 200 million hands before you win. The strategy becomes very simple when you have a 1.6B to .4B stack. One example of a strategy that will work is to call any time you are a 2-1 favorite or better. This may not be the best strategy possible, but it certainly is good enough. If you lose, you revert to the original strategy. If you win, game over.

Additionally, in the event that you do have to wait 200 million hands, and still haven't gotten the matchup you want, you can alter the strategy to include more matchups, for example any XX vs X2 hand. If you get down to 500 million chips, you can wait for XX vs XY where Y < X. But chances are, you won't have to.

[slickpoppa]

Here is the math that I did showing that you can afford to wait for exactly KxKy v. Kz2x or a similar hand:

The absolute best scenario would be KxKy v Kz2x, in which you are a 19:1 favorite. The probability of this exact deal is [(4/52)*(3/51)]*[(2/50)*(2/49)*2] = 1/67,681. However since the blinds are so small compared to your stack, you can afford to wait for this spot. The probability of this deal not occurring after 1,000,000 hands is [1-(1/67,681)]^1,000,000 = 1/2,600,000. After just 2 million hands, the probability of this deal not occurring is already 1/6.8x10^12. But since you only lose $1.5 per hand, even after 2 million hands (and that is about the worst case scenario) you would only have lost .3% of your stack. And if you double up after losing 3 mil, you would still have a commanding 332:1 chip lead and would have to lose 8 all-ins in a row before you would be close to having even stacks again.

[DougOzzzz]

[ QUOTE ]
You simply wait for the matchup again. It is statistically impossible that you will have to wait more than 200 million hands before you win. The strategy becomes very simple when you have a 1.6B to .4B stack. One example of a strategy that will work is to call any time you are a 2-1 favorite or better. This may not be the best strategy possible, but it certainly is good enough. If you lose, you revert to the original strategy. If you win, game over.

Additionally, in the event that you do have to wait 200 million hands, and still haven't gotten the matchup you want, you can alter the strategy to include more matchups, for example any XX vs X2 hand. If you get down to 500 million chips, you can wait for XX vs XY where Y < X. But chances are, you won't have to.

[/ QUOTE ]

Yes, Yes... I am not arguing that it is not beneficial to wait for the ideal matchup. I am simply stating that because the stack sizes are not infinite, that EV is a part of the equation. It may be a very, very, tiny part of the equation, but it exists nonetheless.

The whole point of this is that while your (Wins/Wins+Losses) formula is extremely close to correct, it is not 100% correct.

[TomCollins]

EV is about as significant as analyzing the probability of the end of the world during the match.