Hello.

Was playing with some friends and a game came up where we imagined if you could play a tourney but you get to pick the hands you play, what the best strategy would be. everything face up, just running cards - like this;

Imagine you have 10k in chips. The goal is to double them, to 20k.

You do this by wagering them in whatever poker hand you like, but the pot odds are the hand odds. So, AK vs 22 pays 1:1, while drawing one card to the nut flush pays 4:1, and so on and so on.
You can pick a hand where you are 50:1 dog or you can take a 100:1 lock, it will just be reflected in the pay out, so fair odds for everything.

You can bet 1 of your chips, or you can bet all of them, it's your choice. You can bet as many times as it takes to get to 20k, but once you bust, that's it.

Is there an optimal strategy to this game? Or does it not matter, because the odds are fair? Is it better to to take 4 shots at an 7-1? or 2 shots at a 3-1? One shot at a coinflip? Or run 1-3 3 times?

Then - imagine the goal was not to double your stack, but score the highest chip stack out of 5 people playing the same game - what's the strategy? Or again, does it not matter?

me brain too smooth to know if it matters or not, but i'm curious.

With no time limit you just wait for AA vs A9o and go all in every time, and bet no other hands. AA vs A6o is almost as good. These are your highest heads up equity possible. Doing this has very close to 100% chance to double before bust. But it will take a while as you only win about 5% each time and these deals will be rare.

Sorry - maybe I explained it really bad.

It would be fair odds, and you can pick whatever cards/odds you want.
So your pay off in that situation would only be 1-19 then.

The only thing I know that would be wrong would be taking odds that exceed the 20k needed.
But apart from that, I canÂ’t see any difference, in between taking a coin flip for your whole stack once, or taking a 75% and taking a 66% after, or taking half your stack at 25% twice.

But at the same time, running aces vs a6 15 times will get you to 20k and that sort feels safer than a coin flip.

On the flip side, you could run a strategy where you take an 80% favourite for your whole stack, and if you win, have 12500, then take 2500 as a 3-1 dog, where if you win, youÂ’ve hit the target and if you lose, youÂ’re still alive with 10k, you take 10k and run it at 80% again, rinse and repeat, which would mean youÂ’re not at risk as often.

But mathematically, is there any difference between these strategies if the whole game is fair odds?

I thought you wanted the strategy with the best chance to double before bust. That's what I gave you. Betting on longer odds hands increases the chance to go broke first. If you get to pick the hand instead of waiting for it, that just makes it faster.

There's a lot of difference in strategy unless you have infinite money and time. But then doubling has no meaning.

No, thanks for the response but I think you’ve stopped reading after the first paragraph, and missed the rules of this (hypothetical) game.
Unless of course I’m misunderstanding you. But surely running aa vs a6 20 times the risk of ruin is not 5%, but 50%. Which then makes the proposition no different to simply coin flipping.

I have specified a game where the objective is to double your stack using any odds you want (that pay fair) and using any part of your stack you want, as many times as you like.

And I’m asking if there is an optimal/best way to do it, or if it doesn’t matter, because the odds are all fair and everyone’s expected value is zero, or whether you take 20 pieces of your stack at long shots or where you put your whole stack at risk once to double, is there actually any difference between the strategy.

P

I don't see how you could possibly improve your odds beyond 50%. There is probably a fairly simple mathematical proof. You could make them worse though, by making a bet where winning puts your over 20k.

The second game is more interesting. Can players see others' scores before the game is finished? I would assume no because I think that kind of breaks the game. This definitely can have some strategy. For simplicity, imagine 2 instead of 5 players. If I know my opponent's strategy is to play no hands and hope I gamble and lose, I can win with probability 1 by risking an infinitesimally small amount of money trying to get a fraction of a chip above 10k. Clearly exploitative strategies can yield an edge in this game. As far as the equilibrium strategy, I need to think about it.

Right, amaZing, thanks for that.

My original intuitive thinking is exactly what you said, but I just don’t have the math ability or knowledge for know for sure.

You’re right about knowing scores etc. maybe if it was defined as such:
For simplicities sake - we can say you get two draws - you can stand pat at any time. You know scores before and after after draws but you don’t know what your opponent is drawing to.

It’s even more straightforward than I thought. If you can double more than half of the time, your EV is necessarily positive, because (.5+x)*20000 >10000 for positive x. No series of neutral EV bets can become +EV.

Right, and so in theory, let’s say it was a $10 buy in, and if you double your stack, you get paid out $30 - just playing the game is +ev. Is there a strategy that is better than the others? (Apart from not making them worse, as you mentioned)

Like if feels like taking high percentage plays is better than a bunch of long shots, but expected value is the same, so does it matter? I can’t tell

If the only outcomes are win/lose then the best strategy just maximizes chances of winning by doubling your stack, which is 50%.

The other game is hard to figure out even with only one round and two players. I would generalize it to being able to make any fair wager with a binary outcome instead of just situations that are possible in poker. Restricting it to poker makes it more complicated but not really in an interesting way and doesn't fundamentally change how the game works very much.

For the Nash Equilibrium strategy you want to bet in such a way that you create a probability distribution for your final stack size that is unexploitable. Basically you can just choose any distribution with a mean of 10k and then work out a betting strategy to achieve that distribution. The distribution represents a Nash Equilibrium strategy if there is no other distribution which has a greater stack size more than 50% of the time (for HU with 1 draw). It's not clear to me how to find this strategy but it probably requires calculus. I'm not sure if the NE will be like RPS where it is neutral EV vs any other strategy or like poker where it is +EV against some suboptimal strategies.