A fun (hopefully) gambler's ruin question

[DoomSlice]

How much more money (relatively speaking) would you need to have to ensure a >50% chance to win money if you went all in every time and were called every time your opponent held a 2% advantage over you (51% chance to win) and folded with everything less?

Now to complicate things, how much would you need to have if he had to have a 20% advantage over you (60% chance to win) in order to call?

Assume Player B knows nothing about Player A's cards.

Ex. Player A pushes all in with XXXXX, Player B calls $100 with Q8 (or some hand that holds a 2% advantage over a random hand)

Now factor in the blinds (.5/1) ... and stipulate that Player B only has his initial $100 to play with, and I have now confused myself.

Happy calculating!

[AaronBrown]

I'll start with a simpler version. You and your opponent both get independent numbers drawn from a uniform distribution from 0 to 1. There are no antes or blinds. You go all in every time, your opponent calls with every number greater than X.

Your chance of winning is X*(1 - X)/2 and your chance of losing is X*(1 + X)/2. The other 1 - X of the time your opponent folds and no money changes hands.

Given that your opponent calls, your chance of losing is (1 + X)/2. There is a 50% chance that you will lose 1/[1 + ln(1+X)/ln(0.5)] hands in a row. So you will have a 50% chance of winning if you have 2^N - 1 times as much money as your opponent, and she sets X = 2^(1 - 1/N) - 1.

The table below shows how much money you have relative to your opponent, and what X gives you each a 50% chance of winning.

1 0.0000
3 0.4142
7 0.5874
15 0.6818
31 0.7411
63 0.7818
127 0.8114
255 0.8340
511 0.8517
1,023 0.8661

Your game is made more complex due to the blind and ante, and because the odds of winning at hold'em are not uniform from zero to one. They're approximately Normal with a mean of 50% and a standard deviation of 10%.

[well]

Here's my simplification...

I'll disregard the blinds for now, and I don't think it matters that much.
Furthermore, I will not count comebacks, i.e. once you have less than your opponent, you'll be ruined.

Suppose the opponent will call with any hand that has an equity of X over a random hand.

Then there are three numbers to our interest:

- the probability that he will fold the hand (we will look at this as a tie, t)
- the probability that he will call and win, w
- the probability that he will call and lose, l

And what we want to know, is for what N there is a probability P<.5 that we will lose (at least) N times before winning one.

The probability that the next non-tie outcome will be a win, is w+tw+t^2w+... = w/(1-t) = W.
For a loss, this is l/(1-t) = L.

So, with P=L^N, this yields N> log .5 / log L

With X=.5, N>1.12
X=.51, N>1.15
X=.55, N>1.33
X=.6, N>2.07

For X=.85 (i.e. he calls with Aces and with Aces only), N>27.56.

The higher X is, the less hands your opponent will play, the more relevant the blinds become.
So for the only Aces case, I don't think that is too good an approximation.

Regards,

Well.