Bases for Bankroll Recommendations?

[pzhon]

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Could you show what happens to the RoR if you drop down a limit once your effective BR drops to 50% of starting strength? Rather than playing that limit until busto.

Assume you're playing 2000NL with a 1BB/100 WR and 20BB/100 SD. You can drop to 1000NL where you're 2BB/100, and if you drop to 50% of that roll, on to 500NL with a 4BB/100 WR (no drop down from here, win or you bust).

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First, that SD/100 is quite low for NL.
Second, with those stats, you are winning the same $/hand at each level, so you should normally play in the game with the least variance. Of course, you might expect that your win rate would rise faster in the tougher game.

To determine the actual risk of ruin, you need to determine your plan for moving back up, if any.

Case 1: You never move back up.
Case 2: If you move down to NL $1k, then double up to your original bankroll, you return to NL $2k. If you move down to NL $500 and double up to half of your original bankroll, you move back to NL $1k.

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Roughly the percentages of hitting the floor of each limit multiplied by one another, no? Except that your BR for the higher two limits is halved in the calculation?

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Yes, for Case 1.

Let ROR_n(x) be the risk of ruin if you play at level n forever with a starting bankroll of x.

Then your risk of ruin with this strategy is

ROR_2k(BR/2) * ROR_1k(BR/4) * ROR_500(BR/4)

~ ROR_2k(BR/2) * ROR_2k(BR) * ROR_2k(4 BR)
~ ROR_2k(5.5 BR)
~ ROR_2k(BR)^5.5

So, this strategy has the effect of increasing your bankroll by a factor of 5.5, or raises the risk of ruin to the power 5.5. Just playing at NL $500 in the first place would be a factor of 16 rather than 5.5, since your bankroll is 4 times as many buy-ins, and your win rate is 4 times as great.

Case 2 is more complicated.

What is the probability that you lose everything before reaching n times your initial bankroll? The probability of going to infinity is both 1-ROR(BR), and probability of making it to n*BR, and then failing to bust out.

(1-ROR(BR)) = P(BR->nBR)(1-ROR(n BR))
P(BR->nBR) = (1-ROR(BR))/(1-ROR(n BR))

You can use this to set up a solvable system of equations for the risk of ruin with BR, BR/2, and BR/4, but the answer I get is not worth typing. Returning to the higher stakes games increases your ROR relative to Case 1.

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Cue the chimpanzee to scream

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what's up with you and screaming chimpanzees

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They don't like me? It's hard to tell, since it seems to be a different one each time. I write that so that someone who seriously disagrees will take more effort to distinguish himself from a screaming chimpanzee than is sadly typical, or will look like a joke.

In another forum, someone posted a poll, A or B. It turns out he really wanted people to answer A. I alone voted B, and posted the results of a mathematical model supporting B, and gave some qualitative arguments about why B is better. The responses? "Kill yourself." "Intuition always beats mathematics." I hope we can do better in the 2+2 forums.