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Jezebel
12-06-2003, 05:16 PM
If I calculate the Risk of Ruin (with formulas provided in the past by BruceZ) and figure that my 300BB bankroll has a .5% risk of ruin, then I can assume that there is only a .5% shot that my bankroll will go bust from normal fluctuations. However, lets say that I have a bad run only due to fluctuations not bad play and lose 150 bb. Now I figure my RoR lets say to be 15%. Assuming the game conditions have not changed and I'm still playing optimally, do I still have only a .5% shot of losing my whole roll?

Edit: I think I answered my own question with a little additional thought. Can I assume that with my winrate and SD I have only a .5% chance to lose 300bb, but a 15% chance to lose 150bb?

These aren't my numbers, I was just daydreaming and was wondering how the whole RoR works.

bigpooch
12-06-2003, 06:08 PM
Assuming that your risk of ruin were truly 0.5% for your
initial bankroll, once your bankroll is halved, your risk
of ruin (given the same game conditions, i.e., same SD and
win rate) should now be exp(ln(0.005)/2) = 0.0707 or about
7.07%.

Using the formula B = -exp (-sigma**2/(2*mu)) * ln(r), in
general, if the risk of ruin is initially r for the bankroll
of size B and then for some reason it is reduced to aB where
0&lt;a&lt;1, the new risk of ruin is then exp(a ln(r)).

Nottom
12-06-2003, 06:23 PM
I think you already know the answer but think of it like this.

You have A /images/graemlins/diamond.gifA /images/graemlins/heart.gif in the BB of a NL tourney and the button goes all-in which you happily call. He shows A /images/graemlins/spade.gif9 /images/graemlins/diamond.gif.

This is essentially where you are with a big bankroll