PDA

View Full Version : A slightly unusual Risk-Of-Ruin question

NoSoup4U
08-31-2005, 03:07 PM
Over at another forum, a few of us have embarked on what we call a "silly quest." The rules of this quest are that you have to sit down at the lowest limit at a particular site with 20BBs and play that lmit until you either go broke or win enough to move up to the next limit with 20BBs. You then play with the same rules at that limit until you go broke or move up again. You can never drop limits or add money.

The question we have is how to calculate the risk of ruin over multiple levels of this quest. Let's pretend that the hero is very good and wins 3 bb/100 with an 18 sd/100. How about if he is 2 bb/100?

The Party version of this quest has 7 stages: .5/1, 1/2, 2/4, 3/6, 5/10, 10/20, 15/30. If one of you clever math guys can teach me how it works, I can translate to attempts with more levels.

We're ignoring the fact that the true winrate probably drops at higher levels. Our feeble attempts to calculate the RoR are resulting in extremely high numbers and I'd love for one you gurus to set me on the path to enlightenment.

BruceZ
08-31-2005, 03:37 PM
[ QUOTE ]
Over at another forum, a few of us have embarked on what we call a "silly quest." The rules of this quest are that you have to sit down at the lowest limit at a particular site with 20BBs and play that lmit until you either go broke or win enough to move up to the next limit with 20BBs. You then play with the same rules at that limit until you go broke or move up again. You can never drop limits or add money.

The question we have is how to calculate the risk of ruin over multiple levels of this quest. Let's pretend that the hero is very good and wins 3 bb/100 with an 18 sd/100. How about if he is 2 bb/100?

The Party version of this quest has 7 stages: .5/1, 1/2, 2/4, 3/6, 5/10, 10/20, 15/30. If one of you clever math guys can teach me how it works, I can translate to attempts with more levels.

We're ignoring the fact that the true winrate probably drops at higher levels. Our feeble attempts to calculate the RoR are resulting in extremely high numbers and I'd love for one you gurus to set me on the path to enlightenment.

[/ QUOTE ]

<font color="red">The following is no good. Ignore this post, and see my next post in this thread for the correct method.</font>

To compute the probability that you go broke before your bankroll reaches some amount B, compute the risk of ruin for your initial 20 bb bankroll if you played forever (normal risk of ruin), and subtract the risk of ruin for a bankroll of size B played forever at the same level.

For example, if your bankroll is 20 bb, with a win rate of 3 bb/100, and a SD of 18 bb for 100 hands, then your risk of ruin if you played forever at this level would be exp(-2*3*20/18^2) = 69%. Now once you double your bankroll, your risk of ruin from that point on if you stayed at the same level forever would be (69%)^2 =~ 47.7%. This is because to go broke you would have to lose a 20 bb bankroll twice. You can also get this by using the formula again for a 40 bb bankroll. So the probability that you go broke before you double your bankroll is 69% - 47.7% = 21.3%. That means there is a 1 - 21.3% = 78.7% chance that you will reach the next level.

Compute these probabilities of success for each level, and multiply them all together to get the probability that you reach the highest level, and then 1 minus this would be the probability that you go broke before you reach the highest level.

NoSoup4U
08-31-2005, 04:39 PM
As always, you are the man. We were missing the idea of subtracting your risk of ruin at the upper end, which was resulting in numbers that we knew defied reason. Your method results in numbers that "feel" more accurate.

If you'll indulge me, I think this results in:

.5/1 -&gt; 1/2 .787
1/2 -&gt; 2/4 .787
2/4 -&gt; 3/6 .884 (only need 30 BBs to advance)
3/6 -&gt; 5/10 .849
5/10 -&gt; 10/20 .787
10/20 -&gt; 15/30 .884
Total = 32.3 chance of success. This seems much more like I thought it would be. Maybe even a bit higher than I expected.

However, when I try to adjust for 2 bb/100, something seems wrong. I'm doing:

exp((-2 * 2 * 20) / (18^2)) = 0.781 - exp((-2 * 2 * 40) / (18^2)) = 0.610
which says I have only a 17.1% chance of busting if my earn-rate is lower.

Can this be correct or have I missed something fundamental?

BruceZ
08-31-2005, 06:09 PM
[ QUOTE ]
As always, you are the man. We were missing the idea of subtracting your risk of ruin at the upper end, which was resulting in numbers that we knew defied reason. Your method results in numbers that "feel" more accurate.

If you'll indulge me, I think this results in:

.5/1 -&gt; 1/2 .787
1/2 -&gt; 2/4 .787
2/4 -&gt; 3/6 .884 (only need 30 BBs to advance)
3/6 -&gt; 5/10 .849
5/10 -&gt; 10/20 .787
10/20 -&gt; 15/30 .884
Total = 32.3 chance of success. This seems much more like I thought it would be. Maybe even a bit higher than I expected.

However, when I try to adjust for 2 bb/100, something seems wrong. I'm doing:

exp((-2 * 2 * 20) / (18^2)) = 0.781 - exp((-2 * 2 * 40) / (18^2)) = 0.610
which says I have only a 17.1% chance of busting if my earn-rate is lower.

Can this be correct or have I missed something fundamental?

[/ QUOTE ]

Sorry, I made a mistake. The formula is more complicated than that. Let ror(B) mean the risk of ruin for a B bet bankroll if you play forever, and let r(B') mean the risk of going broke before the bankroll has grown to B'. Then we have:

ror(B) = r(B') + [1 - r(B')]*ror(B')

That is, to go broke with bankroll B, we can go broke before we ever reach B', which has probability r(B'), or we can reach B', which has probability [1 - r(B')], and then lose B', which has probability ror(B'). I was leaving out the factor of [1 - r(B')] before.

Solving for r(B') gives:

r(B') = [ror(B) - ror(B')] / [1 - ror(B')]

where as always, ror(B) = exp(-2*EV*B/SD^2).

So if we start with a 20 bet bankroll, with a win rate of 2 bb/100, and SD = 18 bb for 100 hands, then ror(B) = ror(20) = 69%. Then ror(B') = ror(40) = 47.6%, and the probability that we go broke before we reach 40 is:

r(40) = (69% - 47.6%) / (1 - 47.6%) =~ 40.8%.

So you need to divide the risk numbers you computed earlier by the probability of success. With that correction, the probability of going broke before doubling your bankroll is higher with a win rate of 2 bb/100 than with 3 bb/100, as you would expect.

Note that the probability of reaching the next level is 1 - r(B') or

P(success) = [1 - ror(B)] / [1 - ror(B')].

NoSoup4U
09-01-2005, 05:56 PM
Thanks. This formula results in numbers with no apparent inconsistancies. I would have expected a bigger difference between 2 bb/100 and 3 bb/100, but I bow to the immutable rules of math.