Am I using this ROR Formula Correctly?

[gaming_mouse]

I am using BruceZ's ror forumula for a random walk problem and I want to double check that the formula I am using is the correct one for this type of problem, and also that I have applied it correctly.

BruceZ's ROR Formula Post

Imagine you start with 50 units and play a coin flip in which you win 1.1 units 50% of the time and lose 1 unit 50% of the time.

EV = (1.1)*.5 + (-1)*.5 = .05
Var = ( (1.1)^2*.5 + (-1)^2*.5 ) - .05^2 = 1.1025
SD = 1.05

ror = exp(-2*.05*50/1.1025) = 0.0107

Is this result accurate for the problem described?

Thanks,
gm

[jay_shark]

Winning 1.1 units 50% of the time and losing 1 unit 50% of the time is the same as winning 1 unit 52.5% of the time and losing 1 unit 47.5% of the time .

p=0.525,q=0.475

RoR= (0.475/0.525)^50 ~ 0.006709

[jay_shark]

For further clarification , this assumes that you will play this game indefinitely and that your opponent's bankroll is infinite .

[pzhon]

[ QUOTE ]
Winning 1.1 units 50% of the time and losing 1 unit 50% of the time is the same as winning 1 unit 52.5% of the time and losing 1 unit 47.5% of the time .

[/ QUOTE ]
No. It is similar, but to make that change, you also have to change the amount you start with. That's the main reason you came up with such a different value.

[jay_shark]

Yup , I see the difference now .

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
Winning 1.1 units 50% of the time and losing 1 unit 50% of the time is the same as winning 1 unit 52.5% of the time and losing 1 unit 47.5% of the time .

[/ QUOTE ]
No. It is similar, but to make that change, you also have to change the amount you start with. That's the main reason you came up with such a different value.

[/ QUOTE ]

so pzhon, was my orig calc correct?

[pzhon]

It looks like a correct application of the formula.

One problem is that it is not clear what to do if you end up with, say, 0.2 units. If you count that as busting out, I think that has the effect of decreasing your starting amount by about half of a unit, so you should use a bankroll of about 49.5 units in the formula.

I'm not sure how accurate the formula is for this type of problem. It could be off by 10% of the result or more. I'll check that later.

[pzhon]

[ QUOTE ]

I'm not sure how accurate the formula is for this type of problem. It could be off by 10% of the result or more. I'll check that later.

[/ QUOTE ]
The formula says e^-2000/441 = 0.0107253, if you use a bankroll of 50, and 0.0111722, if you use a bankroll of 49.55 to make up for ending up with an amount from 0 to 0.9 when you bust out.

The actual value is 0.01114212036696421114353026474437004599225433570306...
which is pretty close.

The dominant term for the actual risk of ruin with b units is
1.046022697510633516701401355992 * 0.990957099620702208500186160002^(10b)
where the 0.9909... is the principal root of
((1 + x^21)/2 - x^10)/(x-1). There are 20 roots, but 10 have norm >1, and therefore don't contribute to the risk of ruin, and the other 9 have smaller norms.

According to the approximation with the adjustment by 0.45 units, the dominant term is
1.0416607625275209774 * 0.99097071625143190131 ^ 10b.

Since I knew this would be slightly off asymptotically, I was surprised that it was so accurate.

[jay_shark]

If you want to know the exact probability that you will be cleaned out . (ie , left with 0)

Then you would need to solve this equation for r .

r=0.5 + 0.5^10*r^11

Now take r^50 to get the probability that you lose exactly 50 units .

[BruceZ]

[ QUOTE ]
If you want to know the exact probability that you will be cleaned out . (ie , left with 0)

Then you would need to solve this equation for r .

r=0.5 + 0.5^10*r^11

Now take r^50 to get the probability that you lose exactly 50 units .

[/ QUOTE ]

That equation implies that the only ways to lose a 1 unit bankroll are losing on the first play, or winning 10 in a row, and then losing an 11 unit bankroll. This isn't correct as these are not the only possibilities.