Am I using this ROR Formula Correctly?

[jay_shark]

Right , stupid me .

Any multiple of 10 works.

r=0.5 + [0.5^10r^11 + 0.5^20r^21 + 0.5^30r^31+...]

Notice that [..] is a geometric infinite series with
a=0.5^10r^11
ratio= 0.5^10r^10

[pzhon]

[ QUOTE ]
r=0.5 + [0.5^10r^11 + 0.5^20r^21 + 0.5^30r^31+...]


[/ QUOTE ]
That still contains multiple errors. There is no simple way to fix it.

[BruceZ]

[ QUOTE ]
Right , stupid me .

Any multiple of 10 works.

r=0.5 + [0.5^10r^11 + 0.5^20r^21 + 0.5^30r^31+...]

Notice that [..] is a geometric infinite series with
a=0.5^10r^11
ratio= 0.5^10r^10

[/ QUOTE ]

NO, it's still no good. You are only considering the highly unlikely cases where he wins multiples of 10 in a row, and ignoring the primary cases such as winning less than 10 games and losing 1 more than he wins.

[pzhon]

[ 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.

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

[/ QUOTE ]
The accuracy might be just a coincidence.

That the adjustment by 0.45 units is not quite accurate. You are almost twice as likely to end up with 0.9 as you are to end with 0.

0.0 0.0006948208005020496252
0.1 0.0008859642953569293349
0.2 0.0009940893229263886848
0.3 0.0010712675444494628930
0.4 0.0011330034250735614834
0.5 0.0011857261540454036738
0.6 0.0012326439947740134358
0.7 0.0012755630322889410147
0.8 0.0013155907260234342068
0.9 0.0013534510715240267910

partial Mathematica code:

<ul type="square">
// {t[k]} are the solutions of t^10 = (1+t^21)/2 with norm less than 1
equations[k_] := Table[Sum[a[ii] t[ii]^j, {ii, 10}] == If[j == k, 1, 0], {j, 0, 9}]
solutions[k_] := solutions[k] = NSolve[equations[k]]
bustprob[n_, k_] := Sum[a[ii]t[ii]^n, {ii, 10}] /. solutions[k]
Table[{k, bustprob[500, k]}, {k, 0, 9}][/list]The average amount you have when you bust out is .4988 units, so in some sense a better adjustment was to use a bankroll of 49.5012, but that produces a less accurate estimate of the risk of ruin, 0.0112217, which is even higher.