ROR question

[Eliot]

I think you will get same answer even for n = 13660(instead of 10M), could you please check this ? I choose n=13660 with certain reason.

Is it something to do with the periodicity of rand()? I can re-write with drand48 or a more modern RNG and run it if you wish. I doubt the result will be much different. In the mean time, here are the requested results:

$ ./a.exe
13660, 0.124671
1000000, 0.120268
2000000, 0.120455
3000000, 0.120313
4000000, 0.120312
5000000, 0.120361
6000000, 0.120417
7000000, 0.120509
8000000, 0.120519
9000000, 0.120577
10000000, 0.120642

[James989]

Is it the periodicity of rand()? No problem to re-write it with drand or any other RNG.

No. It is related to "LONG TERM" rounds . . . Normally I use 3 Standard Deviation as "long term" play, it give accurate results most of the time, but this assumption may be wrong, I am not sure.

n = 3^2 * Var/ev/ev = 13660

What is drand ?

Any comments for POST #16 ?

[Eliot]

Any comments for POST #16 ?

In my opinion, choice (b). A simulation will give a more accurate estimate of the ROR for the situation you described.

Surely there is commercial software that does this. Hint. Hint.

By the way, regarding this formula: ROR = e^(-BR * ev/Var). What happens in the gambler's ruin formula above if you set a = 0, x = BR and let b get large?

[James989]

Is it something to do with the periodicity of rand()? I can re-write with drand48 or a more modern RNG and run it if you wish. I doubt the result will be much different. In the mean time, here are the requested results:

$ ./a.exe
13660, 0.124671
1000000, 0.120268
2000000, 0.120455
3000000, 0.120313
4000000, 0.120312
5000000, 0.120361
6000000, 0.120417
7000000, 0.120509
8000000, 0.120519
9000000, 0.120577
10000000, 0.120642

I get the same results ! It's a bit weird, why ROR for n=13660 more than ROR of 10M ? I thought more rounds will give results closer to one Kelly ROR = 13.53% ? No ?

[Eliot]

I get the same results ! It's a bit weird, why ROR for n=13660 more than ROR of 10M ? I thought more rounds will give results closer to one Kelly ROR = 13.53% ? No ?

The left-hand column is the total number of trials, the value "tripno" in your code. It has nothing to do with the number of hands played in any given trip.

Please study my code

[James989]

The left-hand column is the total number of trials, the value "tripno" in your code. It has nothing to do with the number of hands played in any given trip.

Please study my code

13660 is rounds played(n) in a trip, NOT the tripno. I always set tripno = 500,000. Sorry for the confusion.

n = rounds/hands played in a trip.

My simulation results :-
1) n= 2,000, tripno = 500000, ROR = 8.91
2) n= 5,000, tripno = 500000, ROR = 11.56
3) n= 13660, tripno = 500000, ROR = 12.044
4) n= 1000,000, tripno = 500000, ROR = 12.010