Actual results Question for Don

[MJ1]

Don,

I am playing a game that is 6D, S17, DAS, DA2, LS.

BR = 1000 units
EV per 100 rounds = 6 units
SD per 100 rounds = 86.5 units
Lifetime RoR = 19.6%
Probability of reaching goal in 20000 rounds = 72%
Double Barrier RoR for 20000 rounds and goal of 2000 units (this includes the initial 1000 units) = 15%.

So, the probability of either doubling the BR or going broke given the goal and time constraints is 87% (72% + 15%). Wouldn't this imply that there is a 13% chance (100% - 87%) of being somewhere within the interval of -1000 units to 1000 units? If this is so, then how would I figure out the probability of landing somewhere within the following intervals-

-1000 units to 500 units?
-500 units to 0 units?
0 units to 500 units?
500 units to 1000 units?

I'm not sure there is a way to do this. If there is could you walk me through how you ballpark your estimates? I would guess the probability of being from 0 to 1000 units would be greater than being within -1000 to 0 units.

Thanks,
MJ

[DSchles]

There's definitely a way to do it, but I can't do it tonight. I'll look at it for you tomorrow.

Don

[DSchles]

Before I start, let me just verify the conditions. We already know that the TOTAL of all four intervals you give should add to 13%. So, you want to split that larger 13% interval into four sub-groups, as indicated, right?

If so, I'm not sure that we actually can do this with the current calculators, but I'll try to take a look today. Rather busy, so be patient.

Don

[MJ1]

Before I start, let me just verify the conditions. We already know that the TOTAL of all four intervals you give should add to 13%. So, you want to split that larger 13% interval ionto four sub-groups, as indicated, right?

Correct.

Rather busy, so be patient.

No problem. Thanks!

MJ

[DSchles]

As I think more about it, there really isn't a way for the calculators to pinpoint the various end intervals. You were right. But we can make reasonable guesses. As the total is 13%, and as the answers should be skewed to the right, I'd say that the two positive intervals would be at least 7% of the total, if not more, leaving 6% or less for the two negative intervals. Now, we have to split those larger two once again into two more. I'm guessing the same logic might apply, but I wouldn't swear to it.

So let's make the far-left group a little less than 3%, the higher of the two losing groups a little more than 3%, the smaller winning group a little less than 3.5% (but larger than the last group), and the highest group a little more than 3.5%.

If you put a gun to my head: 2.8%, 3.2%, 3.4%, 3.6%. Can't be off by very much.

Don

[MJ1]

Thanks Don, your help is much appreciated. That is just what I was looking for.

MJ1