[edit: nevermind all this i solved it but i was asking myself the wrong question, going to post a follow up question below.]

I'm curious how multiple players playing off the same bankroll affects the variance.

For example using the bankroll required equation,

B = -ln(R)*s^2/(2W)

Where,

B = Bankroll
R = Risk of Ruin
s = Standard Deviation
W = Win rate.

If one player is playing MTTs with:

W = 0.24 (BI)
s = 5
R = 0.01

His bankroll requirements are:

B = -ln(R)*s^2/(2W)
B = -ln(0.01)*5^2/(2*0.24)
B = 320 Buy ins.

But what would happen if his identical twin started playing with the same win rate and variance and tolerance for risk?

My feeling is that because the bankroll formula and RoR assumes playing over an infinite number of hands absolutely nothing happens to the bankroll requirements? It would still be the same. Volume doesn't change RoR.

The reason I'm thinking about this is to think about the bankroll requirements of a staple of backed horses.

If you have 10 players in a staple all playing with the same WR, Variance, Risk tolerance and thus bankroll. The bankroll required is the same as if it was just one player. They just put in 10 times as much volume and volume doesn't affect bankroll requirements.

What am i trying to calculate is how much money needs to be kept behind by the backer of the staple (backers bankroll) to always have enough to reload the accounts in the cases that the $ won from the accounts that are up doesn't cover the dollars lost from the accounts that are down $.

So this is just going to be the RoR formula again but with the total win rates of all the players and the total variance used? If the backer wants only a 1% chance of his entire staple going bust he'll need $x dollars in total. The win rate would be WR * N of horses but what about the variance, how would that be calculated if we knew the variance of each individual player?

Ideally, you would do this if:

1. All the players are about the same level
2. They are +EV for the games they will play in
3. They are playing at separate tables

Assuming each of N players has a bankroll of size x, then playing off the same bankroll is equivalent to 1 player with a bank roll of N*x. Effectively, all that is happening is that you are playing N times as many hands per hour with the other N - 1 players acting as surrogates of yourself. Variance is unchanged even though it might seem like it is greater. Counter-intuitively, each of the players involved have effectively multiplied their BR by N. So this can be a way of moving up in stakes for the impatient.

Yep, thanks for that Gibert. That is where i got my understanding too.

Carrying on with my thinking because i'm finding this fun. [nerd]

N horses all w/ x RoR would mean a total RoR for the staple of x^N. Each one would have to go bust. P(B^N).

RoR[s] = RoR[h] ^ N[h]

So if each horse had a 1% chance of going bust a staple of 5 horses would have 0.01^5 chance of the staple going bust: 1/1000MM.

Working backwards the staple wants a 1 in 10k chance of going bust 10,000 and has 50 horses, what RoR should each horse have.

RoR[s] = RoR[h] ^ N[h]

0.0001 = RoR[h] ^ 50

RoR[h] = 83%

Am i on the right track to answer my question going this way?

So we have 50 horses (N) all given bankrolls (x) that give them a 83% chance of going broke.

Staple bankroll = (N) * x.

To find x for a 83% RoR we need to use the BR formula. Taken some MTT figures, variance = 5, WR = 0.24 (24% ROI)

B = -ln(R)*s^2/(2W)

B = -ln(.83)*5^2/(2 * 0.24)

B = 9.7BIs

9.7BI * 50 horses = 485BIs.

While if this was one person who wanted a 1/10k chance of going broke with the same WR, Variance it BR needed would be:

B = -ln(R)*s^2/(2W)

B = -ln(0.0001) * 5 ^2 / (2 * 0.24)

B = 480BIs.

yep, this method isn't the right way for my answer and i just went around in a circle which i would have noticed if not for the rounding which gave me 485BI instead of 480BI. fell for the same mistake twice.

will continue with this tomorrow and if someone can point me in the right direction i'd appreciate it!