Risk after so many hours

[sp1n-d1zzy]

Does anyone know how to calculate the percentage chance of losing "x" amount after "y" number of playing hours?

Thanks in advance...

 

[assume_R]

Yup, use the following (all in terms of Units):

Y = trip hours
X = trip bankroll
TripEV = EVPerHour * Y
TripVar = VarPerHour * Y
TripStd = sqrt(TripVar)

P(Losing X after Y hours) = N(-(X+ TripEV) / TripStd) + exp(( -2 * TripEV * X / TripVar) * N( (TripEV - X) / TripStd )

Where:
exp(x) is e^x
N(x) is the cumulative normal distribution with mean of 0 and variance of 1, which is defined as the integral from -infinity to x of 1/sqrt(2 * pi) * exp(x^2/2) (see http://en.wikipedia.org/wiki/Normal_distribution). In Excel, for example, NORM.DIST(x, 0, 1, TRUE)

If you want any other equations which take into account a trip goal, let me know.

 

[assume_R]

Empirical Example

As an example:

Unit Size = $10
Win/Hour = $25 (2.5 units)
Trip Length = 10 hours
Variance / Hour = 2300 units^2
Trip Bankroll = $2,000 (200 units)

So:

P(Losing $2,000 after 10 hours) =
N( (-200 - 2.5 * 10) / sqrt(2300 * 10) )
+ exp(-2 * 2.5 * 10 * 200 / (2300 * 10))
* N( (-200 + 2.5 * 10) / sqrt(2300 * 10))

P =
N( -1.48 )
+ exp( -0.43 )
* N( -1.15 )

P =
0.069 + 0.647 * 0.124

P =
0.15

P = 15% of losing $2,000 after 10 hours.