Risk after so many hours

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

Thanks in advance...

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.

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.