# poisson process and time

• Oct 13th 2010, 08:37 PM
yannyy
poisson process and time

A casino has a game that makes payouts according to a Poisson process
with rate 5 per hour. The payout amounts are 1; 2, 3, . . . without limit,
and the probability that any given payout is equal to i is $\displaystyle \frac{1}{2^i}$ Payouts are
independent. Calculate the probability that there are no payouts of 1 or
4 in a given 10 minutes period.
• Oct 17th 2010, 04:11 AM
Moo
Hello,

The number of payouts of i follows a Poisson process with a rate of 5/2^i per hour, since there are ~ 5 payouts an hour and that for each of these, there is a probability of 1/2^i that it is a payout of i.

So we have, for a given i, the probability of having k payouts of i in an amount of time equal to t : $\displaystyle P(N_i(t)=k)=P(N_i(t)=k)=\frac{e^{-\lambda_i t}(\lambda_i t)^k}{k!}$, where $\displaystyle \lambda_i=\frac{5}{2^i}$

Now compute this probability for i=1 and 4, k=0 and t=1/6, since 10 minutes = 1/6 hour.