Originally Posted by

**SlipEternal** There are more sophisticated methods of dealing with 0.16 expected winners. For instance, you can weight the expected winnings by the probability that there is a winner. This would be vastly more complex, as it would require you to evaluate the distribution of the expected number of winners. This would give you more accurate estimates of expected winnings, but it is, again, far more difficult to calculate. For each prize, you multiply the prize money if there is exactly one winner by the probability that the prize will be won by exactly one winner. Then add half the prize money times the probability that the prize will be won by exactly two winners. Etc. This becomes an infinite sum for each prize that is guaranteed to converge (evaluating this type of statistics would likely be easier with a program like R rather than Excel).