Hey we were trying (and failing) to calculate the odds of winning our super bowl pool at least twice based on the below info:

There are 100 squares in the pool
Each quarter of the game there is 1 winner (so 4 winners for the whole 100 square pool, the same person could win all 4 quarters potentially)
The person bought 4 squares.
What is the probability of that person winning with at least 2 squares?
Also what would the probability have been if they had just bought 1 square instead of the 4?

Not very timely, but since no one posted an answer, I thought I would give it a shot.

4 / 100 = .04 or 4% for the first quarter. To get the probability of winning multiple quarter you can multiply the probability of winning one by the same probability for each subsequent quarter: .04 * .04 = .0016 (for two) .04 * .04 * .04 = close to zero.

That would be more accurate for a random selection: think roulette wheel. This problem is more complex, because of the nature of the game and the the scoring. I would say that there are more beneficial numbers to have (i.e., multiples of 3, 7 and 10 versus multiples of 4, 5 or 9). Also, if the first quarter produces a win, then the likely hood of additional wins, due to no more scoring will go up.

I'm sure someone more well versed in the subject could give you a more scientific answer, but that is my stab at it. Hope it helps at least a little.