I thought I have read enough probability question threads here to have a go at the following problem but so far the numbers I am coming up with are absurd to say the least. Nevertheless I think this should not be all that difficult a problem for someone who knows how to approach it.
So:
I play Bingo sometimes and they have this progressive jackpot which to win you have to get all 15 of your number within 31 balls (there are 90 balls). Balls are numbered 1-90 and do not repeat (so a given number can only be drawn once).
I got around 8E-9. The chances of getting the 15 balls in 15 tries is 15!/(90!/15!). The number of ways those can be arranged in 31 slots is 31nCr15. Multiply the two and you get 7.7E-9. That seems low to me, somebody else will correct me before I can check my work I’m sure.
Yeah…does seem low. I was getting some numbers even lower than that but still. While your odds may not be great that seems a bit off the charts difficult (unless that is the point…seems achievable but in reality is super unlikely and whoever runs it makes off like a bandit).
This is a problem to be modeled with the hypergeometric distribution. The probability of winning is [sub]15[/sub]C[sub]15[/sub] * [sub]75[/sub]C[sub]16[/sub] / [sub]90[/sub]C[sub]31[/sub], which is approximately 6.5626e-009. If anyone does win, they’re probably cheating.
Hmm…pretty close to what Snarky Kong was getting too. My attempts at this (albeit wrong) were getting quite low numbers as well and I was thinking they were far too low to be correct.
My answers were in fact incorrect but apparently the odds are rather staggeringly awful just the same. While I do not expect the chances to win to be all that great I would still hope they were within the realm of reasonable possibility. Time to find another game methinks.
Hmm, turns out I wrote down the wrong thing, but then also typed in a wrong number in my calculator. The 15! in my equation should be (90-15)!, if you do that my answer agrees with ultrafilter, that helps the ego.
