# Modeling a promotion with balls and urns

1. Nov 28, 2011

### moscafly

Hi, i'm an online poker player and i need some help modeling a promotion that the site where i play offers. I think it's rigged and intend to prove it but i'm short on skillz.

The promotion is described here and i already made a model but can't solve it. It goes like this:

Supose you have 72 distinguishable urns. every one of them exactly alike except for 18 of them, which are smaller. let's call them "rare" urns.

A gun randomly fires undistinguishable balls towards this urns, and each ball thrown goes always to one of the urns (no missed shots). The urns allow infinite balls to fall on them (they don't get full)

Suppose the odds of throwing the ball to one of the 18 smaller urns is 1 in 350. Lets say urns 1 thru 18 are the "rare" ones.

let r= 1/350 (r stands for "rare" urn)

then the probabilty "p" of the gun throwing to a regular urn must be so that

18r + 54p = 1 solving for "p" it stands that the odds of getting the ball to a normal urn is roughly 1 in 56.9

Now suppose firing the gun costs 10 points.

The question goes like this: How many points on average do you have to spend in order to achieve a 95% probability of getting at least 1 ball in every urn?

Poker site claims that the last winner spent 54,000 points and manage to hit every urn. I call horse***.

Of course, they don't give away the exact odds of hitting a "rare" urn and they might be different kinds of "rares", but from a very small sample in a poker forum, people seem to be hitting "rare" urns spending an average of 3500 points, hence i estimated the odds of getting a ball into one of this urns to be 1 in 350.

2. Nov 29, 2011

### Office_Shredder

Staff Emeritus
If it takes you on average 350 shots to get into one of the 18 urns, then the probability of getting into a specific urn is 1/(18*350).

In such a case the expected number of shots to fill all the rare urns (and hence all the urns at that point I'm going to assume) is

350+(18/17)*350+(18/16)*350+....+(18/1)*350

You can solve this by noting that once you've gotten a ball into k urns, the probability of getting another urn filled on your next shot is (18-k)/18*350, so it takes on average 18/(18-k)*350 shots to fill the k+1st urn. So the expected number of points spent is about 220000

3. Nov 29, 2011

### moscafly

Thanks!

i see you didn't include the 95% assuarance... did you managed to calculate the EV(total points spended until last urn is filled)?
or is it more like a rough estimate?