probabilities

I was at a fundraiser and they had a 50-50 drawing. They were selling blue tickets and red tickets for the same price. If a red ticket were drawn first, you won the cash. Then they would continue drawing until a blue ticket came out, and that person won some baseball tickets. However, if a blue ticket was drawn first, that person won both the cash and the baseball tickets. I bought blue and red tickets in equal amounts. A friend bought the same number of tickets, but all blue. Presuming the same number of red and blue tickets were sold, and presuming the cash and the baseball tickets had the same value, how should the tickets have been purchased?

For the minute, let’s consider the case where you’re just buying one ticket. Let R be the number of red tickets sold, B the number of blue tickets, c the value of the cash, and t the value of the baseball tickets.

If you buy a red ticket, your expected winnings are c/(R + B). If you buy a blue ticket, your expected winnings are t/(R + B) + t/(R + B - k - 1), where k is the number of red tickets drawn after the first before a blue ticket is drawn. If you buy one red and one blue ticket, your expected winnings are c/(R + B) + t/(R + B) + t/(R + B - k - 1). Fortunately, it’s impossible for a red and a blue ticket to be drawn in the same round–that would make the calculations ugly.

There’s one important question left here: what is the expected value of k? It’s a little complicated because you’re presumably removing the red tickets that are drawn rather than replacing them. I think it’s given by the sum over n of nBR(R + B - n)[sup]2[/sup] (1 < n), but I don’t know how to do that one by hand.

I’ll get back to you on what happens when you buy multiple tickets of each color.

Combinatorics is the branch of mathematics studying the enumeration, combination, and permutation of sets of elements and the mathematical relations that characterize their properties. :stuck_out_tongue:

True, but this is a probability problem.

This sounds pretty straightforward to me. The first ticket drawn, regardless of its color, gets the cash. And the first blue ticket drawn gets the baseball tickets. So there is an advantage to blue tickets, but there is no advantage to red tickets. If you place no value at all on the baseball tickets, then the red and blue tickets are equal. If you place any value at all on the baseball tickets, then the blue tickets are unambiguously better than the red tickets, no matter what that value is or how many people buy how many tickets of any color.

If the ‘take’ is less than 1/3rd of the amount collected, then you should buy as many blue tickets as possible, including the total available, as your expected return is greater than the amount spent.

If the take is more than 1/3rd of the total collected, then your best bet is to buy none at all, as to buy tickets, your expected return will be less than the bet.

I concur with Chronos’s analysis.