A restaurant drawing setup


A few weeks ago a restaurant offered my wife and I entry into a weekly drawing for a gift certificate. They mentioned a little twist by which a first-time entrant would have their ticket remain in the pot until they win. I wondered how such a drawing would play out and if it gave us a reasonably higher chance of winning over a normal drawing in which the pot is cleared each week.

I don’t know exactly how this drawing is carried out, but let’s hypothesize the following setup: first-time entrants place a red ticket in the pot and repeat entrants place a blue ticket in the pot. Each week a ticket is drawn from the pot (the winner), and then all of the blue tickets are removed from the pot. Notice that if the blue ticket of a repeat customer is drawn, their red ticket remains in the pot. This would be the case if the restaurant didn’t bother to go through all the red tickets to find the one corresponding to the drawn blue ticket, which seems reasonable if they were doing this by hand.

Imagine that n red tickets and m blue tickets are added each week, and that Ni red tickets have built up from previous weeks. The probability that a red ticket will win the drawing is (Nin) / (Ninm), and the probability that a blue ticket will win is m / (Ninm). If a red ticket wins, Ni+1 = Nin - 1 red tickets are carried over to the next week, otherwise Ni+1Nin are carried over.

Clearly, if n > 1 then the Ni increase to infinity because Ni+1Ni every week. For the case of n = 1, we know that Ni never decreases, and that for each week there is a positive probability that it increases. Since there is always a positive probability that Ni increases, it will do so infinitely many times, and since it never decreases, it goes to infinity.