# Silent Auction

I was at a friend’s party the other day and it got me thinking about silent auctions. For those of you who don’t know, a silent auction is kind of like a draw. There are different prizes and each prize has a bag next to it. You buy tickets and put your tickets in the bags next to the prizes you hope to win. At the end of the night, one ticket is drawn from each bag to decide the winner.

The question I had was this: suppose I want to win a prize. Should I put all my tickets in one bag, or should I distribute my tickets among all of the bags?

To attack this problem I am going to simplify the situation. I want to address the probabilities involved in the two strategies. Now I can already hear some of you saying “but that’s not real life.” I know it’s not real life. However, to understand real life, we are often required to investigate a simpler version of the problem first. Then we can build back in the complexity of real life later.

At my idealized silent auction, each prize is of equal value and desirability. Suppose there are only two prizes, a red gift card and a blue gift card. Further, suppose each bag has 10 tickets in it already and you are the last person putting in tickets. Let’s start by buying 2 tickets.

In strategy one, you put both of your tickets into the bag for the red gift card. Now that bag has 12 tickets in it and 2 of them are your tickets. Hence, basic probability tells us that your chance of winning is 2/12 or 16.7%.

In strategy two, you put one ticket in each bag. Now the red gift card bag has 11 tickets in it and 1 of them is your ticket. Hence, your chance of winning the red gift card is 1/11 or 9.1%. The blue gift card bag has 11 tickets in it and 1 of them is your ticket. Hence, your chance of winning the blue gift card is 1/11 or 9.1%. We then add up these probabilities*, and our chance of winning one of the gift cards is 17.4%.

Clearly 17.4% > 16.7%. Hence, our second strategy seems better. What if we had more tickets of our own, say 10.

In strategy one, you put all of your tickets into the bag for the red gift card. Now that bag has 20 tickets in it and 10 of them are your tickets. So 10/20 or 50.0% chance of winning.

In strategy two, you put 5 tickets in each bag. Now the red gift card bag has 15 tickets in it and 5 of them are your tickets. So 5/15 or 33.3% chance of winning. The blue gift card bag has 15 tickets in it and 5 of them are your tickets. So 5/15 or 33.3% chance of winning. We add up these probabilities**, and our chance of winning one of the gift cards is 55.5%.

Strategy two seems to be gaining significant ground. What if there were 3 prizes? I guarantee you that strategy two is better. I’ll leave this situation as an exercise for the reader to verify.

Here is my intuitive explanation as to why strategy two is better. The problem with dumping all of your tickets into a single bag is that it dramatically increases the number of tickets in the bag. All of your tickets have to fight against each other to try to win the same prize. When all of your tickets are spread out in different bags, they don’t have to compete with one another and you gain a higher probability of victory.

There you have it. If you want to optimize your chances of winning a prize at a silent auction, distribute your tickets evenly among all the prizes. And when you win yourself that big 60’ TV, remember to thank math.

*To add up these probabilities we need to use a bit of sophisticated math. We don’t just add up the percentages of each event occurring.

Think of finding this probability as finding the area of two overlapping circles. We can take the area of event A and add it to the area of event B. But that means that we double counted the area where the two circles overlap. Therefore, we have to subtract off this area,

Below is a diagram of how the “addition” works.

We use the following formula:

Probability (A or B) = Probability (A) + Probability (B) – Probability (A and B)

17.4% = 9.1% + 9.1% – 0.8% (0.8% is the probability of winning both prizes, 1/11 * 1/11 = 1/121)