What is the chance of getting a prize on a single trial? If we're working with a population of boxes in which an 1/8 of all boxes have prizes inside, then there is a 1/8 chance of getting a prize on a single hit. That means that every 8^th box, on average, has a prize. But if the boxes are not distributed uniformly, then our experience will never literally be that in every 8^th box we get a prize, right? (as in, "no, no, no, no, no, no, no, YES, etc.") So say we did not get a prize on our first try -- what is our chance of getting a prize on our next try? For a person who just walked up and joined us at the candy store, it would seem as though, again, we have a 1/8 chance. But to us, it's not quite the case. We know we missed out on that 1/8 chance on the last box, so we lucked out in the 7/8 chance of not getting it on the first try. We're thinking, "Oh, I lucked out, so maybe I'll get the prize now, on my second try." That is, we're thinking that we were in the 7/8 unlucky group before AND we'll be in the 1/8 group now. That is, we hope our chances are 7/8 * 1/8. So for us, there is a 7/64 chance of getting the prize now, on the second try. And 7/8 * 1/8 is clearly a smaller chance as compared to 1/8, the chance that the newly arrived onlooker assumes we have. The onlooker would probably find this all very peculiar. The onlooker would say, "Hey, but you have the same chance of 1/8 each time." Alas, when we're paying for candy box after box, we don't treat these incidents as independent. We care for how many boxes we're buying. For us, these first and second chances together make a single dependent compound outcome. So for us, the perspective is different, and our calculations should be, accordingly, different.

What are our chances of getting the prize on the third try?

[last updated August 17, 2004]