This week the question is: Can You Tip the Dominoes?

You are placing many, many dominoes in a straight line, one at a time. However, each time you place a domino, there is a 1 percent chance that you accidentally tip it over, causing a chain reaction that tips over all dominoes you’ve placed. After a chain reaction, you start over again.

If you do this many, many times, what can you expect the median (note: not the average) number of dominoes placed when a chain reaction occurs (including the domino that causes the chain reaction)? More precisely, if this median number is M, then you would expect to have placed fewer than M dominoes at most half the time, and more than M dominoes at most half the time.

And for extra credit:

You’re placing dominoes again, but this time the probability of knocking each domino over and causing a chain reaction isn’t 1/100, but rather 10^–k, where k is a whole number. When k = 1, the probability of knocking over a domino is 10 percent; when k = 2, this probability is 1 percent; when k = 3, this probability is 0.1 percent, and so on.

Suppose the expected median number of dominoes placed that initiates a chain reaction is M. As k gets very, very large, what value does M/10^k approach?

Highlight to reveal (possibly incorrect) solution:

Program 1, 2.

Calculating using probabilities: Assuming I simply start all these lines at the same time. After placing 1 domino in each line, 1/100 of them will tip over. After placing the 2nd domino in each remaining line, 1/100 of these will tip over. And so on, until more than half have tipped over. I write a short program to keep track of the numbers. The answer is 69.

Simulation: I write a short program to simulate a lot of lines. Same answer.

And for extra credit:

Program.

I change to first program a little, so I can go through probability 1/10, 1/100, 1/1000 etc. Apparently M/10^k goes towards 0.69314718, which looks like ln(2).