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Problem 10

Expected number of dice rolls

Due: January 19
Points: 2

The previous problem talks about a common way to choose who goes first among \(n\) players of a game:

  1. If \(n=1\), there is just one player remaining so they go first.
  2. Otherwise, if \(n\ge 2\), each player rolls 2 dice and takes the sum.
  3. Any players who do not have the highest sum drop out; they won't be going first.
  4. Set \(n\) to the number of players who tied for the highest, and return to Step 1.

For \(n=1\) up to \(n=10\), calculate the expected number of total dice rolls that \(n\) players will make in order to decide who goes first. For example, with \(n=3\), you know the expected number is at least 6 because every player starts out by rolling 2 dice. But if there's a tie, there's a chance that more rolls will happen. So the expected value for \(n=3\) will be somewhat greater than 6.

Hint: In fact, the expected number of rolls when \(n=3\) is \[\frac{3888}{575} \approx 6.76174.\]

(You can come up with a formula for this if you want, but I only need the values for \(n=1\) up to \(n=10\), accurate to at least 4 decimal places.)