This week the question is: How Many Dice Can You Roll the Same?

… To get started, you roll all 10 dice—whichever number comes up most frequently becomes your target number. In the event multiple numbers come up most frequently, you can choose your target number from among them. At this point, you put aside all the dice that came up with your target number. …

Now, consider a simplified version of the game in which you begin with three total dice (call it “THREEZI”).

On average, how many dice will you put aside after first rolling all three?

And for extra credit:

Let’s return to the original game of TENZI, which has 10 dice.

On average, how many dice will you put aside after first rolling all 10? (What if, instead of 10 dice, you have N dice?)

Læs mere: #ThisWeeksFiddler, 20241004

Highlight to reveal (possibly incorrect) solution:

Program.

Let’s call the 3 dice rolls a, b and c.

• There are 3 possible cases:
  • 3 dice are put aside. This means a = b = c. This again means a is something with p = 1, and b and c are the same, each with p = 1/6. So p(3) = (1/6)² = 1/36.
  • 2 dice are put aside. This means e.g. a = b != c. This again means a is something with p = 1, b is the same with p = 1/6, and c is different with p = 5/6. Furthermore, I could have chosen the different one in 3 different ways. So p(2) = (1/6) * (5/6) * 3 = 15/36.
  • 1 die is put aside. This means a != b, a != c and b != c. This again means a is something with p = 1, b is something else with p = 5/6, and c is something else again with p = 4/6. So p(1) = (5/6) * (4/6) = 20/36.
• A quick check: 1 + 15 + 20 = 36. The sum of our probabilities is 36/36 = 1. Good.
• The average we’re looking for is 3 * p(3) + 2 * p(2) + 1 * p(1) = 3 * 1/36 + 2 * 15/36 + 1 * 20/36 = (3 + 30 + 20)/36 = 53/36 = 1 17/36, about 1.472.
• A quick check: my program says the same.

And for extra credit:

Program 1 and 2.

I figure out a way to extend the previous program to run with more dice. Mainly by programming the dice as a number in base 6. With 10, the average number of dice set aside is 3.445… In general, with N dice, on average I set aside about √N dice.