Ponder This Challenge - February 2026 - Blot-avoiding backgammon strategy

This puzzle was suggested by Dan Dima, who based it on a puzzle presented by Stan Wagon who attributes it to Erich Friedman - thanks all!

We consider a solitary backgammon-like game. The game board consists of an infinite sequence of locations, marked by $0,1,2,\ldots$. At the beginning, there are 5 disks ("men") at location 0.

On every turn, two dice are rolled. For the sake of simplicity, we assume the dice only have two equally probable outcomes: 1 and 2.

If rolling two dice results in two different values, the player moves two men forward according to the numbers on the dice (the same man can be moved twice).

If the dice are the same, each die is used twice, so the player can move four men according to the number of the dice (the same man can be moved several times).

A situation where a man is alone at a location is called a blot. The goal of the player is to avoid blots; the game ends after a turn in which a blot was created.

A simple strategy to avoid blots is as follows: If the dice turn out different - too bad, a blot will be created. Otherwise - pick two men and move each of them twice. It can be shown that this strategy gives 1 as the expected number of dice rolls in the game, except for the last roll of the dice.

Your goal: Find the expected number of rolls when playing with an optimal strategy. Note that this can be given as a rational number, but you can also supply it as a decimal number with 6 digits of precision.

A Bonus "*" will be given for solving the puzzle, but instead of the case of dice that can only produce results of 1 or 2, do it for standard dice that produce values in the range of 1, 2, 3, 4, 5, 6 with uniform probability.