Blog
probability – IBM Ponder This puzzle for February 2026
We consider a solitary backgammon-like game. The game board consists of an infinite sequence of locations, marked by
0
,
1
,
2
,
…
0,1,2,…. 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.
My question : What is the utility of this question when blots are avoidable only in doubles. The game continues as long as you roll doubles and ends on the first non-double. Such a thing is not feasible right? I am certain IBM Research would not provide a non meaningful question but I also cant help get out of the non-meaningfulness of the question.