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.

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  Solution

  The numercial solutions are approximately

  This riddle is a classical use case for using discrete time Markov chains to model a combinatorial game. Since there are 5 men and blots are not allowed, the possible states consist of either the 5 men together, or two groups of 2 and 3 at some distance. The locations on the board are irrelevant; only the distance between the groups matter, so if the distance can be bounded we have only a finite number of possible states, which makes the problem relatively easy to deal with using standard methods.

  Denote by $S_n$ the state "the group of 2 men is $n$ ahead of the group of 3 men". For example, $S_3$ describes the situation "3..2", $S_0$ means all the 5 men are together, and $S_{-2}$ "2.3" (dot being empty loation). A key observation is that if $S_k$ was reached for $k\ge 9$, the game reduces to the "simple strategy" of "different dice - lose; same dice - move two men"; we can only move the two men on the front, making the gap larger. Since this is not the optimal strategy, those $S_k$ states need not be reached at all.

  In all the states $S_k$ with $k<0$, i.e. states where the group of 3 is ahead of the group of 2, for all the dice results one must move the group of 2 (moving only men from the group of 3 results in a blot for every possible outcome), so the resulting state will have index larger than $k$. It can be seen that $S_{-8}$ is the minimal state that can be reached (from $S_8$, if the two dice are 2 and we move one men 8 steps). An exhustive check of all possibilites show that the only relevant states are $S_{-8}, S_{-6}, S_{-4}, S_{-3}, S_{-2}, S_{-1}, S_{0}, S_{1}, S_{2}, S_{3}, S_{4}, S_{6}, S_{8}$, meaning we can make do with a 13-state markov chain.

  Write $E_k$ for the expected number of dice throws given we start at $S_k$. Now it is possible to write recursive equations for the $E$ values. Consider the relatively complex case of $E_4$, i.e. the state "3...2". In this case, the possible outcomes are:

  1. If the two dice differ (probability $1/2$) - we lose. Since we don't count the losing move, this gives a value of 0 (the only state where two different dice don't lose is $S_3$).
  2. If the two dice are 1 (probability $1/4$) we have two optsions. First, we can move one men 4 steps, reaching "2...3", and from this state we have additional $E_{-4}$ expected moves. Second, we can move the two leading men 2 steps each, and then have an additional $E_{6}$ expected moves. So in this case the optimal strategy yields $1+\text{max}(E_{-4}, E_{6})$.
  3. If the two dice are 2 (probability $1/4$), we can either move the leading two men 4 steps (reaching $E_8$) or move the whole 3-people group; two move 2 steps and 1 moves 4, so we reach "2.3" i.e. $E_{-2}$. So in this case the optimal strategy yields $1+\text{max}(E_{-2}, E_{8})$.

  Putting all these together, we have

  $E_{4} = \frac{1}{4}(1+\text{max}(E_{-4}, E_{6}))+\frac{1}{4}(1+\text{max}(E_{-2}, E_{8}))$

  Since we don't know yet the max value in each case, this means we can have four different strategies upon reaching $S_4$. There are not many states where different options are possible, so the total number of possible stategies remain low. A systematic solutions solves the resulting set of linear equations for each possible strategy, determining be optimal one. It can be seen that $E_0=\frac{38643}{37723}\approx 1.024388304218$.

  The solution of the bonus, while similar, is much more complex. This did not prevent solvers from finding a rational solution. David F.H. Dunkley presented the solution

  $1+\frac{666945281308130947393378290610337684171436121400818033523}{3145854282951008317812602474580783762466266380786580139075}\approx 1.21200768418379967$

Solvers

• *Lazar Ilic (30/1/2026 2:15 AM IDT)
• Alex Fleischer (30/1/2026 5:44 PM IDT)
• Sean Carmody (31/1/2026 4:30 AM IDT)
• Stephen Ebert (31/1/2026 9:11 PM IDT)
• *Jean-François Hermant (31/1/2026 9:38 PM IDT)
• *Bert Dobbelaere (1/2/2026 4:30 PM IDT)
• *Prashant Wankhede (2/2/2026 3:48 AM IDT)
• *Sanandan Swaminathan (2/2/2026 7:44 PM IDT)
• *Paul Lupascu (2/2/2026 8:03 PM IDT)
• *King Pig (2/2/2026 10:19 PM IDT)
• *Ankit Aggarwal (3/2/2026 12:01 AM IDT)
• *Alper Halbutogullari (3/2/2026 1:18 PM IDT)
• *Daniel Bitin (3/2/2026 4:37 PM IDT)
• Stéphane Higueret (3/2/2026 7:36 PM IDT)
• *George Jiri Spitalsky (3/2/2026 11:16 PM IDT)
• *Guangxi Liu (4/2/2026 3:31 AM IDT)
• *Bertram Felgenhauer (4/2/2026 6:28 PM IDT)
• Lorenz Reichel (4/2/2026 9:29 PM IDT)
• *Jan Ondras (4/2/2026 11:03 PM IDT)
• *Pitiwat Chimplee (5/2/2026 6:07 AM IDT)
• *Martin Thorne (5/2/2026 7:53 PM IDT)
• *Doug Coleman (6/2/2026 1:58 AM IDT)
• Quentin Higueret (6/2/2026 10:25 AM IDT)
• *Latchezar Christov (6/2/2026 1:54 PM IDT)
• *Chris Howe (6/2/2026 11:32 PM IDT)
• *Titos Betsos (7/2/2026 3:27 PM IDT)
• *Christopher Marks (7/2/2026 9:20 PM IDT)
• Shouky Dan & Tamir Ganor (8/2/2026 6:43 PM IDT)
• *Rethna Pulikkoonattu (9/2/2026 2:03 AM IDT)
• *Naftali Peles (9/2/2026 11:17 PM IDT)
• Wolf Wackeroth (10/2/2026 8:06 PM IDT)
• *Peter Moser (10/2/2026 10:50 PM IDT)
• Gary M. Gerken (11/2/2026 6:35 AM IDT)
• *Kang Jin Cho (11/2/2026 8:58 AM IDT)
• John Tromp (11/2/2026 1:02 PM IDT)
• *Francisco Rodríguez-Carretero Roldán (11/2/2026 2:25 PM IDT)
• *Micah Day-O'Connell (12/2/2026 12:40 AM IDT)
• *Robert Berec (12/2/2026 3:08 AM IDT)
• *Armin Krauss (12/2/2026 9:28 PM IDT)
• Rahid Zaman (13/2/2026 1:47 AM IDT)
• *Daniel Chong Jyh Tar (13/2/2026 7:21 PM IDT)
• Juergen Koehl (15/2/2026 1:29 PM IDT)
• *Patrick Deumann (15/2/2026 11:32 PM IDT)
• Fakih Karademir (16/2/2026 9:48 AM IDT)
• *Jack Saleeby (16/2/2026 8:59 PM IDT)
• *Vladimir Volevich (17/2/2026 5:20 AM IDT)
• Shirish Chinchalkar (18/2/2026 8:14 AM IDT)
• *Dieter Beckerle (18/2/2026 8:40 AM IDT)
• *Mathias Schenker (20/2/2026 3:10 PM IDT)
• *Ahmet Yuksel (20/2/2026 5:10 PM IDT)
• *Travis Crew (21/2/2026 10:15 PM IDT)
• *Franciraldo Cavalcante (24/2/2026 6:17 AM IDT)
• Gumpina Rama Krishna Chaitanya (24/2/2026 9:17 AM IDT)
• *Justin Mazenauer (24/2/2026 6:21 PM IDT)
• Eran Vered (24/2/2026 11:03 PM IDT)
• *Erik Hostens (25/2/2026 10:51 PM IDT)
• *Tim Walters (26/2/2026 1:54 AM IDT)
• *Nyles Heise (26/2/2026 11:32 PM IDT)
• *David F.H. Dunkley (27/2/2026 1:34 AM IDT)
• *Jackson La Vallee (27/2/2026 8:23 PM IDT)
• Kennedy (28/2/2026 6:49 PM IDT)
• *Motty Porat (28/2/2026 7:40 PM IDT)
• *Todd Will (1/3/2026 8:21 PM IDT)
• *Evan Semet (2/3/2026 5:03 PM IDT)
• Gürkan Koray Akpınar (2/3/2026 7:39 PM IDT)
• Radu-Alexandru Todor (3/3/2026 2:08 AM IDT)
• Karl D’Souza (3/3/2026 2:50 PM IDT)
• Karl Mahlburg (4/3/2026 2:49 AM IDT)
• Reiner Martin (4/3/2026 3:28 AM IDT)
• *Zoltan Haindrich (4/3/2026 12:37 PM IDT)
• *Li Li (5/3/2026 2:27 PM IDT)
• David Greer (8/3/2026 12:00 AM IDT)
• Marco Bellocchi (9/3/2026 8:20 PM IDT)