An entry for Fortnightly Topic Challenge #40, have fun with non-Chess puzzle! ;)

Three young children: Red, Green, and Blue; were playing a classic game of Snakes and Ladders, when suddenly their mom called them to have a dinner.

When they came back to their room, they forgot whose turn it was; even they were not sure either about the order of the play!

This was the last condition of the board, including the die. Could you tell them whose turn it is and the order of the play?

Unfinished Snakes and Ladders

Some notes and clarifications:

  • They put their markers at tile 1, at the beginning.
  • They will win if their markers land exactly on tile 100. They are using the "bouncing-back" version (roll 3 will move the marker from tile 99 to 98).
  • As long as they roll 6, they will get an extra roll after moving the marker.
  • The ladders should be used to go to a higher number only, and the snakes are used to go to a lower number. For the "interrupted" spaces (the squares being described by @Bass in his first comment below), both are immediately used (e.g. just as landing on 39 will bring you to 25, landing on 20 will bring you to 25 too).
  • If someone lands on a square where there is already a player, it doesn't send him back to square 1.
  • 1
    $\begingroup$ There are a couple of non-obvious things about this board. I'm guessing the straight ones are the ladders, and that a snake's head is always on a higher number than its tail. There are squares, however, that have one snake's tail, and another snake's head. How do these work? If you slide down the upper snake, will you immediately also slide down the lower one? There are also similarly interrupted ladders in there. $\endgroup$
    – Bass
    Oct 23, 2018 at 6:15
  • 2
    $\begingroup$ @Bass Ah right, I should clarify that the purple parallel-lines are ladders and the orange ones are snakes. The ladders should be used to go to higher number, but the snakes are used to go to lower number. For the interrupted ones, both are immediately used (e.g. landing on 39 will bring you to 25, same as landing on 20 will bring you to 25 too.) $\endgroup$
    – athin
    Oct 23, 2018 at 7:19
  • $\begingroup$ In the version of the game I used to play with my kids, if you land on a square where there is already a player, you sent him back to square 1. Does it apply here ? $\endgroup$
    – Evargalo
    Oct 23, 2018 at 7:26
  • 1
    $\begingroup$ @Evargalo No, it doesn't apply here $\endgroup$
    – athin
    Oct 23, 2018 at 8:07
  • 2
    $\begingroup$ Ah; My version was chutes and ladders. With that perspective the diagram is a little easier. $\endgroup$
    – Joshua
    Oct 23, 2018 at 15:26

3 Answers 3


The other guys already got the answer. (Maybe not with completely airtight arguments, but nevertheless.) The explanations would be a lot easier to follow if they had pictures, so here's one:



  • Red square: after turn 1, you are in one of these.
  • Green square: after turn 2, you are in one of these, or in some red square (except square 2).
  • Red circle with R: if Red was the last player to move, Red's turn started at one of these. (Because of the 4 visible as the last throw.)

From there, the solution is pretty simple:

1. Green has had either one or two turns.
2. Therefore, both Red and Blue have had at most three turns.
3. Therefore, because no green square overlaps with a letter R, Red was not the last to move.
3.1 There is no way to get to Red's current position within two turns, so Red has taken at least three turns. Combined with point 2, Red has taken exactly three turns.
3.2 Since Red has already taken three turns, but wasn't the last player to play, someone else must have also played three turns. It can only be Blue.
4. Because Red has played three turns, Green has taken 2 turns, rolling first 1, then 2.

So the only remaining things to check are that

5. Red really can get to its current square in 3 turns (yes, starting the third turn at 66, 62, 60 or 56.)
6. Blue can get to its current spot in 3 moves, ending with a 4 (yes, starting the third turn at 58, 60, 65, or indeed 77)

So finally, the answer is

It's Green's turn, and after that it's time for the fourth round, with Red playing next.

  • $\begingroup$ rot13(Jul vfa'g vg Oyhr'f 3eq ghea? Oyhr vf ba n fcnpr gung pna or ernpurq va 2 gheaf. V'z pregnva lbh unir na rkcynangvba, ohg V qba'g frr vg va gur nafjre. Creuncf V zvffrq fbzrguvat be lbhe nafjre pbhyq hfr n yvggyr rynobengvba?) (I hate these rot13 things, but there's no other way to spoiler comments.) $\endgroup$
    – jpmc26
    Oct 23, 2018 at 21:23
  • $\begingroup$ @jpmc26, Well, the last one to play tossed a 4, and it wasn't Green or Red, so.. $\endgroup$
    – Bass
    Oct 23, 2018 at 22:02
  • $\begingroup$ Yup, this is the full solution, congrats! :D -- As @jpmc26 noticed, you may add an explanation why ROT13(Oyhr pna'g or ba gur frpbaq ghea). It's because ROT13(vs Erq vf gur bayl bar jub nyernql ba gur guveq ghea, Erq vf gur ynfg crefba gb cynl; naq vg'f n pbagenqvpgvba orpnhfr gur ynfg crefba ebyyrq 4 naq vg'f vzcbffvoyr sbe Erq). Nevertheless, have a green tick! :) $\endgroup$
    – athin
    Oct 23, 2018 at 23:54

It is:

Green's Turn
The Turn order is: Red, Blue, Green


Blue was the child who just rolled the 4, you know this because Green is only 3 spots from the beginning and if Red had rolled a 4 he would've started the last turn from an impossible position on top a chute
Because of Green's position he must have only taken 1 or 2 moves from the beginning of the game. Both Red and Blue's position is impossible to reach in 1 turn, as even with the roll again on 6 rule, they get trapped in an infinite loop with the slide from 31 to 7.
There are 9 effectively different starting positions for turn 2, a unique one for if their last roll landed them on one of the escape ladders or their last die roll. It is impossible to reach Red's current position or Blue's known starting position from any of these 9 starting positions, meaning that Red has already taken 3 turns, Blue has just taken his 3rd turn and Green must go next.

  • 4
    $\begingroup$ couldn't red have been on 95 the previous turn (atop a ladder) $\endgroup$ Oct 23, 2018 at 5:35
  • $\begingroup$ He couldn't have gotten to 95 in 1 turn though, so that might have been his starting spot from the last turn but it was on the third turn. $\endgroup$ Oct 23, 2018 at 5:57
  • 1
    $\begingroup$ There are six possible starting positions for Red's last turn, even with the restriction that the last roll was a four. I think you need to show that none of them can be reached in 2 moves. Blue's last move's "known starting position" isn't unique either. $\endgroup$
    – Bass
    Oct 23, 2018 at 13:22
  • 2
    $\begingroup$ The die is loaded man. $\endgroup$
    – Joshua
    Oct 23, 2018 at 15:27
  • 1
    $\begingroup$ Yeah, I didn't include my image because I didn't have it done in nearly as nice a way as Bass did (I filled each tile with what possible T1 turns could've gotten where). I think another smart solution would be to just prove the infinite loops here that get 6 rolls stuck, the first one on 31-7 is obvious, but all you need is to prove you need to make it past two of those. Great puzzle regardless, really enjoyed it $\endgroup$ Oct 24, 2018 at 5:11

Assuming the order of players to roll the dice is R B G,
the next player after they come back from dinner will be



On the first round:
R rolled 6-6-6-6-5 => ends in 30.
B rolled 6-6-6-6-5 => ends in 30.
G rolled 1 => ends in 2.
On the second round:
R rolled 6-6-5 => ends in 66.
B rolled 6-6-3 => ends in 77.
G rolled 2 => ends in 4.
On the third round:
R rolled 6-6-6-6-6-3 => ends in 99.
B rolled 6-6-4 => ends in 77 again.
Last roll was 4 and now it is G's turn.

  • 1
    $\begingroup$ This is one possible game resulting in the given position, so you get one possible next player, and one possible starting order. You still need to add retrograde analysis to show that any possible game that leads to that position always has those same properties. $\endgroup$
    – Bass
    Oct 23, 2018 at 13:33
  • $\begingroup$ Oh, okay. Sorry. Don’t really know how to answer these kinds lol $\endgroup$
    – ImongMama
    Oct 23, 2018 at 16:54
  • $\begingroup$ No problem whatsoever, there's been a first retrograde analysis problem for everybody. (In the very limited sense of "everybody who's ever tried to solve a retrograde analysis problem", which has to amount to at least a one-millionth part of the world's human population..). And you got the final answer right anyway :-) $\endgroup$
    – Bass
    Oct 23, 2018 at 17:07
  • $\begingroup$ Yep, recalling @Bass comment, you may still have to show other configurations are invalid. Good luck for the retrograde analysis journey and have a +1! :D $\endgroup$
    – athin
    Oct 24, 2018 at 0:05

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