Readers, I have a question: how many unique games of chutes and ladders can you have on a particular board? It’s a question that I believe cannot be answered, but I would love to hear your answers, or at least your methodology for trying to reach an answer.

The question stems from me and my wife talking about the forthcoming Pandemic Legacy board game. Like Risk Legacy, it will involve a series of playthroughs, each of which has consequences for future playthroughs because the events mean the cards and the board are permanently modified. At the end of between 12 and 24 playthroughs, I explained, owners should be left with a uniquely configured board. As a writer you will understand why I then immediately took issue with my own terminology: while potentially large, the various possible combinations of game events and changes to the board and cards is surely a finite number.

That prompted my wife to pose the question of how many unique games — that is a specific sequence of moves — could be had on a snakes and ladders (aka chutes and ladders) board. She believed the calculation must be possible given the game is so simple. Here’s my thinking which, given my leaning to linguistics rather than mathematical aptitude, may well be flawed:

Such a question is certainly viable in chess. While the variety of pieces and the various nuances such as castling and Queening make it a complicated mental arithmetic sum, there is surely an answer. Any move creates a specific configuration of the board, from which there’s only a set range of allowable moves, each of which in turn creates a new configuration. The cycle isn’t endless as some configurations are checkmate and the end of the game. So such a calculation must be possible, though it seems most people who’ve tried to carry it out have given up.

As for chutes and ladder, any answer could of course vary depending on the specific board set-up, though the principle should be the same.

If we ignore the chutes and ladders themselves, there’s very clearly a possible calculation. Mathematically-skilled readers will no doubt be able to offer the formula, but it would simply involve working out every combination of numbers between one and six that either add up to 100 or where the last number takes the total above 100.

That’s for one player of course. For the two player game, we’d presumably need to multiply this by a similarly-calculated number of the possible combinations that add up to 99 or less (and where the number of turns for either side at each stage is either equal or the starting player’s is bigger by one.)

It’s the addition of the chutes and ladders that to me makes the question impossible. It’s clearly possible to make a series of rolls that winds up with them falling down a chute or chutes and returns them to an earlier position. For example, in the image above, a player could be on square 7, throw a six, move to square 13, throw a three and fall down the snake from square 16 to square 6, then throw a one and return to square 7.

In turn, it’s possible (if less likely) that the player could repeat this series of rolls and thus repeat the exact same advance/fall down chute cycle. It’s also possible this could happen three times. And it’s conceivable that both players could repeat the necessary series of rolls three times in a row.

(In chess, this would be the end of the complexity given the common rule that a player faces their turn with the board in the same position three times, the game is a draw.)

The possibilities in chutes and ladders of course go to four repeated series of rolls for both players, and beyond. At first glance it might seem as if we can extend this sequence to the — literally infinitesimal — possibility of both players being stuck in an infinite loop by repeatedly throwing the same series of numbers. And any series of consecutive numbers that ends in infinity must logically have an infinite number of entries, thus making the answer to the original question “there’s an infinite number of different chutes and ladders games.”

In practice however, this doesn’t necessarily answer the question: not that is, if you take it to mean the number of different completed games. A completed game must have had a finite number of moves. The problem as far as I can figure is that you can  map out any possible game involving a repeated loop — for example that both players repeated the same advance/chute loop a trillion times before one got the luck of the die and broke away. However, in the manner of a small child badgering their exasperated parents with questions about large numbers, you can then immediately conclude that there’s a longer game possible in which the players repeat the loop a trillion and one times before the sequence of rolls changed. These would be different games in just the same way as a game where somebody falls down a particular chute once is different to one where they fall down that chute twice.

So if that’s right, there’s no possible answer to the number of unique games possible: any numerical answer can be proven wrong by adding one. That only leaves the answer of “infinity minus one” which most mathematicians consider not only non-existent as a number, but a meaningless or impossible concept.

But that’s just my take. What’s yours?