Candyland has a variation for older players where you draw two cards each turn and then choose one to use.

Obviously you would normally choose whichever card moved you the furthest down the board. But is there any situation where you would pick the other card? Maybe to take advantage of better draws later on?


When playing the two-card variant, both cards will end up in the discard pile at the end of the turn. The only choice is which of the two colours to advance to.

There is one obvious case where choosing the card with less movement distance is the better choice: Liquorice.

Suppose one card X moves the player to a liquorice space, and the other card S stops short. X is 100% going to skip the player's next turn (or worse depending on the rule variant). S has at most 100% to skip, if that colour is drawn again next turn, but this has been made less likely by the discard of the first card, and generally the presence of other colours still in the draw pile further reduces the likelihood of repeating X in the next turn.

If playing with the rule whereby Liquorice/Dot is cancelled by drawing the same colour again, then this probability has been reduced by the removal of the original X that put the player there.

The net result is that taking S will never put the player in a worse position than taking X, but is very likely to give a better position at the end of the turn following the skipped turn.

It is therefore always better to avoid the Liquorice space, when the choice exists.

  • I don't know the details of the game, but it seems likely that it's possible that sometimes, depending on the deck, you could move further in two turns by taking the loss of a turn. If you're 4 spaces from the liquorice and you can move 1 or 4, taking the 4 might be better. Taking a 1 might leave you likely to get a 1-3 the next turn, meaning you're still further behind. Unless you mean S is always 1 less than X. – Samthere Jan 6 '17 at 9:51
  • By the third turn, I think S can be no worse, and probabilistically is expected to be better. The rest is the negligible tail on the distribution; to make use of it you'd have to be counting cards perfectly and at that point you may as well decide the game based on one card each as bother calculating each turn. – Nij Jan 6 '17 at 10:08
  • What's the maximum distance you can go in a single turn in Candy Land (or more specifically, what could be the highest X compared to an S that moves 1)? If this X is low enough, then S will be better. However, as it decreases, the deck limits it. For example, if there were only 3 colours, and X were 3 and S was 1, you only have a 1/3 chance of S being better than X. Similarly, if X is large and near its maximum, and S is near its minimum, X could become preferable. – Samthere Jan 6 '17 at 10:26
  • There are six colours, each with (almost) equal distribution in the deck. On the board, I believe that no colour repeats without all other colours occurring first, but need to check. Without that assumption it becomes much more particular to where exactly on the board a player is. I've also assumed the simple Liquorice variation ("miss 1 turn") rather than the original ("do not move until you draw the current colour again") since I don't currently have a simulation for the game. – Nij Jan 6 '17 at 10:49
  • So if you have a 1 and a 6, and the 6 will land you on a skip space, the only way for taking the 1 to be better is to draw a 6 on the next turn (and getting 5 is very bad). Not counting the cards drawn from the deck, taking the 6 and skipping a turn is better 5/6 of the time. If it were 2 vs 5, taking the 5 and skipping a turn is better 1/2 of the time, with a slight preference for taking the 5 due to the possibility of having to take a 3 after the 2, resulting in an extra turn lost. – Samthere Jan 6 '17 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.