Winning Deduckto #

For the holidays, my kids got a game called Deduckto. It’s a cute game for 8 years and older, where each player has their own mystery suspect which consists of a suspect (like a pig), a disguise (a mustache), and a location (the library). The goal of the game is to be able to identify all three attributes.

The game works by playing cards that are also a suspect, disguise, and location, and putting each one into piles of “yes”, where there is one or more attributes shared by a card, or “no” where none of the attributes are shared.

The game is meant for children and building their reasoning ability. And for my kids, it’s been a great way for them to exercise that skill. I usually play more cooperatively, where I talk with my kids about what the possibilities are, and help them find relationships they didn’t before, like the fact that a card in the “no” pile eliminates all of the attributes as possibilities.

So optimal play, I’m sure, won’t be the goal for most. That said: what is the most optimal way to play? Here’s my thoughts.

TLDR #

Try to get as many pairs of cards that have a single common attribute in the “yes” pile as possible. The strategy is:

If possible, play a card with only one common attribute with a single “yes” card you already have.
If you have no such card, play a card that shares nothing in common with any of your yes or no cards, to find a new “yes” card. Between the two, prefer yes (since something that matches cards you already have), then no.

Why #

The optimal play is the one that eliminates the most possibilities.

Getting a “no” will at most eliminate three possiblities: a suspect, a disguise, and a location.

Getting a “yes” can, at best, validate three attributes. But in most cases, it will validate one. One way that is done by finding a pair of cards that are disjoint in all but one attribute, but both are yes.

If you find a common attribute, you immediately eliminate all the other choices for that feature. That means you can eliminate up to 6 other possibilities in a single turn! Much more effective than eliminate 6 possibilities with 6 cards.

Getting a pair of yes cards would leave one of two possibilities:

The common attribute matches the secret suspect (e.g. they both are at the “library”).
The two cards match for a different reason, and each on a separate attribute (e.g. for a (fox,bandana,library) and a (bear,mustache,library), fox and mustache are matching or bear and bandana are matching).

Either could be validated by having a third matching card that shares attributes with one or the other. And so on.

Among the cards, you will have the following, in order of preference to play.

card(s) that share 2 attributes, each with one or more yes cards, if the attributes are not yet validated.
card(s) that share 1 attribute with only one yes card, that does not already share an attribute with another yes card.
card(s) that share 1 attribute with two other yes cards.
card(s) that share 0 attributes with any yes cards.
all other cards.

The reason is:

1: This will validate multiple attributes simultaneously. Although rare, it is an ideal play.

2: A single attribute, common with a single yes card, will help identify a unique attribute of your suspect.

3: This will help confirm that the attribute shared by the other 2 cards is indeed the card that is unique to the suspect.

3: The above work to validate attributes your suspect may have. But if you can’t do that, then it’s best to find new “yes” cards. The best way to do that is to play unique cards as much as possible, disjoint to “yes”es, but ideally also “no”s as well.

Anything else is effectively useless, giving no information.