# Quantum Go Fish

Quantum Go Fish is a game I learned while I was at Berkeley. Scott Morrison reintroduced it in the Berkeley ecosystem when I was there; he says he thinks he learned it from Dylan Thurston. My enthusiasm for the game has been directly buttressed by the enthusiasm of David Penneys.

The basic nature of the game is that the players start with no information about any of the cards, and they reduce this uncertainty by asking and answering questions. Paradoxically, it's a complete information game.

# How to play

Any number of players can play the game. Each player starts with 4 "cards" (traditionally simply the fingers on one of their hands). Nobody knows the identity of any of the cards to begin with. The players then take turns playing. When it is your turn, you may ask a player for a particular suit. If that player has any cards of that suit, she gives you one of them. The players and the game follow the following rules:

1. There are as many suits as there are players. Traditionally, the suits are strange things. You get to invent the name of a suit by being the first player to ask for cards of that suit.
2. There are precisely 4 cards of each suit.
3. A player may only ask for a suit which she already has a copy of. This means that if you provably do not have any ducks, then you may not ask another player for ducks. If you might have ducks, then you may ask for ducks. By asking for ducks, you have determined that you do have at least one duck.
4. If you are asked for a particular suit which you might have, you may decide whether you do not have that suit (in which case you can never ask for that suit thereafter) or whether you do have that suit (in which case you must give one card of that suit to the player who asked for it).

## How to win

A player can win in one of two ways:

1. If at the end of your turn (on your turn, you ask a question and get an answer), the hands of all players are determined, you win.
2. If you provably have all four cards of any one suit, and no other player has won (based on the previous condition), you win.

## How to lose

If a paradox occurs (e.g. if it turns out there are five or more of some suit in the game), then the players have allowed some illegal move to occur at some point, so everybody loses.

# A sample game

Suppose there are three players, unimaginatively named A, B, and C. Then there are three suits, which we will for simplicity call 1, 2, and 3. When the game starts, the three players each have four cards which could be anything. The state of the game is

A has ????
B has ????
C has ????

A goes first.

A: B, do you have any 1's?

At this point, we know that A must have a 1, so the state of the game is

A has 1???
B has ????
C has ????

Since it is possible that B has a 1, and it is possible that B does not have a 1, B gets to choose how he answers.

B: Yes, I have a 1. Here you are.

B gives one 1 to A. Now A has at least two 1's. B may still have another 1. So the state of the game is

A has 11???
B has ???
C has ????

Now it's B's turn.

B: C, do you have any 1's?

By asking the question, it has been determined that B has a 1, so the state of the game is

A has 11???
B has 1??
C has ????

C: Yes, I have a 1. Here you are.

Now B has at least two 1's, and A has at least two 1's, so all the 1's are accounted for. This means that the remaining cards cannot be 1's. So the state of the game is

A has 11(not 1)(not 1)(not 1)
B has 11(not 1)(not 1)
C has (not 1)(not 1)(not 1)

Now it's C's turn. Note that since C does not have any 1's, she cannot ask for 1's.

C: A, do you have any 2's?

The state of the game is now

A has 11(not 1)(not 1)(not 1)
B has 11(not 1)(not 1)
C has 2(not 1)(not 1)

A: No, I don't.

The state of the game is now

A has 11(not 1,2)(not 1,2)(not 1,2)
B has 11(not 1)(not 1)
C has 2(not 1)(not 1)

But there are only three suits, so as soon as a card is not a 1 or a 2, it must be a 3. Therefore, the state of the game is

A has 11333
B has 11(not 1)(not 1)
C has 2(not 1)(not 1)

But there are only four 3's total, so at least one of B's cards must be a 2. Similarly, at least one of C's undetermined cards must be a 2, so the state of the game is

A has 11333
B has 112(not 1)
C has 22(not 1)

We see that A's answer has really reduced the amount of uncertainty in the game! Now it's A's turn. Note that A has no 2's, so he cannot ask for 2's.

A: C, do you have any 3's?

Since we already knew that A had 3's, this question does not change the state of the game.

C: No, I don't.

This means that C's (not 1) is not a 3, so it must be a 2, which means that B's (not 1) must be a 3. So at the end of A's turn, the state of the game is

A has 11333
B has 1123
C has 222

Since all the hands are determined, A wins.

# A physical deck for the game

Quantum Go Fish is traditionally played by trying to keep track of the state of the game entirely mentally, often while inebriated. Though that's fun, I came up with a physical deck for playing the game which automatically does a lot of the bookkeeping for you, freeing your brain cycles up to strategize.

The basic observation is that each player has some cards which are determined, and some which are not, and the cards which are undetermined are undetermined in exactly the same way. In other words, a hand might look like 1(2 or 3)(2 or 3)3, but it cannot look like 1(2 or 3)(1 or 3)3.

With this in mind, the physical deck for 6 players (you can replace 6 by $n$ if you wish) consists of 6 packets, each packet consists of

• 6 cards numbered 1 through 6, and
• 4 paperclips

When the game starts, each player places the 6 cards face up in front of him, and has 4 paperclips in hand. The paperclips represent the player's hand, and the cards represent the possible suits of the paperclips. Read the previous sentence again. When the suit of a paperclip becomes determined, you attach it to the card corresponding to its suit. When a player cannot have any paperclips of a given suit, he turns the card corresponding to that suit face down.

Suppose I ask a player for a 5. If there is a paperclip attached to my 5 card, it means we already knew that I had a 5, so nothing changes. If I do not have a paperclip attached to my 5 card, I attach one of the paperclips I have in hand to the 5 card.

Suppose another player asks me for a 5. If I have a paperclip attached to my 5 card, then I remove that paperclip from the card, give it to the other player, and she attaches it to her 5 card. If I do not have a paperclip attached to my 5 card (but I have at least one paperclip in hand), then I either

• say "yes, I have a 5," give one of my paperclips to the other player, and she attaches it to her 5 card, or
• say "no" and turn my 5 card face down.

Note that not all events (attaching a paperclip or turning a card face down) are triggered by a question or an answer. For example, in the sample game above, all the interesting deductions we were able to make after A said he didn't have any 2's would have to be worked out logically. If a player works something out, she announces it to the other players, paperclips are attached, cards are turned over, and the game continues.