## Teaser 2903: What’s my card?

**From The Sunday Times, 13th May 2018** [link]

Four expert logicians play a game with a pack of six cards, numbered from 1 to 6. The cards are shuffled and each draws a card, holding it in front of him so that he cannot see his own number but can see the numbers held by the others. The winner is the player who can first deduce what number he is holding.

Each player in turn is asked to announce the sum of two numbers that he can see on the other cards. One game started with Alf announcing 6. After a long pause, no-one claimed victory, so Bert then announced 5. There was a significant pause, but before Charlie spoke, someone claimed victory.

Which cards did Alf and Bert hold, and which cards remained on the table?

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## Jim Randell 12:38 pm

on17 June 2019 Permalink |I took the “long pause” to mean that no-one can immediately declare victory, and each player can take the fact that no-one can immediately declare victory into account. And even with this additional information no-one can declare victory, and this too can be taken into account, and so on until the process provides no further information to any of the players.

I took the “significant pause” to mean that no-one can immediately declare victory, and each player can take this fact into account.

This Python program starts by consider all possible

P(6, 4) = 360ways the cards can be dealt, and then reduces the possibilities as each piece of new information is taken into account.It runs in 84ms.

Run:[ @repl.it ]Solution:Alf has 3. Bert has 5. The cards remaining on the table are 2 and 6.Starting from 360 possibilities the possible hands are reduced as follows:

The two remaining possibilities are (A=3, B=5, C=1, D=4) and (A=3, B=5, C=4, D=1). We don’t know from the information given which order C and D are holding 1 and 4, but each player can see at least one of the cards, so they would know the exact hand dealt, and presumably all four of them would declare victory simultaneously.

Here is my manual solution:

A says “6”, so can see 1+5 or 2+4 (3 and 6 cannot participate in any pair that sum to 6)

If A is holding one of these cards (1, 2, 4, 5) they must be seeing the sum that doesn’t involve their own card. B, C, D can see which card A is holding, so would know which of the sums A is seeing, and so the people who can only see one half of that sum would immediately be able to declare their number (as the other half of that sum). This doesn’t happen.

Hence A must be holding 3 or 6. And the other card (6 or 3) must still be on the table, as if one of B, C, D were holding it the other two could deduce their cards immediately.

So B, C, D are holding three cards from 1, 2, 4, 5.

B says “5”, so can see 1+4 or 2+3.

If A is holding 6, then 3 is still on the table, so B can see 1+4 from C and D. C and/or D can immediately declare their cards as soon as B says “5”. This doesn’t happen.

So A can deduce he is holding 3 (and can declare victory, after the pause where neither C nor D declare victory).

B must be able to see 1+4 (from C and D) or 2+3 (from A’s 3 and one of C and D holds the 2).

If B is holding 1 or 4, then he must be able to see 2+3, so whichever of C and D cannot see the 2 must be holding it and can declare immediately. This doesn’t happen.

Similarly if B is holding 2, then he must be able to see 1+4 from C and D, so C and D can declare their cards immediately. This doesn’t happen.

So B can deduce he is holding 5 (and can declare, after the pause where neither C nor D declare victory).

This leaves C and D holding two cards from 1, 2, 4, and the other is still on the table (along with 6).

If C and D are holding 1 and 2, then when B says “5” A would immediately know his card was 3 (which goes with the 2 to make 5). This doesn’t happen.

If C and D are holding 2 and 4, then when B says “5” A would immediately know his card was 3 (which goes with the 2 to make 5). This doesn’t happen.

So C and D are holding 1 and 4, and can declare (after the pause where A does not declare victory). 2 and 6 remain on the table.

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## John Crabtree 3:00 pm

on21 June 2019 Permalink |Let a declared value be D.

If the non-declarers have cards >= D or = D/2, two players can immediately determine their cards.

If two pairs of cards add to D, two players can immediately determine their cards.

After A’s call, only A can have and as 1 + 5 = 2 + 4 = 6 must have, 3 or 6

After B’s call, A must have 3 and, as 1 + 4 = 2 + 3 = 5, B must have 5.

After B’s call, A knows immediately that he has 3 unless he can see C + D = 5 = 1 + 4.

The cards on the table are 2 and 6.

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