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Number picking III. 15 but not a race

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You play a game with a friend.

Nine poker chips numbered 1, 2, 3, ..., 7, 8, 9 lie on the table.

Players take turns drawing a chip.

The first player to own three chips that total 15 wins.

Do you choose to play first or second?

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Posted · Report post

first. i'd take the 5.

then its very difficult for player 2 to counter anything i do.

for example, if player 2 pick 3, i pick 1 then player 2 has to pick 9 to prevent me from winning.

player 2 already has the 3, so not much he can do to get 15.

player 2 picking 8 or 7 is the "best" he can do and player 1 can easily counter by picking the other one.

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Posted · Report post

As first player you pick 5.

As second player if I pick an even number I can prevent you from winning.

e.g.:

You Me

5 2 (or 4 or 6 or 8)

6 4* Or you play 4. Then I play 6* and you must play 7*, etc. If you play 8 I play 4 or 6.

9* 1*

7 3*

8* --

* = forced.

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Posted · Report post

maybe player 2 does have the advantage.

if player 1 picks 5, player 2 picks 8, player 1 picks 7.

player 2 picks 3 player 1 has to pick 4 but can't pick it as it puts him over.

if player 1 starts 7 player 2 starts 8,

maybe player 1 2, forcing player 2 to take 6, then take 1, and now player 2 has the upper hand?

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Posted · Report post

I don't see a winning strategy. All the ways to get a sum of 15 from 3 digits can be represented by a magic square with the even numbers in the corners that is symmetric to the following example:

492

357

816

Making it effectively a game of tic tac toe, in which, if both players play ideally, there is no winner. However, assuming non-ideal play, the first player has 5 marks to the second player's 4 so he has more ways to win in case the second player makes a mistake, so I would play first...

...and get my opponent really really drunk ;P.

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