First to Fifteen

[bonanova]

Nine white poker chips have the numbers 1, 2, 3, ..., 8, 9 written on them, one number on each chip. You and a friend alternately select a chip, with the aim of being the first to draw three numbers that sum to 15. You may play first or second. Do you have a winning strategy?

Edited January 29, 2013 by bonanova
Add condition the three chips total 15 to win

[CaptainEd]

I'm assuming that the chips are taken out of circulation once selected.

I choose to go first and

5

, my opponent chooses N (not equal to 5), and I choose 10-N

[bonanova]

I'm assuming that the chips are taken out of circulation once selected.

I choose to go first and

5

, my opponent chooses N (not equal to 5), and I choose 10-N

Excellent.

And yes the chips are taken out of circulation once selected.

But ... I neglected to mention one other condition.

The person wins whose total is 15 after picking up three chips.

I changed the OP to include that condition.

[CaptainEd]

I see, I was solving a different problem anyway, thinking that we were driving the combined total to 15.

Let's see if I've got it right now.

Each person is trying to achieve his/her own total of 15. And each person has to achieve his/her sum using three chips.

[Molly Mae]

Choose 8.

The second player must take 7.

Choose 6

The second player must take 1.

Choose 5

The second player cannot stop your next move of taking either 4 or 2.

I see now, however, that this isn't the problem we're trying to solve.

Edited January 29, 2013 by Molly Mae

[CaptainEd]

I can't guarantee a win, but I can prevent the other player from winning.

O (Other) draws N, I draw anything, O draws M, I draw 15-M-N, if available. If not, O can't draw it either.

[bonanova]

Choose 8.The second player must take 7.Choose 6The second player must take 1.Choose 5 The second player cannot stop your next move of taking either 4 or 2. I see now, however, that this isn't the problem we're trying to solve.

Right. The problem properly sated is to be the first to draw three numbers that total 15.

[bonanova]

I can't guarantee a win, but I can prevent the other player from winning.

O (Other) draws N, I draw anything, O draws M, I draw 15-M-N, if available. If not, O can't draw it either.

Ah, but the game can go beyond that point, as you note in your spoiler title.

[CaptainEd]

Choose 8.The second player must take 7.Choose 6The second player must take 1.Choose 5 The second player cannot stop your next move of taking either 4 or 2. I see now, however, that this isn't the problem we're trying to solve.

Right. The problem properly sated is to be the first to draw three numbers that total 15.

Sorry to be dense. Is this true: I win if I am the first to have drawn M chips (M>=3), exactly 3 of which total 15?

[bushindo]

Nine white poker chips have the numbers 1, 2, 3, ..., 8, 9 written on them, one number on each chip. You and a friend alternately select a chip, with the aim of being the first to draw three numbers that sum to 15. You may play first or second. Do you have a winning strategy?

Here's a winning strategy

Pick to be the first player and choose 5 as the first number. Let the second player's first choice be S.

If S is 1, 9, 2, 6, or 7, then player 1's 2nd, 3rd, and 4th moves should be among the following set (3,4,8)

If S is 3, 4, or 8, then player 1's 2nd, 3rd, and 4th moves should be among the following set (2, 6, 7). The key is that after playing S, player 2 would be constrained to block player 1's move from getting 15.

Example. Let say player 1 choose 5, and player 2 chooses 2. The game would go as follows

Player 1- 1st choice: 5

Player 2- 1st choice: 2

Player 1- 2nd choice: 3

Player 2- 2nd choice: 7 (have to choose 7 to prevent a player 1 win)

Player 1- 3rd choice: 4

Player 2- 3rd choice: 6 (have to choose 6 to prevent a player 1 win)

Player 1- 4th choice: 8 (WIN- 3 + 4 + 8 = 15 )

[Prime]

Nine white poker chips have the numbers 1, 2, 3, ..., 8, 9 written on them, one number on each chip. You and a friend alternately select a chip, with the aim of being the first to draw three numbers that sum to 15. You may play first or second. Do you have a winning strategy?

Here's a winning strategy

Pick to be the first player and choose 5 as the first number. Let the second player's first choice be S.

If S is 1, 9, 2, 6, or 7, then player 1's 2nd, 3rd, and 4th moves should be among the following set (3,4,8)

If S is 3, 4, or 8, then player 1's 2nd, 3rd, and 4th moves should be among the following set (2, 6, 7). The key is that after playing S, player 2 would be constrained to block player 1's move from getting 15.

Example. Let say player 1 choose 5, and player 2 chooses 2. The game would go as follows

Player 1- 1st choice: 5

Player 2- 1st choice: 2

Player 1- 2nd choice: 3

Player 2- 2nd choice: 7 (have to choose 7 to prevent a player 1 win)

Player 1- 3rd choice: 4

Player 2- 3rd choice: 6 (have to choose 6 to prevent a player 1 win)

Player 1- 4th choice: 8 (WIN- 3 + 4 + 8 = 15 )

In this example, doesn't 2nd player win on his 3rd move having (2,7,6)?

Edited January 29, 2013 by Prime

[bushindo]

Nine white poker chips have the numbers 1, 2, 3, ..., 8, 9 written on them, one number on each chip. You and a friend alternately select a chip, with the aim of being the first to draw three numbers that sum to 15. You may play first or second. Do you have a winning strategy?

Here's a winning strategy

Pick to be the first player and choose 5 as the first number. Let the second player's first choice be S.

If S is 1, 9, 2, 6, or 7, then player 1's 2nd, 3rd, and 4th moves should be among the following set (3,4,8)

If S is 3, 4, or 8, then player 1's 2nd, 3rd, and 4th moves should be among the following set (2, 6, 7). The key is that after playing S, player 2 would be constrained to block player 1's move from getting 15.

Example. Let say player 1 choose 5, and player 2 chooses 2. The game would go as follows

Player 1- 1st choice: 5

Player 2- 1st choice: 2

Player 1- 2nd choice: 3

Player 2- 2nd choice: 7 (have to choose 7 to prevent a player 1 win)

Player 1- 3rd choice: 4

Player 2- 3rd choice: 6 (have to choose 6 to prevent a player 1 win)

Player 1- 4th choice: 8 (WIN- 3 + 4 + 8 = 15 )

In this example, doesn't 2nd player win on his 3rd move having (2,7,6)?

Shucks, back to the drawing board.

[CaptainEd]

Yes, now you guys have made it pretty clear

If Player A plays 5 and B plays 6, I think B has time to respond to A's forces, or has time to force A to respond, resulting in a draw.

There are 8 3-chip combinations that sum to 15: 168, 159, 267, 258, 249, 357, 348, 456.

Of these, the most useful chip is 5, as it has 4 combinations, while the other chips only pertain to 2 or 3.

So A chooses 5. This knocks out those 4 combinations from B's possibilities.

If B chooses 6, this knocks out 3 combinations from A's possibilities.

If A chooses 4, this knocks out all but two of B's possibilities. Even so, as far as I can tell, B can prevent A from creating a fork.

I can't give an exhaustive list, or a principle/algorithm that is easy to follow.

[Prime]

All possible game variants may be traversed by a computer program. I suspect the result would be a draw. A smarter computer program would utilize a "forced move" concept. A simpler program could be the old mini-max algorithm.

[bonanova]

Yes, now you guys have made it pretty clear

If Player A plays 5 and B plays 6, I think B has time to respond to A's forces, or has time to force A to respond, resulting in a draw.

There are 8 3-chip combinations that sum to 15: 168, 159, 267, 258, 249, 357, 348, 456.

Of these, the most useful chip is 5, as it has 4 combinations, while the other chips only pertain to 2 or 3.

So A chooses 5. This knocks out those 4 combinations from B's possibilities.

If B chooses 6, this knocks out 3 combinations from A's possibilities.

If A chooses 4, this knocks out all but two of B's possibilities. Even so, as far as I can tell, B can prevent A from creating a fork.

I can't give an exhaustive list, or a principle/algorithm that is easy to follow.

This is correct.

Can you draw a picture?

[CaptainEd]

I'm not happy with my pictures, they do not capture my tictactoe feeling

The only 3-chip combinations that add to 15:

The A and B below indicate that, at this stage of the game, both A and B have this possibility available to them. As a player chooses one of the chips, the other player loses that possibility.

1 1 2 2 2 3 3 4

6 5 6 5 4 5 4 5

8 9 7 8 9 7 8 6

A A A A A A A A

B B B B B B B B

Note that there are four combinations with 5, three combinations with each of 2,4,6,8, and two combinations with each of 1,3,7,9

(1) 5, A chooses 5, because it has the most combinations

1 1 2 2 2 3 3 4

6 5 6 5 4 5 4 5

8 9 7 8 9 7 8 6

A A A A A A A A

B B B B

(2) B chooses 6, (it knocks out 3 of A's combinations--I believe 4 would have been equivalent)

1 1 2 2 2 3 3 4

6 5 6 5 4 5 4 5

8 9 7 8 9 7 8 6

A A A A A

B B B B

(3) A chooses 4, because it knocks out half of B's possibilities, while advancing two of A's own possibilities

1 1 2 2 2 3 3 4

6 5 6 5 4 5 4 5

8 9 7 8 9 7 8 6

A A A A A

B B

At this point, A has 54 and B has 6.

B had better not choose 7, appealing though it be, as A is forced to choose 2, which creates a fork, with A threatening both 258 and 249. I believe the choice of 2 has the similar failure.

B's better choice is 8.

(4) B chooses 8, threatening 168

1 1 2 2 2 3 3 4

6 5 6 5 4 5 4 5

8 9 7 8 9 7 8 6

A A A

B B

(5) A has no immediate win, so is forced to choose 1, threatening 159

1 1 2 2 2 3 3 4

6 5 6 5 4 5 4 5

8 9 7 8 9 7 8 6

A A A

B

A has 145, B has 68

(6) B is forced to choose 9

1 1 2 2 2 3 3 4

6 5 6 5 4 5 4 5

8 9 7 8 9 7 8 6

A

B

A has 145, B has 689

(7) A chooses 7, threatening 357 and blocking B's last combination (267)

(8) B chooses 3, blocking A's last combination (357)

-----

After the first three moves (A5, B6, A4), I think it is clear that both players can force a draw (or can mess up and lose)

One could question whether one of the three first moves could be replaced by more powerful ones. But each move seems to have a good rationale...

Edited January 30, 2013 by bonanova

[bonanova]

Great. You win.

But what I really meant was can you draw a magic 3x3 picture?

[CaptainEd]

Great. You win.

But what I really meant was can you draw a magic 3x3 picture?

Oh, woops! Uh, no...especially on a day when I can't even code a spoiler

[bonanova]

8 1 6

3 5 7

4 9 2

Does that help?

[CaptainEd]

Rather easy when you put it that way...