19 replies to this topic

[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 by bonanova, 29 January 2013 - 05:10 PM.
Add condition the three chips total 15 to win

[CaptainEd]

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

Spoiler for my choice and strategy

I choose to go first and

Spoiler for my selection is

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.

Spoiler for my choice and strategy

I choose to go first and

Spoiler for my selection is

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]

Spoiler for I (don't) have a winning strategy

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 by Molly Mae, 29 January 2013 - 04:54 PM.

[CaptainEd]

Spoiler for now it feels like tic tac toe

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]

Spoiler for I (don't) have a winning strategy

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]

Spoiler for now it feels like tic tac toe

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]

Spoiler for I (don't) have a winning strategy

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

Spoiler for

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 )