1. Jul 14, 2008

### golfchic2006

A coin game requires:
1. Ten coins in one pile
2. That each player takes one, two, or four coins from the pile at alternate turns.
3. That the player who takes the last coin loses.

I. When Austin and Brooks play, Austin goes first and Brooks goes second,
II. Each player always makes a move that aloows him to win when possible; if there is no way for him to win, then he always makes a move that allows a tie when possible.

Must one the two men win? If so, which one?

2. Jul 14, 2008

### ssjSolidSnake

Two men can win?

Brooks takes 4

Austin takes 2

Brooks takes 1

Austin takes 1

Brooks takes 2

3. Jul 14, 2008

### golfchic2006

There can be a tie. But look at number 3. The player who takes the last coin loses. Would that mean that Brooks looses according to your answer?

4. Jul 14, 2008

### CRGreathouse

The first player always wins in this case. The correct starting move is to take 4 coins.

5. Jul 14, 2008

### robert Ihnot

The first player has the option to win by taking an even number. If A takes 2, then B gets nowhere by taking an even number, as gone into below, so B will take only 1. This leaves now 7 and A then takes 1, leaving B with 6, and B must take 2, leaving 4. In this case, A again takes 1.

6. Jul 15, 2008

### CRGreathouse

If the first player takes 2 the second player can also take 2, which forces a win for the second player. The only correct starting move is to take 4.

7. Jul 15, 2008

### robert Ihnot

CRGreathouse: If the first player takes 2 the second player can also take 2, which forces a win for the second player. The only correct starting move is to take 4.

Well, if A takes 2, B takes 2, then A can take 2, leaving 4. So A wins!

Last edited: Jul 15, 2008
8. Jul 15, 2008

### CRGreathouse

No, B responds by taking 3, forcing A to take the last and thus lose.

9. Jul 15, 2008

### robert Ihnot

CRGreathouse: No, B responds by taking 3, forcing A to take the last and thus lose.

PLEASE! You are not allowed to take 3.

10. Jul 15, 2008

### uart

Hi CRGreathouse. I think you misread the question, 3 is not an option.

"2. That each player takes one, two, or four coins from the pile at alternate turns."

Edit : Robert just beat me to it. :)

11. Jul 18, 2008

### CRGreathouse

Ah, sorry, missed that. I did solve the problem where {1, 2, 3, 4} rather than {1, 2, 4} were legal.

I still don't follow. Can't B then force a loss for A by taking 2?

A: 10 -> 8
B: 8 -> 7
A: 7 -> 6
B: 6 -> 4
A: 4 -> 3
-----
B: 3 -> 1
A: 1 -> 0

It looks like leaving a player with a number of coins equal to 1 mod 3 at each turn forces a win.

12. Jul 18, 2008

### tiny-tim

You can make this rigorous:

If player A leaves 1 (mod 3), what can player B then leave (mod 3)?

Can player A always get back to 1 (mod 3), next move?

If so, then he can always leave 1 (mod 3), until eventually he leaves exactly 1.

13. Jul 20, 2008