Friday Puzzler -- Game to 15

Re: Friday Puzzler -- Game to 15

Larry Barrett

Your chance of winning is greatest if you go first.

There are 8 combinations of 3 of the 1-9 numbers that add to 15:

1 5 9, 1 6 8

2 4 9, 2 5 8, 2 6 7

3 4 8, 3 5 7

4 5 6

Of course, the three numbers can be drawn in any order, along with any other numbers. But note that the winning combinations contain at least one odd number.

(This is rather obvious since the sum must be an odd number, 15).

At the start, there are 5 odd numbers (1,3,5,7,9) and 4 even numbers (2,4,6,8) and your odds of drawing an odd number are 5/9 if you draw first.

Re: Friday Puzzler -- Game to 15

Alex Y

Interesting, Larry. I believe you are right if the numbers are drawn at random, In fact, I created a crude spreadsheet, and found that #1 won about 60% or the time, #2 won about 20% of the time, and no-one won about 20% of the time. I'm not sure I believe those numbers, and haven't done any qc on my spreadsheet to verify.

However, the intent of the puzzle was that each player gets to choose which of the remaining numbers to take at each turn. Sorry I didn't state that clearly.

Does that change your answer?

Clarification

Alex Y

Both players see the numbers remaining as well as those they and their opponent have taken, and at each turn the player chooses which of the remaining numbers to take.

Re: Friday Puzzler -- Game to 15

Larry Barrett

Your clarification makes a much more interesting game.

I think the eight combinations I listed are still the only winning combinations. And I think the winner must have either 3 odd numbers or one odd and two even numbers (and possibly more in either case). So it seems to me that the first player has the best chance of putting together either of these cases. So far, in the test cases I have looked at, it seems that if both players play rationally the game will end in a draw.

On the right track

Part 2

Alex Y

Larry has made several observations that are correct:

1) There are exactly 8 combinations of three numbers 1-9 that add up to 15, which Larry listed.

2) Any combination that adds to 15 has either 1 or 3 odd numbers.

3) It seems like going first would be advantageous -- and it is if the players are not skilled at this game.

4) Test cases ended up with a draw if the players both followed their best strategy.

#4 turns out to be true in general, not just the test cases Larry tried. The remaining challenge (part 2) is to give a demonstration that is closer to a proof than test cases.

Big Hint

Alex Y

Arrange the number cards that the players are to take like this:

8 1 6

3 5 7

4 9 2

Re: Big Hint

Larry Barrett

Do you want to be x or o?

Doesn't matter :-)

Re: Doesn't matter :-)

Larry Barrett

What is the insight in this problem that would lead one to arrange the nine digits in this pattern (or to select digits in a similar way), and thus see that it doesn't matter who goes first?

Re: Doesn't matter :-)

Alex Y

Good question. I guess it is that in a 3x3 "magic square", the rows, columns, and major diagonals all add to 15. In general, the rows, columns and diagonals of an nxn magic square add to

n*(n^2+1)/2.

Why one would have that bit of info available for recall is beyond me. :-)