[Guest]

There are these 2 kids -- let's say A and B, and there is a doughnut. Since both of them wants the doughnut, and doesn't wish to share, they agreed they will play a game, and let the winner have the doughnut.

The game is this number game where both of them shall count together from "1" to "20" with each one saying their numbers alternatively. Whoever counts "20" WINS the doughnut. Here are the rules:

1. Kid A, since he is younger, will start the game.

2. Each person can say either 1 number or 2 adjacent numbers together as they wish (For E.g. A can count as "1" or "1 and 2").

3. Once started, each person should say the immediate next number (or next two, if they wish). They can't skip any number in between (For E.g. if A started as "1", B can say either "2" or "2 and 3").

4. The person who says "20" wins the doughnut.

So, the game started, and Kid A counted "1 and 2".

The question is -- Who gets the doughnut, and what strategy should each follow to win it?

Good luck..

Peace!

[Guest]

If played correctly, A should get the donut.

Basically, you want to go first and take the "1 and 2". The key is to leave your turn on the numbers 2, 5, 8, 11, 14, and 17 as the game progresses. Based on your opponent's decision, you will advance 1 number if he (or she) chooses to advance 2 numbers; you advance 2 numbers if the opponent chooses to advance 1 number.

In this manner, you pin your opponent, the B kid, to respond after you've chosen 17. You then collect your donut and move on to find coffee to with your prize.

[Guest]

Kid A will allways get the doughnut:

He said "1 and 2" - a perfect start. Now all he has to do is follow the numbers 5, 8, 11, 14 and 17 (every third number, which should not be a problem...) - he has to "count" these numbers. At the end, Kid B has a choice to take 18 or 18 and 19, but in both cases kid A "counts 20" (third (again) number from 17).

Example:

A: 1, 2

B: 3

A: 4,5

B: 6,7

A: 8

B: 9

A: 10, 11

B: 12, 13

A: 14

B: 15

A: 16, 17

B: 18

A: 19, 20

[Guest]

kid A wins, whoever says 17 wins, because then the other person has to say 18 or 18,19, in which case the response would be 19,20 or 20. So using that logic, the winning path of numbers requires kid A to have his last number he says on his turn to be 5, 8, 11, 14, 17, 20. If he slips up, then B can win taking the same path.

[Guest]

B could get the doughnut if he mimicks A and B says 17

[Guest]

The key to winning is always stopping 1 number short of those numbers divisible by 3. Or, in other words, forcing your opponent to say all the numbers divisible by 3. To ensure this, you must start by saying "1,2" as boy A did. After that, the key is to do exactly opposite of what your opponent does. If he advances 2, you advance 1, if he advances 1, you advance 2. That will make the sequence increase by 3 every time, stopping short of the next integer that is divisible by 3. This strategy forces your opponent to say "18" or "18,19" which allows you to say either "19,20" or just "20" (again, stopping short of 21, which is divisible by 3)

This works because the only thing that is controllable by anyone playing the game (other than the one who starts) is to make the sequence advance by the same amount each time, which is 3.

If you do not start the game, there is no way of ensuring a win. If however, your opponent does not understand the strategy, and allows you to say a number short of those divisible by 3, you can from that point forward lock in the win.

[Guest]

Good going, guys.. Would appreciate if some of you use "spoiler" to tag the answer..

Thanks!