Problem of the Week

In the game of "Odds and Evens" two players each put out either one or two fingers. One player wins if the total number of fingers is odd, the other player wins if the total is even. This game is fair since both players have an equal chance of winning.
Now suppose three people are playing. They each hold out one or two fingers. One player wins if the total is four, another if the total is five, and the third wins if the total is either three or six.
Is the game fair? If not, who has the best chance of winning?

Benedict's Solution:

Assuming players put out fingers randomly -

1 + 1 + 1 = 3
1 + 1 + 2 = 4
1 + 2 + 1 = 4
1 + 2 + 2 = 5
2 + 1 + 1 = 4
2 + 1 + 2 = 5
2 + 2 + 1 = 5
2 + 2 + 2 = 6

Player one (4) 3/8 ...Player two (5) 3/8...Player 3 (3 or 6) 2/8 UNFAIR!

In addition, players one and two can fix the game by holding out one or two finges, respectively, increasing their own odds to 1/2. Since player three's plight relies entirely on other players - they need to agree with him - so he can not help his misfortune.

Can you beat the odds?
In the game below, you control the player with the red sleeves.
You win a point each time the total number of fingers is either 3 or 6.
Click the buttons to your left or right to put out either one or two fingers. (click again to prepare for the next round)
The game keeps a running tab of all points won.
To clear all points and start again, click on the red button in the bottom left corner.