You’re playing a game of cornhole with your friends, and it’s your turn to toss the four bean bags. For every bean bag you toss onto your opponents’ board, you get 1 point. For every bag that goes through the hole on their board, you get 3 points. And for any bags that don’t land on the board or through the hole, you get 0 points.
Your opponents had a terrible round, missing the board with all their throws. Meanwhile, your team currently has 18 points — just 3 points away from victory at 21. You’re also playing with a special house rule: To win, you must score exactly 21 points, without going over.
Based on your history, you know there are three kinds of throws you can make:
For each bean bag, you can choose any of these three throws. Your goal is to maximize your chances of scoring exactly 3 points with your four tosses. What is the probability that your team will finish the round with exactly 21 points and declare victory?
There are 8 ways to get 3 more points in 4 throws:
1-1-1-0 1-1-0-1 1-0-1-1 0-1-1-1
3-0-0-0 0-3-0-0 0-0-3-0 0-0-0-3
The aggressive throw has the best chance for getting 3 points, the conservative throw has the best chance for getting 1 point, and the wasted throw has the best chance for getting 0 points.
The only time you can strategically go over 3 points if you have 1 or 2 points, and your bean bag goes in the hole.
I compared two different strategies for getting 3 points.
1-1-1-0 is $(0.8)(0.8)(0.8)(1.0) = 0.512$
1-1-0-1 is $(0.8)(0.8)(0.1)(0.8) = 0.0512$
1-0-1-1 is $(0.8)(0.1)(0.8)(0.8) = 0.0512$
0-1-1-1 is $(0.1)(0.8)(0.8)(0.8) = 0.0512$
$0.512 + 3(0.0512) = 0.6656$
3-0-0-0 is $(0.4)(1.0)(1.0)(1.0) = 0.4$
1-1-1-0 is $(0.3)(0.8)(0.8)(1.0) = 0.192$
1-1-0-1 is $(0.3)(0.8)(0.1)(0.8) = 0.0192$
1-0-1-1 is $(0.3)(0.1)(0.8)(0.8) = 0.0192$
0-3-0-0 is $(0.3)(0.4)(1.0)(1.0) = 0.12$
0-1-1-1 is $(0.3)(0.8)(0.8)(0.8) = 0.1536$
0-0-3-0 is $((0.3)(0.3)(0.4)(1.0) = 0.036$
0-0-0-3 is $(0.3)(0.3)(0.3)(0.4) = 0.0108$