Life is a Test: A Series of Pony Logic Puzzles

by Brony_of_Brody

The Answer 37

0-0: Favour Hoof, which wins 52% of the time, compared to Double Horn (0%), Horn (24%) and Wings (24%).
0-1: Favour Hoof, which wins 55% of the time, compared to Double Horn (0%), Horn (21%) and Wings (25%).
1-0: Favour Hoof, which wins 40% of the time, compared to Double Horn (33%) and Wings (27%).
1-1: Randomise equally between Hoof, Wings and Double Horn.

When you're playing a regular game of Hoof, Horn, Wings, you should randomise equally between the three with equal probability, because any other strategy will be exploited by a savvy opponent: if, for example, you favour Wings, the opponent will know to play Horn and you will lose in the long run.

A similar train of thought applies here. First off though, you should probably have noticed that Double Horn basically is Horn except better, for all intents and purposes when you're in the lead. After one win, it's never optimal to use just Horn again: there's no reason not to use what is effectively Horn with more win-conditions when you have one win in the bag, and you won't benefit from single Horn's insta-win condition at this point. So when you're tying 1-1, you cycle between Hoof, Double Horn, and Wings with equal probability.

What if you're winning 1-0? Smart opponents will definitely know that it's now a contest of Hoof, Double Horn and Wings, because Horn is a strictly dominated strategy for you. They will then pick from the four options, as usual.

If you play Double Horn against Horn or Wings, you win. If you play Wings against Horn, you lose the match thanks to the insta-win con. Let's call the probability of winning the entire match X. X is 1 if you win this round, and 0 if you play Wings against Horn. And of course, if you're 1-0 up and the opponent instead wins with anything that isn't Horn, it's 1-1 and the odds of winning are 50%, assuming you're both playing optimally.

There are some Xs unaccounted for however, and that's odds of winning the match if you tie: Wings against Wings, for example.

Let's say I play Hoof with some probability H, Double Horn with probability Dh, and Wings with probability W. No matter what the opponent throws out, playing according to this strategy should deliver you your optimal 1-0 win strategy of X. So, for example, if your opponent goes with Hoof, we get an equation like this:

(H)(X)+(Dh)(0.5)+(W)(1)=X

If they throw Hoof: assuming you put out Hoof too, you win with probability X; in those times you throw Double Horn, you win with probability 0.5; and Wings gives you a sure win with probability 1. The same holds true for the other three things the opponent can throw at you, which gives us a system of equations:

(H)(1)+(Dh)(X)+(W)(0.5)=X
(H)(1)+(Dh)(1)+(W)(0)=X
(H)(0.5)+(Dh)(1)+(W)(X)=X

A similar equation can therefore be derived from this, where you're playing Hoof with probability H₂, Double Horn with probability Dh₂, Horn with probability h₂, and Wings with probability W₂. So if you play Hoof, for example, I know that:

(H₂)(X)+(Dh₂)(1)+(h₂)(1)+(W₂)(0.5)=X

The solution to this big system of equations gives X≈0.73. It also gives that you should play Hoof, Double Horn and Wings with probabilities 0.40, 0.33 and 0.27, respectively. The opponent should play Hoof, Double Horn, Horn and Wings with probabilities 0.55, 0, 0.21 and 0.25, respectively.

Plays for when the score is 0-0 can be calculated similarly. i'll spare you the maths, but both of you should play Hoof 52% of the time, compared to Double Horn (0%), Horn (24%) and Wings (24%).

Game's stacked in favour of Hoof, huh?