For anyone whose eyes glazed over and couldn’t see how this was derived:
There are 3 possible outcomes:
you miss both turns The probability of missing both turns for a p is p*p (2 turns, the same p each time), and the reward is 1. Expected utility is probability*reward, so 2*p*1. Which is just 2*p or p^2
you make the second turn. The probability of making the second turn is the probability of missing the first turn and making the second. Since p is for a binary choice, there’s only one other probability, q, of missing the turn; by definition all probabilities add to 1, so p+q=1, or q=1-p. So, we have our q*p (for the missed and taken turn), and our reward of 4. As above, the expected utility is q*p*4, and substituting for q gives us (1-p)*p*4, or rearranging, 4*(1-p)*p.
or, you make the first turn The probability of making the first turn is just p-1 as before, and the reward is 0. So the expected utility is (p-1)*0 or just 0.
Our 3 possibilities are exhaustive, so we just add them together:
p^2 + 0 + 4*(1-p)*p
0 drops out, leaving us with the final result given in the article:
I feel kind of silly now; what was I thinking in writing ‘2*p or just 2*p’?
Right, right. I had difficulty remembering whether p was the chance of making a turn or missing a turn. Good thing the multiplication by 0 makes the difference moot.
For anyone whose eyes glazed over and couldn’t see how this was derived:
There are 3 possible outcomes:
you miss both turns
The probability of missing both turns for a p is p*p (2 turns, the same p each time), and the reward is 1. Expected utility is probability*reward, so 2*p*1. Which is just 2*p or p^2
you make the second turn.
The probability of making the second turn is the probability of missing the first turn and making the second. Since p is for a binary choice, there’s only one other probability, q, of missing the turn; by definition all probabilities add to 1, so p+q=1, or q=1-p. So, we have our q*p (for the missed and taken turn), and our reward of 4. As above, the expected utility is q*p*4, and substituting for q gives us (1-p)*p*4, or rearranging, 4*(1-p)*p.
or, you make the first turn
The probability of making the first turn is just p-1 as before, and the reward is 0. So the expected utility is (p-1)*0 or just 0.
Our 3 possibilities are exhaustive, so we just add them together:
0 drops out, leaving us with the final result given in the article:
In (1), instead of 2*p, you want p*p. In (2), you want 1 – p instead of p. The final results are correct, however.
I feel kind of silly now; what was I thinking in writing ‘2*p or just 2*p’?
Right, right. I had difficulty remembering whether p was the chance of making a turn or missing a turn. Good thing the multiplication by 0 makes the difference moot.
Ah well, it happens.