then the expected payoff is p^2^+4(1-p)p

For anyone whose eyes glazed over and couldn't see how this was derived:

There are 3 possible outcomes:

1. 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
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.
3. 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:

p^2 + 4*(1-p)*p