Sunday, November 22, 2009

Going for two

Sunday I was listening to the Giants-Falcons game on the radio when the Falcons, trailing by 14, scored a touchdown with about six minutes left in the game. Playing for overtime the Falcons kicked the extra point. They scored another touchdown but then lost in overtime. There was a case for playing to win in regulation by going for the two point conversion after the first touchdown. Suppose we assume the Falcons will score another touchdown and the Giants will not score again in regulation. Suppose the chance of converting a two point conversion try is p and the chance of making an extra point try is q and that if the game goes into overtime each team has a 50% chance to win. Then what is the relationship between p and q that determines whether it is better for the Falcons to go for the two point conversion or try to kick the extra point after their first touchdown?


  1. They should go for 2 points if p > (p-q)^2.

    Their probability of winning given that they go for 2 is p*q + p(1-q)(0.5) + (1-p)(p)(1-0.5). Their probability of winning given that they go for 1 is q*q*0.5 + (1-q)(p)(0.5). If the 1st expression is greater than the 2nd, then p > (p-q)^2.

    (Note that in getting the 2nd expression I've assumed that they go for 1 after their 2nd TD if they successfully get an extra point after their 1st TD. This means p < 2q. If p > 2q then p > (p-q)^2 so this assumption does not affect the result.)

    As Bob Costas said last night during the Bears game, the probability of winning in OT is a lot closer to 50% than most football fans might prefer given that it is so heavily dependent on the coin flip: the team that gets the ball first is at a strong advantage due to the high probability of getting a field goal on their 1st drive.

  2. The paranthetical should say "this means p < 0.5q" instead of "2q." Also, I should note that in the trivial case where p = 1 and q = 0 you should go for it.

  3. It should be noted that, if p > q and the Falcons are aware of this unlikely situation, then it would be foolish for them ever to go for 1, and their probability of winning is p*(p + (1-p)/2) + (1-p)*p/2 = p. This means that when p = 1 and q = 0 they unequivocally "should go for it", because then they would win with probability 1, not (as indicated by the formula applicable for p < q) only 1/2.

  4. p >= q : go for 2, w == pr{win} = p
    q/2 < p < q : go for 1, w = 0.5p(1+q)
    p <= q : w =
    if go for 1, 0.5p(2+q-p)
    if go for 2, 0.5(q^2-qp+p)
    so er go for the greater of these:-)

    Sorry, no time to finish the 3rd case or reconcile w/ other solns. Coupla comments:

    1) the simplifying assumptions are appreciated but kinda unrealistic: if one could be sure of scoring again while holding opponents scoreless in regulation, why not in OT?
    2) as you might expect, this kind of thing has been analyzed to death in the real world, e.g. search for
    football commentary "Two-Point Conversion Chart"

  5. 1) oops, my third case should have been prefaced "p <= q/2", not "p <= q"
    2) thanks for the fun puzzle

  6. The simplifying assumptions were intended to isolate the cases where the decision to go for one point or two points matters. Obviously if Atlanta doesn't score again it won't matter and if New York scores again it probably won't matter.

    It appears in the NFL currently q is about .99 and p is about .45. This article has an interesting chart showing how PAT percentage has increased over the last 35 years.

  7. Yup, the assumptions made the puzzle a puzzle. And thanks for the info about p and q.

  8. The criteria in the first comment is correct as far as it goes. However if we assume q > p (else it is clear you should go for the two points) then we may continue sqrt(p) > q-p or p+sqrt(p) > q which seems a bit more elegant. When q = 1 we have p > .381966 which suggests going for it as the two point conversion rate appears to be about .45. Of course this all depends on the chances in overtime being even.