[QUOTE=IrishTim;34252]
And CB, if you're ever bored on a rainy day, feel free to "bore" us with the math. :)[/QUOTE]
Ok, remember you asked for it ... I'm sure with some thought, it is something that you should have no problem deriving ...
Anyone, please feel free to check the logic and math. I did this hurriedly on the back of an envelope.
Goal: maximize Team A's win probability where Team A is up by 15 pending the XP.
Assumptions: Let P2xA be the two pt successful conversion rate for Team A. Let P2xB be the 2 pt rate for Team B (opponent). Assume the 1 pt successful conversion rate for both teams is 98%.
Prob Matrix for Team A
[B]Result after XP .... 1ptXP ..... 2ptXP[/B]
Up 15 ... 2% ....1-P2xA
Up 16 ... 98% ...0%
Up 17 ... 0% ....P2xA
Now Team B will [B]try to tie [/B]Team A in the following manners (not very likely or optimal at all to try to win outright), better to tie and go to overtime.:
Prob Matrix after Team A XP for Team B
[B]Result after XP .... options[/B]
Down 15 ... 1TD + Onside kick(Succ) + 2ptXP + 2ndTD + 1ptXP to tie
Down 16 ... 1TD + Onside kick(Succ) + 2ptXP + 2ndTD + 2ptXP to tie
Down 17 ... 1TD + Onside kick(Succ) + 1ptXP + 2ndTD + 1ptXP + 2nd Onside(Succ) + FG to tie
Now let's make the assumption that the common probabiity term for "1TD + Onside kick(Succ) + 2ndTD" (defined as PBc) is indeed almost the same for all 3 cases. Probably a very good assumption since this is central to being able to tie the game with little time left. Any failure will almost certainly result in Team B losing the game outright.
Let P15B be the probability of Team B, down 15, tying, P16B and P17B similar definitions. Similarly, let P1A be the win % for Team A trying the 1ptXP and P2A be the win % for trying a 2ptXP. Further let's assume that once the game is tied, the win % for each team is 50%.
So:
P15B = PBc * P2xB * 98%
P16B = PBc * P2xB * P2xB
P17B = PBc * 98% * 98% * PB2nd_onside * PBFG
Now let's put this stuff together:
P1A = 2% *(1 - 50%*P15B) + 98% *(1-50%*P16B) and
P2A = (1-P2xA) *(1 - 50%*P15B) + P2xA *(1-50%*P17B)
Simpifying, we get:
P1A = 1 - 0.01*P15B - 0.49*P16B
but P16B is only a factor of P2xB smaller (roughly 44% say) than P15B so the dominant term in the equation above is 0.49%*P16B
also
P2A = 1 - (1-P2xA)*50%*P15B - P2xA*50%*P17B
Now P17B is extremely small compared to P15B because a 2nd onside kick might be a 30% prob and a third score of a FG is maybe a 1% prob since almost all time would be gone. (This is common sense, a 3 score cushion is virtually insurmountable)
Thus P2A approx = 1 - 0.5*(1-P2xA)*P15B and
P1A approx = 1 - 0.49*P16B
so to get the best win % we want to identify the smaller (case 2 vs case 1) of
0.5*(1-P2xA)*P15B vs 0.49*P16B
Divide each by P15B to get:
0.5*(1-P2xA) vs 0.49*P16B/P15B or
0.5*(1-P2xA) vs 0.49*(P2xB/0.98)
Assume P2xA = P2xB = 44% and we have
0.5*0.56 vs 0.49*(0.44/0.98) or
0.28 vs 0.22
Ergo, to a first approximation, Case 1 (kicking the XP) gives a smaller Team B win% and greater Team A win%.