Wednesday, May 02, 2012

We Get Letters: OKC Edition

SdM writes:
Dallas lost to OKC last night by 1 pt and I think it's bc they can't do math. Replay: 9 secs left: Nowitski gets a free throw and gives Dallas the lead. 1 secs left: Durant makes a 15 ft shot and gives OKC a 1pt lead. 0 secs left: Dallas is unable to get up the floor and shoot. Assume everyone on the floor makes 75% of their free throws and is 50% from the floor. If there's an OT, winner is determined by a coin toss. What do you do if you're Dallas at the 9 second mark?

SdM's answer: Dallas should have fouled, even though they had the lead, with 7 or 8 seconds left. That's obviously stupid, so I told SdM that was stupid. He pathetically tried to defend himself.

Four cases: 1. OKC misses both free throws. Dallas up 1 and has ball with ~8 secs left. 2. OKC makes 1 and misses 1. Unless OKC gets an offensive rebound off a free throw, Dallas has ~ 8 seconds to (2a) win the game on a last shot. If (2b) they miss, they go to OT and have a 50/50 chance. 3. OKC makes both, takes the lead by 1, but Dallas gets the last shot with ~8 secs left. Assuming teams make 50% of their field goals and 75% of free throws, ...

So, take those probabilities as right.

Here is what I came up with:
prob of case #1: .0625 (if two throws are independent)
prob of case #2: .375 (two ways to make one and miss one
prob of case #3: .5625 (again, independent)

case #4: what they actually did. If 50% from the floor is right, Dallas had a 50% chance of winning, with what they did, NOT fouling. No chance of overtime; they win, or lose, depending on whether Durant hits the shot with 1 second left.

If they fouled, Dallas has the following chances (assuming overtime is a coin flip):
A. Case #1, Dallas wins (assume OKC fouls, or not, but there's tool little time left for anything) p(DalWin)=.0625
B. Case #2, Dallas makes basket from floor p(DalWin)=.375*.5= .1875 Case #2, Dallas misses basket from floor, wins in overtime, p(DalWin)=0.09375 Case #2, Dallas misses basket from floor, loses in overtime
C. Case #3, Dallas .5 chance of taking last shot and winning, p(DalWin)=.5 If I have this right, and assuming OKC does not get off reb'd on it's foul shot, and assuming OKC does not foul Dallas, even in case #1, it looks to me like Dallas's chances of winning, if they foul, are .8375, whereas if they don't foul their chances of winning are 50-50.

Is this right? That's not even close, Dallas should foul. Maybe SdM is not so pathetic after all.

And what matters is just that Dallas' chances of winning if they don't foul are only 50-50, and if they do foul their chances are 50-50 to win, EVEN IF OKC makes both free throws, the worst case scenario. But is it really .8375 if you foul, compared to .50 if you don't? I must have missed something.


poisonboy said...

I could be wrong, but it looks like you used the conditional probability in case 3, while in case 1 and 2 you multiplied it out to get the unconditional so that you could add them up.

P(Win|Case 3) = 0.5, but P(Case 3)*P(Win|Case 3) = 0.5 * 0.5625 = 0.28125. That is the number you should use as P(DalWin) for case 3.

P(Win|Foul) = 0.4375

Mungowitz said...

that looks right, thanks for catching the mistake!

but that still leaves the prob difference pretty big in favor of the "foul" strategy.

JPom said...

I'm studying for finals and am doing enough math, but how much does it change if you lower the chances of hitting a free throw to 40% or 35%? Pressure is real, 9 seconds is not that much time, and you're starting in a set offense where the defense can be set. Much harder than transition, especially with nerves playing a factor. What's stopping OKC from fouling with 8 seconds? And Dallas with 7 seconds? Etc. Where does it not make sense? I think you have to lower the chances of making a shot from 50% as time decreases to paint a more accurate picture.
Even still, your bro might have a point.

John Thacker said...

You have .0625*1 + .375*(.75) + .5625*(.5) = .625. Except, of course, that if Dallas is right to foul up 1 with 8 seconds left, then OKC should do so up 1 with 7 seconds left. So

P(DW) = .0625*1 + .375*(.75) + .5625*(1-P(DW))
1.5625*P(DW) = .90625
P(DW) = .58

However, this also ignores that OKC should foul on their end.

Note that about 20% of free throws missed are rebounded in the NCAA. The NBA might be similar.

If you put in the 20% chance to rebound a missed FT (only opportunity on the second one):

Case 1: There's a 80% chance that Dallas rebounds. 20% of the time we'll have the same situation as before.

Case 2: 20% of the time OKC will offensive rebound and be the team with the 75% chance of winning.

P(DW) = .0625*(.8 + .2*P(DW)) + .375*(.8*.75 + .2*.25) + .5625*(1-P(DW))

P(DW) = .005 + .0125*P(DW) + .225 + .01875 + .5625 - .5625*P(DW)

1.55P(DW) = .81125
P(DW) = .5233

Closer, and ignoring that OKC may choose to foul on a tied game.

John Thacker said...

You really can't ignore that 20% of missed free throws are offensive rebounded.

Eric said...

I think it fails at the 50% assumption. Redo the math with shooting percentages in the last 10seconds, and player specific. The real LeBron might be a 50% shooter, but at crunch time he was below 10% at one time.

The rebound % doesn't ad value either as often the shooting team doesn't even line the key, choosing to keep players in the back court ready for defense on end of the game free throws.