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View Full Version : The Spurs had a 96% chance to win this game



Uriel
06-19-2013, 01:46 AM
Kawhi Leonard and Manu Ginobili each missed a free throw down the stretch, either of which would have iced the game. Kawhi shot 82.5% from the FT line this season, while Manu shot 79.6%. The probability that both players would miss a free throw is only 4.0425%, meaning that the Spurs' chances of winning this game was 95.9575%.

:depressed

Darius Bieber
06-19-2013, 01:47 AM
Sports doesn't take statistics to win games, it takes heart.

EIC
06-19-2013, 01:49 AM
Interesting, but you don't account for the two rebounds the Spurs missed. I would like to see some statistical odds that they fail to make either defensive board, and miss both those free throws. The odds have to be less than 1%.

Budkin
06-19-2013, 01:50 AM
As if it didn't hurt enough already b

dbreiden83080
06-19-2013, 01:51 AM
Games are not won with a calculator..

dg7md
06-19-2013, 01:51 AM
Impossible odds come up for us 90 percent of the time.

This is what it is like being a Spurs fan.

cjw
06-19-2013, 02:11 AM
I thought I'd never feel worse than losing my six figure suicide pool (under 5% of entries left) when Detroit had like a 98% chance of beating Andrew Luck and the Colts this past year.

Can't attach a price to tonight though. If you give Lebron two shots at it, he'll come through in the clutch eventually. And friggen Ray Allen. Can he please retire?

NRHector
06-19-2013, 02:12 AM
Pop fucked up 100 %

Mal
06-19-2013, 02:17 AM
I blame those missed rebounds after James bricks....
And Pop surely overcoached this one.

BatManu20
06-19-2013, 02:18 AM
God hates the Spurs. It's the only logical explanation for all of this.

mudyez
06-19-2013, 02:56 AM
Kawhi Leonard and Manu Ginobili each missed a free throw down the stretch, either of which would have iced the game. Kawhi shot 82.5% from the FT line this season, while Manu shot 79.6%. The probability that both players would miss a free throw is only 4.0425%, meaning that the Spurs' chances of winning this game was 95.9575%.

:depressed

U have to account for the made freethrows as well (I think one by KL and 3 by Manu). e.g. Manu's 3/4 is pretty much his 79.6%.

Uriel
06-19-2013, 02:59 AM
U have to account for the made freethrows as well (I think one by KL and 3 by Manu). e.g. Manu's 3/4 is pretty much his 79.6%.
Gambler's Fallacy. Each of Manu's individual free throws is a separate event, whose probability of going in is independent of other events that preceded it.

Capt Bringdown
06-19-2013, 03:02 AM
Feel bad for Leonard, but ...damn

mudyez
06-19-2013, 03:13 AM
Gambler's Fallacy. Each of Manu's individual free throws is a separate event, whose probability of going in is independent of other events that preceded it.

So you just take out 2 missed FT's (out of 6) and assume that they go in, without accounting for the other?

If so: You can also say Bron missed 15 shots and usally makes 49%...We got lucky he didn't make 7 more adn we lose by 14 in regulation.

CitizenDwayne
06-19-2013, 03:14 AM
Kick a man when he's down, why dontcha?

eric365
06-19-2013, 03:18 AM
God hates the Spurs. It's the only logical explanation for all of this.

God hates the bobcats not the spurs

Uriel
06-19-2013, 03:20 AM
So you just take out 2 missed FT's (out of 6) and assume that they go in, without accounting for the other?

I recommend you do some reading on Logic so you can understand why this statement is fallacious.

http://www.nizkor.org/features/fallacies/gamblers-fallacy.html

mudyez
06-19-2013, 03:34 AM
I recommend you do some reading on Logic so you can understand why this statement is fallacious.

http://www.nizkor.org/features/fallacies/gamblers-fallacy.html

I know what your trying to tell me without reading this (it's not like I didn't study this ^^). But can you please explain whats the difference between your statement and my Bron counter statement?

mudyez
06-19-2013, 03:43 AM
I'm just saying:

There is a 50,4% chance Manu makes 3FT's.
And there is a 41,6% chance Manu makes 3 and KL makes one.

That said: We had a pretty big chance (58,4) do not make one of the made FT's.

Taking the 6 FT's the usal outcome probalby would be one more make (since Manu is at about his %tage and KL is below), but the 96% are very falseleading.

InRareForm
06-19-2013, 04:04 AM
@WagerMinds 5h
The Heat, down by 5 with just :28 remaining, are now +5000 to win the game outright. $100 bet wins you $5,000. Any takers?

travis2
06-19-2013, 06:06 AM
Kawhi Leonard and Manu Ginobili each missed a free throw down the stretch, either of which would have iced the game. Kawhi shot 82.5% from the FT line this season, while Manu shot 79.6%. The probability that both players would miss a free throw is only 4.0425%, meaning that the Spurs' chances of winning this game was 95.9575%.

:depressed

Actually, the probability of each missing one given those individual probabilities is 9.38%. Calculated as follows:

Ptot = [2C1*p1*(1-p1)][2c1*p2*(1-p2)] = [(2)(0.825)(0.175)][(2)(0.796)(0.204)]

2C1 is the combination of 2 things taken 1 at a time, pn is the probability of a successful free throw for person n. Product of 2 binomial distributions.

Still a small chance of losing, but a 90% chance is normally not betting odds, statistically speaking.

therealtruth
06-19-2013, 07:09 AM
Manu should have at least focused on hitting his free throws. He played a pretty bad game. The Spurs didn't take advantage enough of the Heat sticking to Danny Green. They should have allowed Manu to create more stuff knowing the help wouldn't come. It's on Pop to make the adjustments now. He just chocked the game away. Just one more rebound. Like Riley says no rebounds no rings.

Kobulingam
06-19-2013, 07:41 AM
Clearly the OP does not have a PHD in math

ffadicted
06-19-2013, 07:43 AM
Not to mention tbh that somehow a loose rebound bounces around and lands in the hand of the best 3 point shooter of all time lol

This fuggin game

Kobulingam
06-19-2013, 08:05 AM
Actually, the probability of each missing one given those individual probabilities is 9.38%. Calculated as follows:

Ptot = [2C1*p1*(1-p1)][2c1*p2*(1-p2)] = [(2)(0.825)(0.175)][(2)(0.796)(0.204)]

2C1 is the combination of 2 things taken 1 at a time, pn is the probability of a successful free throw for person n. Product of 2 binomial distributions.

Still a small chance of losing, but a 90% chance is normally not betting odds, statistically speaking.

Each miss one, in this case, means each miss at least one. Basically he should have summed all these up (assume q=1-p):

1) Prob of each player missing exactly 1 FT (your calculation above): 2*2*p1q1p2q2
2) Prob of only one player missing 2FT = q1q1p2p2 + p1p1q2q2
3) Prob of one player missing 2 FT and the other missing 1FT: 2*q1q1p2q2 + 2*p1q1q2q2
4) Prob of both players missing all their free throws : q1q1q2q2

CORRECTION: 1, 3, 4 summed up give probability for each player misses at least one. Adding 2 gives the real desired probability of "at least 2 free throws missed"

travis2
06-19-2013, 08:14 AM
Each miss one, in this case, means each miss at least one. Basically he should have summed all these up (assume q=1-p):

1) Prob of each player missing exactly 1 FT (your calculation above): 2*2*p1q1p2q2
2) Prob of only one player missing 2FT = q1q1p2p2 + p1p1q2q2
3) Prob of one player missing 2 FT and the other missing 1FT: 2*q1q1p2q2 + 2*p1q1q2q2
4) Prob of both players missing all their free throws : q1q1q2q2

True. Of course, if we are looking at this set of possibilities, it's easier to calculate the probability of nobody missing anything and subtracting from 1:

P[at least 1 miss] = 1 - P[no miss] = 1 - [2C2*p1*p1][2C2*p2*p2] = 1 - [(1)(0.825)(0.825)][(1)(0.796)(0.796)] = 1 - 0.431 = 0.569

So yeah, the probability was pretty good someone was going to miss something.

Kobulingam
06-19-2013, 08:31 AM
True. Of course, if we are looking at this set of possibilities, it's easier to calculate the probability of nobody missing anything and subtracting from 1:

P[at least 1 miss] = 1 - P[no miss] = 1 - [2C2*p1*p1][2C2*p2*p2] = 1 - [(1)(0.825)(0.825)][(1)(0.796)(0.796)] = 1 - 0.431 = 0.569

So yeah, the probability was pretty good someone was going to miss something.

What we need here is P[at least 2 misses] = 1 - P[no miss] - P[player A misses 1] - P[player B misses 1]

jstep13
06-19-2013, 08:35 AM
nerd alert!

travis2
06-19-2013, 08:39 AM
What we need here is P[at least 2 misses] = 1 - P[no miss] - P[player A misses 1] - P[player B misses 1]

Rewrite this as P[at least 2 misses] = 1 - P[no miss] - P[player A misses 1, player B makes both] - P[player A makes both, player B misses 1]

1 - 0.431 - 0.183 - 0.221 = 0.165

Definitely not betting odds...