gth816f

03-20-2010, 08:08 PM

Quick answer: 0.118 % chance of shooting 24/25 or 25/25, assuming I did this right.

Long answer:

Okay, this came up in the other thread: what was the probability of us actually shooting so well from the free throw line last night? Well, at first I was annoyed because I couldn't remember how to do it. But then I remembered(I think), and once I remembered I just had to do it.

Obviously when it got to the end, making/missing some of these free throws would have affected if we got the next ones, so we'll just operate based on the assumption that we were going to get that exact number of free throws from those specific players.

Someone check my math, because it's probably wrong...but here goes:

Lawal: 4 shots, .525 FT%

Favors: 2 shots, .629 FT%

Shumpert: 8 shots, .720 FT%

Rice Jr.: 1 shot, .529 FT%

Miller: 4 shots, .804 FT%

Oliver: 2 shots, .704 FT%

Peacock: 4 shots, .791 FT%

Basically, we should be able to figure out the probability of hitting 24/25 overall by figuring out the probability of missing the first then hitting 24 straight, missing the second and hitting the rest of the 25, etc., adding those 25 probabilities together, then adding in the probability of hitting all 25. We'll work with threesignificant digits and use conventional rounding.

So we might as well make the first step the easiest one. The chances of us hitting all 25 is simply the probabilities of each guy hitting all of his shots multiplied together. That is: .525^4 * .629^2 * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 9.31 * 10^-5.

Now it gets a litle dicier. The basic math is still the same but there's a separate step for each shooter. I'll write out the first step to show what I'm doing. What we want to find out is the chance of Lawal hitting three out of four free throws. The probability of him missing the first then hitting three straight is .475 * .525^3.

Since multiplication is commutative, the chance of him making three out of four in each other order(miss second, miss third, miss fourth) is the same number. That means the overall probability of him hitting three out of four free throws is .475 * .523^3 * 4. Now we multiply that by the chance of the remaining players hitting every one of their free throws: (.475 * .523^3 * 4) * .629^2 * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 3.33 * 10^-4

We have to do this for each player:

Lawal: (.475 * .523^3 * 4) * .629^2 * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 3.33 * 10^-4

Favors: .525^4 * (.629 * .371^1 * 2) * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 1.09 * 10^-4

Shumpert: .525^4 * .629^2 * (.280 * .720^7 * 8) * .529 * .804^4 * .704^2 * .791^4 = 2.90 * 10^-4

Rice Jr. = .525^4 * .629^2 * .720^8 * (.471 * .529^0 * 1) * .804^4 * .704^2 * .791^4 = 8.29 * 10^-5

Miller: .525^4 * .629^2 * .720^8 * .529 * (.196 * .804^3 * 4) * .704^2 * .791^4 = 9.08 * 10^-5

Oliver: .525^4 * .629^2 * .720^8 * .529 * .804^4 * (.296 * .704^1 * 2) * .791^4 = 7.83 * 10^-5

Peacock: .525^4 * .629^2 * .720^8 * .529 * .804^4 * .704^2 * (.208 * .791^3 * 4) = 9.79 * 10^-5

Whew. So now we have the probabilities of every possible combination of makes and misses that would get us to 24/25 or better. Now we just have to add them up!

Probability of us hitting 24/25: 9.31 * 10^-5 + 3.33 * 10^-4 + 1.09 * 10^-4 + 2.90 * 10^-4 + 8.29 * 10^-5 + 9.08 * 10^-5 + 7.83 * 10^-5 + 9.79 * 10^-5 = .00118, or 0.118%.

Done! Now you people who actually know math can rip it apart and tell me I wasted the last hour of my life.

Long answer:

Okay, this came up in the other thread: what was the probability of us actually shooting so well from the free throw line last night? Well, at first I was annoyed because I couldn't remember how to do it. But then I remembered(I think), and once I remembered I just had to do it.

Obviously when it got to the end, making/missing some of these free throws would have affected if we got the next ones, so we'll just operate based on the assumption that we were going to get that exact number of free throws from those specific players.

Someone check my math, because it's probably wrong...but here goes:

Lawal: 4 shots, .525 FT%

Favors: 2 shots, .629 FT%

Shumpert: 8 shots, .720 FT%

Rice Jr.: 1 shot, .529 FT%

Miller: 4 shots, .804 FT%

Oliver: 2 shots, .704 FT%

Peacock: 4 shots, .791 FT%

Basically, we should be able to figure out the probability of hitting 24/25 overall by figuring out the probability of missing the first then hitting 24 straight, missing the second and hitting the rest of the 25, etc., adding those 25 probabilities together, then adding in the probability of hitting all 25. We'll work with threesignificant digits and use conventional rounding.

So we might as well make the first step the easiest one. The chances of us hitting all 25 is simply the probabilities of each guy hitting all of his shots multiplied together. That is: .525^4 * .629^2 * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 9.31 * 10^-5.

Now it gets a litle dicier. The basic math is still the same but there's a separate step for each shooter. I'll write out the first step to show what I'm doing. What we want to find out is the chance of Lawal hitting three out of four free throws. The probability of him missing the first then hitting three straight is .475 * .525^3.

Since multiplication is commutative, the chance of him making three out of four in each other order(miss second, miss third, miss fourth) is the same number. That means the overall probability of him hitting three out of four free throws is .475 * .523^3 * 4. Now we multiply that by the chance of the remaining players hitting every one of their free throws: (.475 * .523^3 * 4) * .629^2 * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 3.33 * 10^-4

We have to do this for each player:

Lawal: (.475 * .523^3 * 4) * .629^2 * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 3.33 * 10^-4

Favors: .525^4 * (.629 * .371^1 * 2) * .720^8 * .529 * .804^4 * .704^2 * .791^4 = 1.09 * 10^-4

Shumpert: .525^4 * .629^2 * (.280 * .720^7 * 8) * .529 * .804^4 * .704^2 * .791^4 = 2.90 * 10^-4

Rice Jr. = .525^4 * .629^2 * .720^8 * (.471 * .529^0 * 1) * .804^4 * .704^2 * .791^4 = 8.29 * 10^-5

Miller: .525^4 * .629^2 * .720^8 * .529 * (.196 * .804^3 * 4) * .704^2 * .791^4 = 9.08 * 10^-5

Oliver: .525^4 * .629^2 * .720^8 * .529 * .804^4 * (.296 * .704^1 * 2) * .791^4 = 7.83 * 10^-5

Peacock: .525^4 * .629^2 * .720^8 * .529 * .804^4 * .704^2 * (.208 * .791^3 * 4) = 9.79 * 10^-5

Whew. So now we have the probabilities of every possible combination of makes and misses that would get us to 24/25 or better. Now we just have to add them up!

Probability of us hitting 24/25: 9.31 * 10^-5 + 3.33 * 10^-4 + 1.09 * 10^-4 + 2.90 * 10^-4 + 8.29 * 10^-5 + 9.08 * 10^-5 + 7.83 * 10^-5 + 9.79 * 10^-5 = .00118, or 0.118%.

Done! Now you people who actually know math can rip it apart and tell me I wasted the last hour of my life.