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# Free-throw probability

Our friend and Cleveland Cavalier, LeBron James, asks Sal how to determine the probability of making 10 free throws in a row. Hint: the answer is surprising! Created by Sal Khan and LeBron James.

## Want to join the conversation?

• Wouldn't making the first free throw increase his FT%? Say his initial FT% was 75% and he made the first shot, increasing his FT% to 76%, you would multiply 75% by 76%, rather than doing 75%^2.
• We are assuming that 75% is his true free-throw percentage (the true probability of him making a free throw) and doesn't change from shot to shot. That number cannot be directly known, so we estimate it by looking at his history of making free throws. This is a reasonable thing to do because LeBron has taken thousands of FTs. His true ft% won't change from shot to shot. Our estimate of it might in the future, but for the sake of a question like this one, we assume that our current estimate is the true FT% for the entire experiment.

You are touching on a very important point in statistics. For the most part, we can only estimate a true parameter by looking at samples.
• I"m confused as to why we are taking 75% of 75% if the free throws are independent events.
• I also had a hard time grasping why he was taking 75% of 75% and so on. For some reason none of the other answers to this question in the comments clicked for me. Then I realized this example is exactly the same as figuring out the probability of flipping a coin 3 times and getting HEADS all three times. The math is the same, but Sal is presenting a different way of looking at the problem.

Ok, let's look at if we were calculating the probability of flipping a coin 3 times and getting HEADS all 3 times (flipping HEADS 3 times IN A ROW).

P(HHH) is the same as P(H) * P(H) * P(H)

When flipping a coin, the probability of flipping HEADS is 1/2. So then...

P(HHH) = P(H) * P(H) * P(H) = 1/2 * 1/2 * 1/2 which is the same as (1/2) ^ 3

If we go back to the free throw example, if F represents making a free throw, our problem would look like this:

P(FFFFFFFFFF) = P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F)

The probability of Lebron making a free throw is 75% or .75, so then...

P(FFFFFFFFFF) = P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) = .75 ^ 10

I hope this helps others with the same question. Forgive me if you didn't need the breakdown as I have done above, but for me, that was how I was able to make sense of the problem. If my explanation does not help, I suggest re-watching the Coin Flipping Probability video as that was helpful for me.
• How much greater would it be if it were 90%?
• 0.3486784401 rounded to the nearest hundredth=0.35x100=35%
• This solution assumes that each free throw is an independent event. Once Lebron makes a shot I would argue that his probability of sinking the second one is higher than 75%. He knows the feel of the ball, what the arc should look like etc. I think the question that Sal answers is what is the probability of Lebron sinking his first free throw shot in 10 consecutive games which would be a more independent set of events.
• if you do 10 free throws and your hands get tired doesn't that mess up the calculations
• This calculation is based on the assumption that you will have equal chances of success in each throw. If later throws have lower chances of success (because of tired hands or otherwise), the calculation will be different, but similar method can still be applied.
In short, yes, the calculation will not correctly represent the chances of making 10 free throws in a row.
• What is the significance of the parentheses at ?
• Putting the base number within parentheses is a common convention with exponents especially when we have a unit or a fraction. This is more important with other units because 10cm^2 is different from (10cm)^2. The first one is ten centimeters squared; the second is one hundred centimeters squared.
• at , how could you multiply 75% *75%?
• 75% can be treated as 0.75. So you can simply multiply 0.75 by 0.75 to get the answer.
• How does this work if as you are shooting you are atcually increasing your accuracy?
• The percentage calculated by Sal is based on Lebron's average percent of making a free throw. He is not taking into factor that Lebron is getting better with each shot, just using plain math. Now, to find out the percentage given that Lebron is getting better with each shot, you would have to find out how much better he is getting with each shot, take into factor age, etc... Honestly, I think if you try to figure that out, it's making the original problem more complicated than it should be.
• Why is the probability of missing 10 in a row, if the chance of making it is 75%
(1 - 75%)^10
and not
1 - (75%)^10 ?
or is it the same thing?
• (1-0.75)^10 = (0.25)^10 correct
Both those expressions are not the same
1-(0.75)^10 = 0.94 ...
(1-0.75)^10 = 0.00000095 ...
totally makes a difference
• how come a 60% free throw percentage has a greater chance of making 10 in a row than 75%?
• It doesn't; the calculations he performed in the video were:
P(10 free throws in a row @ 75%) ≈ 6%
P(5 free throws in a row @ 60%) ≈ 8%