# Free throw interview question

I recently had an interview question that posed the following... Suppose you are shooting free throws and each shot has a 60% chance of going in (there is no "learning" effect and "depreciation" effect, all have the some probability no matter how many shots you take).

Now there are three scenarios where you can win $1000 1. Make at least 2 out of 3 2. Make at least 4 out of 6 3. Make at least 20 out of 30 My initial thought is that each are equally appealing as they all require the same percentage of free throw shots. However when using a binomial calculator (which this process seems to be) the P (X > x) seems to be the highest for scenario 1. Is this due to the number of combinations? - – Fillet Jul 17 at 13:14 ## 5 Answers It's an intuitive result especially if you consider the extreme case of having to score 100% to win; you're more likely to score 3/3 than 30/30. - This answer doesn't address that if your chance of making the shot is more than 67%, then you would rather the 20/30 case. This answer seems to suggest that you always take the first choice. – Cruncher Jul 17 at 12:51 I agree and believe Mathmo123's answer is more worthy of acceptance. – lemon Jul 17 at 15:02 The result is linked to the Law of large numbers, which basically states that the more trials you do of something, the closer you will get to the expected probability. So after 10,000 trials, I would expect to be proportionally closer to 6000 than being close to 60 after 100 trials. The point here is that in all these cases the proportion is$\frac23$- i.e. greater than the expected probability of 60%. Since for larger numbers we will be closer to the expected probability of 60%, that means that we are less likely to be above$\frac23$. If you were to change the parameters slightly - and ask what is the probability of success in more than 55% of cases, for example, then you'd see the complete opposite happening. - Rather, the required proportion in each case is exactly 2/3. If option 3 required, say 19/30, and option 1 required 3/3, the computation would be more difficult, even though all of those are greater than the expected probability of 60%. – DanTilkin Jul 17 at 15:32 typo corrected. – Mathmo123 Jul 17 at 15:42 I like this answer - simple and accurate. Also has some relation to the Central Limit Theorem in statistics. The more trials in a sample, the closer you get to a normal distribution - i.e. closer to actual probability. – Jim Beam Jul 17 at 17:23 I'd be careful about the use of the word "closer" in your first paragraph - you expect your percentage to be closer to$60$% in the$10,000$trial case, but you'd expect your actual number of successes to be closer to$60$in the$100$trial case than to$6000$in the$10,000$trial case. Standard deviation grows like$\sqrt{n}$. – MartianInvader Jul 17 at 21:09 @MartianInvader - I've changed the post to stress that point by specifying that by "close" I mean "proportionally close". Do you think that's sufficient? Can you suggest a better wording that doesn't overcomplicate too much – Mathmo123 Jul 17 at 21:48$P(\#1) = \binom{3}{2}\cdot{(\frac{60}{100})}^2\cdot{(\frac{40}{100})}^1+\binom{3}{3}\cdot{(\frac{60}{100})}^3\cdot{(\frac{40}{100})}^0 = 0.648P(\#2) = \binom{6}{4}\cdot{(\frac{60}{100})}^4\cdot{(\frac{40}{100})}^2+\binom{6}{5}\cdot{(\frac{60}{100})}^5\cdot{(\frac{40}{100})}^1+\binom{6}{6}\cdot{(\frac{60}{100})}^6\cdot{(\frac{40}{100})}^0 = 0.54432P(\#3) = \sum\limits_{n=20}^{30}\binom{30}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{30-n} = $feel free to do the math yourself... In general, you need to prove that the following sequence is monotonically decreasing:$A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$I think it should be pretty easy to do so by induction... - What's the downvote for??? – barak manos Jul 17 at 16:53 The question didn't ask for help with the calculations, but for intuition as to why this occurs – Mathmo123 Jul 17 at 16:54 @Mathmo123 Where did the OP ask for intuition? – Quinn Culver Jul 18 at 13:47 "Is this due to the number of combinations?" - i.e. why does this result occur. Maybe intuition was the wrong word. – Mathmo123 Jul 18 at 13:49 By the way, "monotonously decreasing" means "decreasing in a boring way"; you mean "monotonically decreasing". (This is a common mistake among non-native speakers.) – David Richerby Jul 18 at 16:31 Because your expected number of successes is less than the number required in order to win, you need to get a bit lucky in order to succeed. That's much harder for large samples: if you had to get 2/3 of a thousand throws, you would be in very bad shape; whereas even getting 1/1 out of a single throw is a 60% chance. So the intuitive explanation is: you lose in the long term, so you need variance to win. The smaller the sample size, the bigger the impact variance has. - Precisely speaking, each free throw is a Bernoulli trial with probability of success$p = 0.6$. Hence: • the expected value of a single throw is$p$and the variance is$p(1-p) = 0.24$. • the sample proportion of successes out of$n$trials has expected value$p$, but the variance is$p(1-p)/n$This means that although the expected proportion of successes stays constant, the variance decreases as a function of$n$. This intuitively tells us that the probability of observing a sample proportion at least as large as$\frac23$is a decreasing function of the sample size$n$: as the variance of the sampling distribution gets smaller, more probability mass is concentrating around the mean. Since the mean is less than$\frac23$, it becomes less likely to observe outcomes in excess of$\frac23\$.

Conversely, it becomes more likely to observe outcomes in excess of a value that is smaller than the mean.

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