# probability $2/4$ vs $3/6$

Recently I was asked the following in an interview:

If you are a pretty good basketball player, and were betting on whether you could make $2$ out of $4$ or $3$ out of $6$ baskets, which would you take?

I said anyone since ratio is same. Any insights?

• Are you betting you will make exactly that number, or that you wll make at least that number? It changes the answer. – Ross Millikan Feb 16 '14 at 15:31
• Betting for "exactly" is silly, as you can always miss on purpose. – Bach Feb 16 '14 at 15:34
• @HansZauber Unless there is the possibility to fail at missing :) – Hagen von Eitzen Feb 16 '14 at 15:43
• This is a great, unintuitive question. – Thomas Ahle Feb 16 '14 at 17:33
• @Thomas Ahle is the question unintuitive or the answer? – miracle173 Feb 16 '14 at 22:18

# Depends on how good you are The explanation is intuitive:

• If you are not very good (probability that you make a single shot - p < 0.6), then your overall probability is not very high, but it is better to bet that you'll make 2 out of 4, because you may do it just by chance and your clumsiness has less chance to prove in 4 than in 6 attempts.

• If you are really good (p > 0.6), then it is better to bet on 3 out of 6, because if you miss just by chance, you have better chance to correct yourself in 6 attempts.

The curves meet exactly at p = 0.6.

## In general, the more attempts, the more of real skill reveals

This is best illustrated on the extreme case: With more attempts, it is almost binary case - you either succeed or not, based on your skill. With high N, the result will be close to your original expectation.

Note that with high N and p = 0.5, the binomial distribution gets narrower and converges to normal distribution.

## Everything here just revolves around binomial distribution,

which tells you that the probability that you will score exactly k shots out of n is

$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$

The probability that you will score at least k = n/2 shots (and win the bet) is then

$$P(X \ge k) = \sum^{n}_{i=k} \binom{n}{i} p^i (1-p)^{n-i}$$

# Why the curves don't meet at p = 0.5?

Look at the following plots: These plots are for p = 0.5. The binomial distribution is symmetric for this value. Intuitivelly, you expect 2 of 4 or 3 of 6 to take half of the distribution. But if you look especially at the left plot, it is clear that the middle column (2 successful shots) goes far beyond the half of the distribution (dashed line), which is denoted by the red arrow. In the right plot (3/6), this proportion is much smaller.

If you sum the gold bars, you will get:

P(make at least   2 out of    4) = 0.6875
P(make at least   3 out of    6) = 0.65625
P(make at least 500 out of 1000) = 0.5126125


From these figures, as well as from the plots, is apparent that with high N, the proportion of the distribution "beyond the half" converges to zero, and the total probability converges to 0.5.

So, for the curves to meet for low Ns, p must be higher to compensate for this: P(make at least   2 out of    4) = 0.8208
P(make at least   3 out of    6) = 0.8208


Full code in R:

f6 <- function(p) {
dbinom(3, 6, p) +
dbinom(4, 6, p) +
dbinom(5, 6, p) +
dbinom(6, 6, p)
}

f4 <- function(p) {
dbinom(2, 4, p) +
dbinom(3, 4, p) +
dbinom(4, 4, p)
}

fN <- function(p, from, max) {
#sum(sapply(from:max, function (x) dbinom(x, max, p)))
s <- 0
for (i in from:max) {
s <- s + dbinom(i, max, p)
}
s
}
f1000 <- function (p) fN(p, 500, 1000)

plot(f6, xlim = c(0,1), col = "red", lwd = 2, ylab = "", main = "Probability that you will make ...", xlab = "p (probability you make a single shot)")
curve(f4, col = "green", add = TRUE, lwd = 2)
curve(f1000, add = TRUE, lwd = 2, col = "blue")
legend("topleft", c("2 out of 4", "3 out of 6", "500 out of 1000"), lwd = 2, col = c("green", "red", "blue"), bty = "n")

plotHist <- function (n, p) {
plot(x=c(-0.5,n+0.5),y=c(0,0.41),type="n", xaxt="n", xlab = "successful shots", ylab = "probability",
main = paste0(n/2, "/", n, ", p = ", p))
axis(1, at=0:n, labels=0:n)
x <- 0:n
y <- dbinom(0:n, n, p)
w <- 0.9
#lines(0:4, dbinom(0:4, 4, 0.5), lwd = 50, type = "h", lend = "butt")
rect(x-0.5*w, 0, x+0.5*w, y, col = "lightgrey")
uind <- (n/2+1):(n+1)
rect(x[uind]-0.5*w, 0, x[uind]+0.5*w, y[uind], col = "gold")
}

par(mfrow = c(1, 2))
plotHist(4, 0.5)
abline(v = 2, lty = 2)
arrows(2-0.5*0.9, 0.17, 2, 0.17, col = "red", code = 3, length = 0.1, lwd = 2)
plotHist(6, 0.5)

f4(0.5)
f6(0.5)
f1000(0.5)

par(mfrow = c(1, 2))
plotHist(4, 0.6)
plotHist(6, 0.6)

f4(0.6)
f6(0.6)

• Why is 0.6 the tipping point instead of 0.5? I would naturally assume 0.5 would be the point at which the curves meet. Why is this not the case? – 1110101001 Feb 17 '14 at 1:33
• @user2612743 this is a great question! Please see the updated post. – Tomas Feb 17 '14 at 9:37
• As a sidenote, in real life, $p$ might be a function of $N$, as after 1000 trials, the player is likely to be fatigued and perform worse than when they started. – Jam Feb 21 '18 at 11:14

The probability of you getting at least half increases with the number of shots. E.g. with a probability of 2/3 per shot the probability of getting at least half the baskets increases as below.

Edit it is important to point out that this only holds if by a "pretty good basketball player" you mean your chance of making a basket is somewhat better than evens (in the range 0.6 to 1 exclusive). This is shown very clearly in Hagen von Eitzen's answer. An intuitive way of looking at this is that it's like a diversification effect. With only a few baskets, you could get unlucky, just as you might if you tried to pick only a couple of stocks for an investment portfolio, even if you were a good stock picker. You increase the number of baskets -- or stocks -- and the role of chance is reduced and your skill shines through.

Formally, assuming that

• each throw is independent, and

• you have the same probability $p$ of scoring on each throw

you can model the chance of scoring $b$ baskets out of $n$ using the binomial distribution

$$\mathbb{P}(b \text{ from } n) = \binom{n}{b} p^{b}(1-p)^{n-b}$$

To get the probability of scoring at least half of the $n$ baskets, you have to add up these probilities. E.g. for at least 2 out of 4 you want $\mathbb{P}(2 \text{ from } 4) + \mathbb{P}(3 \text{ from } 4) + \mathbb{P}(4 \text{ from } 4)$.

• I guess you're making the assumption that a pretty good player has a better than half chance of making a shot. Reasonable, but just in case it was false the same reasoning would encourage you to go for 2/4. – Ben Millwood Feb 16 '14 at 15:41
• @BenMillwood that's a very interesting point indeed! However the question does state the assumption that the player is good. – TooTone Feb 16 '14 at 15:42
• In contrast a "pretty good" goalkeeper in soccer saves fewer than 50% of penalties (and a pretty good penalty taker scores more than that). So there is a general knowledge component to the question :-) – Steve Jessop Feb 16 '14 at 16:12
• One simple intuitive way to understand this is that: the more shots you make the more likely you are to "get closer to" your "real" shooting percentage. That may possibly help! :) – Fattie Feb 17 '14 at 9:56
• @laxxy it's because there are a small number of shots and they are hit-or-miss, i.e. the probability distribution is discrete, so there are step-changes in the cumulative probability as you consider the chance of making more shots. As the number of shots increases the distribution becomes more like a continuous distribution, so the steps become more like a ramp. In this comment I have summarised parts of this answer, which does a very good job of explaining the intuition and the details behind your question. – TooTone Feb 17 '14 at 12:55

It depends. If your probability to miss in a single try is $p$ (which should be low if you are a "pretty good" basketball player), then the probaility of making less than two out of four baskets (i.e. to lose the first kind of bet) is $$p_2=p^4+4p^3(1-p)=p^3(4-3p)$$ and for less than three out of six (i.e. to lose the second bet) is $$p_3=p^6+6p^5(1-p)+15p^4(1-p)^2=p^4(10p^2-24p+15).$$ We have $p_2<p_3$ iff $$p^3(4-3p)<p^4(10p^2-24p+15)$$ i.e. $$0<10p^6-24p^5+18p^4-4p^3=p^3(1-p)^2(10p-4).$$ In other words: The "2 out of 4" bet is to be preferred when $0<p<\frac2{5}$ and "3 out of 6" is to be preferred when $\frac{2}{5}<p<1$. For $p\in\{0,\frac2{5},1\}$ the bets are equivalent.

• I may be mistaken, but shouldn't the third coefficient in $p_3$'s expression be 15, instead of 30? (where 15 is (6 choose 4)). Also, shouldn't the second to last sentence be the other way around? When p is small (i.e., the player is "good"), the player should choose the "3 out of 6" option, right? (See wolframalpha.com/input/… ) – Detached Laconian Feb 16 '14 at 16:43
• The conclusion is the opposite of what the argumentation says. – durum Feb 16 '14 at 16:58
• I make it 0.4 too. – TonyK Feb 16 '14 at 18:32
• @TuringMachine yes, I was using my own definition of $p$. On Hagen's definition it is indeed 0.4. – TooTone Feb 16 '14 at 22:21
• @HagenvonEitzen Cool. :) One more thing though: I think the conclusion is still backwards; namely, the "3 out of 6" is to be preferred when $0 < p < 2/5$. – Detached Laconian Feb 17 '14 at 0:07

I would want to know what my long term average is.

If I make 333 out of 999 over the course of a season, I know that the more shots I take, the more certain I am that regression to the mean will make it less likely that I'll hit a ratio or 2:3 or better. I stand a better chance of two or three lucky shots out of three than I do 4 or more out of 6, and certainly better than 40 or more out of 60. More shots is bad because good luck can't save me from my long-term average.

On the other hand, if I make 750 out of 1000 in a season, regression to the mean works for me if I take more shots. Getting 1 or none out of 3 wouldn't be any surprise, but 3 or less out of 6, less so and 39 or less out of 60 would be a bad day indeed.

Assuming you have a $50/50$ chance of missing or making it (though it could be anything else really, the calculations would change accordingly), we can view both scenarios as binary permutations, as a means of figuring out the answer. Now in english:

Lets say you have $4$ balls (this is the first case), you would be required (in order to succeed) to make $2$ shots. Lets define: when you make a shot - that ball was a red ball. When you miss, that ball was a blue ball. Now it doesn't matter in which order you make your shots right? You could make the first two, then miss two, make one, miss one, make one, miss the last, etc. All that matters is that you have two blue balls and two red balls.

So the question is how many ways can you arrange $2$ red balls and $2$ blue balls? Well this is just a combination: namely $\binom{4}{2}$ (reads $4$ choose $2$): i.e. the answer is $6$. But lets see, there are other outcomes! For example you could be making all the shots swell (so $4/4$) but I guess we only wan't ($2/4$) so the chance that we get the outcome we desire is: (the $2$ red $2$ blue ball permutations)/(total possibilities). i.e $\binom{4}{2}/(\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4})$. which is: $1/4$. So for the first case (given that missing and making the shot share the same probability), you making $2$ out of four shots, is precisely going to happen $1/4$ of the time.

Similarly, for the $3/6$ we have: $\binom{6}{3}/ (\binom{6}{0}+\binom{6}{1}+…) = 0.3125$. Notice $0.25 < 0.3125$ So clearly: you would have a better chance of making precisely $3/6$ shots than $2/4$ shots. An intuitive explanation behind this is that since $6$ shots are more than $4$, if you mess up a bit on the $6$ shot case, it's more likely that you can still make up for it, than with $4$ shots. Of course, if your question was: making at least $3/6$ shots versus making at least $2/4$ shots, then the calculations would be a bit different. How bout you find out what your odds are in that case?

Btw, if the $\binom{n}{k}$ business is new to you, I suggest you look up permutations and combinations (and their link to probability)!

• Exactly 3 of 6 or at least 3 of 6 is a distinction without a difference, since in the former case, you can simply shoot until you get your 3, and then any shots you have left, you intentionally throw the ball nowhere near the hoop. – Matthew Najmon Feb 17 '14 at 19:50
• I left my answer intentionally in that form as a means of promoting OP to solve the more "realistic" case. – Just_a_fool Feb 17 '14 at 20:20
• Your computation for $n=4$ is wrong. You should be getting $p_4=3/8=6/2^4>5/2^4=20/2^6=p_6$ instead as expected. The more "options" you have, the less likely you are to hit the very specific number. – A.S. Feb 18 '16 at 7:53

more intuitively than the other answers:

let's say you succeed 60% of the time in average. due to the law of large numbers, the more you shoot the more likely your success frequency will be to approach this ration 0.6.

shooting once is all or nothing. shooting infinitely many times is the certainty of success.

Same goes for the opposite. If your real success rate is 0.4, you'd better try the least possible shoots.

The expectation of your average success remains the same, but not the variance/std deviation.

Assume you make your shots with probability $p$ independent of each other. Then, the probability of making $k$ shots out of $n$ is $\binom{n}{k} p^k (1-p)^{n-k}$ (That is, the number of shots you make out of $n$ baskets follows a Binomial(n,p) distribution).

The best free throw shooter in the NBA currently is Steve Nash, with $p=0.9041$. Calculate the probabilities for $n=4, k=2$ and $n=6,k=3$ and compare. You want the choice which gives you the higher probability.

octave:85> bincoeff(4,2)p^2(1-p)^2

ans = 0.045105

octave:86> bincoeff(6,3)p^3(1-p)^3

ans = 0.013036

So, you're better off taking the 2 out of the 4 shots in this model, if you're Steven Nash.

The problem is slightly different when you need to make at least 2 out of 4 shots versus at least 3 out of 6 shots. In this case, you need to sum the tail probabilities and compare them: The probability of making at least $k$ shots out of $n$ under our model is $\sum_{i=k}^n \binom{n}{i}p^i (1-p)^{n-i}$. I will leave you to calculate which one you're better off with in this case ($n=4,k=2$ and $n=6,k=3$ once again) [ The wording of the problem made it sound like you needed the exactly 2 out of 4 shots or exactly 3 out of 6 shots when I first read it ].

• This assumes the task is to make exactly the number given, not at least the number given. – Ross Millikan Feb 16 '14 at 15:31
• And that's exactly whats in the answer? Thats the binomial PMF , not the CDF. Edit: Ah, I see what you're saying. I'll update the answer. – Batman Feb 16 '14 at 15:32
• The "break even" point is at $p=\frac{21}{25}$. – Hagen von Eitzen Feb 16 '14 at 15:32
• @Hagen: it's $p=\frac35$ if you do the sums right. – TonyK Feb 16 '14 at 18:33
• Exactly 3 of 6 or at least 3 of 6 is a distiction without a difference, since in the former case, you can simply shoot until you get your 3, and then any shots you have left, you intentionally throw the ball nowhere near the hoop. – Matthew Najmon Feb 16 '14 at 22:20

According to the law of large number ( https://en.m.wikipedia.org/wiki/Law_of_large_numbers ). the more events you consider the closer will be the outcome to the expected result. In your case the expected outcome depends on actuall ability of you and your opponents in basketball playing. so if you know that you are truely a better performer to your opponents you should choose 3/6. otherwise 2/4.