# Die is rolled seven times. Which is the most likely outcome?

A six sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the most likely outcome of the experiment?

(A) Three green faces and four red faces.

(B) Four green faces and three red faces.

(C) Five green faces and two red faces.

(D) Six green faces and one red face.

My answer: Six green faces and one red face.

Why I think so :

1.The probability of getting a green face (4/6) is more than the probability of getting a red face (2/6).

2.The number of throws doesn't matter because the second throw (or consecutive throws) is (are) independent of the previous throws.

1. More green faces means a more likely outcome (hence option D).

But the answer is option C and the explanation as : Considering uniformly distributed outcomes, we get 4 greens and 2 reds in six throws. Then in one more throw, green is more likely. So, option (C) is correct.

Why are red faces being counted in when it's a matter of most likely outcome?

• I would advise using the binomial distribution rather than intuitive handwaving. – Lord Shark the Unknown Aug 12 '18 at 20:08
• +Lord Shark the Unknown, the question asks for a 'most likely outcome' which is why I think binomial distibution is unnecessary. – Ryu Aug 12 '18 at 20:10
• Why was this downvoted? – Shaun Aug 12 '18 at 20:13

While at first glance your reasoning seems to make sense ("green faces are more likely .. so the more green faces, the more likely that is"), it can easily be shown that there must be something fishy with it.

Take it to its logical conclusion: suppose you throw that dice $1000$ times .. which is the most likely outcome? You can get $0$ green faces, .. or $1$, or $2$ ... all the way up to $1000$. According to your reasoning, out of these $1001$ possible outcomes, getting all $1000$ green faces is the most likely.

OK, but think about it ... not getting any red face?!? That's bordering on absolutely incredible! Getting a red face is not that much more unlikely ... indeed, what if you had a coin that was slightly biased (say, $50.1$%) towards heads, and you flip it $1000$ times ... don't you think it would be crazy to think that the most likely outcome is to get all heads?!

Now, in your case, we have $\frac{2}{3}$ vs $\frac{1}{3}$ ... so sure, there is a bias towards getting green faces ... but again not all that great of a bias ... and indeed getting all $1000$ green faces when flipping $1000$ times should be really, really unlikely.

Indeed, just using your common sense, the probability of getting $1000$ green faces is certainly way smaller than $1$ in $1001$, and given that there are $1001$ possible outcomes, it should be clear that the most likely outcome is in fact not the outcome of getting all $1000$ times a green face when flipping that coin $1000$ times.

OK, so your reasoning does not work. But why not? Where does your reasoning go wrong?

Well, you forget about the fact that there is only and exactly $1$ way to get all green faces: you need to get a green face every time! However, there are $1000$ ways to get $999$ green faces, and $1$ red face, as the red face can be the first throw, or the second, or the third .... So, already you can see that getting $999$ green faces and $1$ red face would be far more likely than getting $1000$ green faces.

So far the intuitions. Mathematically what is going is this. Look at the binomial formula:

$$P(X=N)={M \choose N} \cdot p^N \cdot (1-p)^{N-M}$$

OK, so sure, getting a green face is more likely than getting a red face, i.e. $p > 1-p$, and so if you just look at the part:

$$p^N \cdot (1-p)^{N-M}$$

then that indeed will be higher, the higher $X$ is, and indeed will be highest when $X=M$

However, this does not mean that you are most likely to get 1000 green faces, because you forget about the other part

$${M \choose N}$$

which is the number of ways to get the outcome. And again, for $X=M$, there is only one way ... but for smaller $X$'s there can be many more ways.

So, you get an interesting interplay: yes, the higher $X$ is, the higher the right part of the formula, but if $X$ gets too high, then the left part will shrink.

Now, if you do the math, it turns out that the most likely outcome is an outcome that reflects the probabilities $p$ and $1-p$, i.e. about two thirds green and one third red. So, options A and D are a bit 'out of whack' in their respective proportions, and it would have to be between B and C ... which one? The explanation provided gives the answer

• +Bram, If I change the question and and bring it down to a singe throw, one green face would be a 'more likely outcome' because p(green)>p(red) at any moment. How I am looking at this is I'm taking the 'most likely outcomes' at single throws and combining the result in case of seven throws. – Ryu Aug 12 '18 at 20:23
• @Ryu Yes, I see that ... but if you look at my example: there is only 1 way to get 600 green faces .. there are many more ways to get 400 green faces and 200 red faces ... that's the difference. – Bram28 Aug 12 '18 at 20:25
• +Bram, you mean 600 green faces? – Ryu Aug 12 '18 at 20:38
• @Ryu Sorry, I changed the numbers ...please read my updated post ... I am trying to explain, conceptually/intuitively, as well as mathematically, where your reasoning goes wrong. – Bram28 Aug 12 '18 at 20:45

Think about what your argument would mean if the die was rolled $1\,000\,000$ times and you had $500\,001$ options from $499\,999$ to $999\,999$ green faces.

In general we can say that the most likely outcome of a series of events is not the series of the most likely outcome of the individual attempts.

In this case: As a third of the die's face are red, we would expect about a third of the rolls to come out "red".

• +Henrik, if I asked which one is more likely and gave you two options : 1. 999999 green faces and one red face 2. 999990 green faces and 10 red faces Which one would be a more likely outcome? – Ryu Aug 12 '18 at 20:16
• Both options are very unlikely, but the second is slightly less unlikely. (and if you couldn't see that yourself, I suggest you paint a die and start rolling) – Henrik Aug 12 '18 at 20:36

Strictly speaking you have a binomial distribution $\mathcal{B}in(n, p)$, where $n=7$ and $p = \frac 2 3$. The options you need to compare are:

(A) $$C_7^3 \left(\frac 2 3\right)^3 \left(\frac 1 3\right)^4$$ (B) $$C_7^4 \left(\frac 2 3\right)^4 \left(\frac 1 3\right)^3$$ (C) $$C_7^5 \left(\frac 2 3\right)^5 \left(\frac 1 3\right)^2$$ (D) $$C_7^6 \left(\frac 2 3\right)^6 \left(\frac 1 3\right)^1$$

Obviously, you don't need to compute all these numbers: First of all, you can simply ignore $3^7$ in the denominators. From options (A) and (B) using the property $C_n^k = C_n^{n - k}$ we get that the answer in (B) is twice as much as in (A), so (A) is not the largest. Further, we need to compare $C_7^4 = 35$ and $2C_7^5 = 2 \cdot 21 =42$. So, (B) isn't the largest either. $2C_7^6 = 14 < C_7^5 = 21$ yielding the correct answer of being (C).

• To find the largest, it suffices to compute the ratios between consecutive terms. That's less work than computing the terms individually. – Lord Shark the Unknown Aug 12 '18 at 20:19
• yep, i was editing my answer. – pointguard0 Aug 12 '18 at 20:23

The expectation is $7\cdot\dfrac 46=4.666\cdots$ greens, which is closest to $5$ (the distribution is unimodal).

A six-sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the most likely outcome of the experiment

The probability of rolling a green is $p=\frac{4}{6} =\frac{2}{3}$, the probability for rolling a red is $p=\frac{2}{6}=\frac{1}{3}$ They are complements of eachother.

Let $X\sim Bin(n,p)$ be a binomial distributed random variable. We let $k$ be the number of successes of rolling greens. Then $X \sim Bin(7,\frac{2}{3})$

the mass function is given by

$$f(k,7,\frac{2}{3}) =Pr(X=k) = \binom{n}{k} p^{k}(1-p)^{n-k}$$

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

going through these

$$Pr(X=3) = \binom{7}{3} (\frac{2}{3})^{3} (\frac{1}{3})^{4}$$ $$Pr(X=4) = \binom{7}{4} (\frac{2}{3})^{4} (\frac{1}{3})^{3}$$ $$Pr(X=5) = \binom{7}{5} (\frac{2}{3})^{5} (\frac{1}{3})^{2}$$ $$Pr(X=6) = \binom{7}{6} (\frac{2}{3})^{6} (\frac{1}{3})^{1}$$

Simplifying them all $$Pr(X=3) =35 \frac{8}{27} \frac{1}{81}$$ $$Pr(X=4) =35 \frac{16}{81} \frac{1}{27}$$ $$Pr(X=5) = 21 \frac{32}{243} \frac{1}{9}$$ $$Pr(X=6) = 7 \frac{64}{729} \frac{1}{3}$$ then we have $Pr(X=3),Pr(X=4)$ are visually much more likely working them out

$$Pr(X=3) =35 \frac{8}{27} \frac{1}{81} \approx .12$$ $$Pr(X=4) =35 \frac{16}{81} \frac{1}{27} \approx .25$$

Edit: it's actually after an edit

$$Pr(X=5) = 21 \frac{32}{243} \frac{1}{9} \approx .30$$

Offhand $Pr(X=5)$ makes the most sense as $\frac{1}{3}$ is to a low power however the binomial coefficient out weighs it.

• $7-5$ does not equal $5$ (see your $Pr(X=5)$). – Lord Shark the Unknown Aug 12 '18 at 20:23
• ahh hah oops will edit – Shogun Aug 12 '18 at 20:26