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You toss a fair die three times. What is the expected value of the largest of the three outcomes?

My approach is the following:
calculate the probability of outcome when $\max=6$, which is
$$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{no }6) = 1 - (5/6)^3$$ and then calculate the probability of outcome when $\max=5$, which is $$P(\text{at least one $5$ of the three rolls & $5$ is max}) = 1 - P(\text{no $5$ & $5$ is max}) = 1 - (4/6)^3.$$

I wonder if this approach is right.

Em.
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Jovoszhou
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4 Answers4

36

More generally let's find $E[X_\max]$ and $E[X_\min]$ where $X_\max$ and $X_\min$ are the largest and smallest outcomes among $n$ independent rolls of an $s$-sided die with sides numbered from $1$ to $s$.

Define $X_i$ to be the random variable whose value is $1$ if $X_\max\ge i$ and $0$ otherwise; then $$X_\max=\sum_{i=1}^sX_i$$ and $$E[X_\max]=E\left[\sum_{i=1}^sX_i\right]=\sum_{i=1}^sE[X_i]=\sum_{i=1}^sP(X_i=1)=\sum_{i=1}^sP(X_\max\ge i)$$ $$=\sum_{i=1}^s[1-P(X_\max\lt i)]=\sum_{i=1}^s\left[1-\left(\frac{i-1}s\right)^n\right]=s-s^{-n}\sum_{i=1}^s(i-1)^n=\boxed{s-s^{-n}\sum_{i=1}^{s-1}i^n}.$$ Since the outcome of a single roll is distributed symmetrically about its mean $\frac{1+s}2$, we have $E[X_\min]+E[X_\max]=1+s$, so that $$E[X_\min]=1+s-E[X_\max]=\boxed{s^{-n}\sum_{i=1}^si^n}.$$

When $s=6$ and $n=3$ we have $$E[X_\max]=6-6^{-3}(1^3+2^3+3^3+4^3+5^3)=6-6^{-3}\binom62^2=6-\frac{25}{24}=\boxed{\frac{119}{24}}$$ and $$E[X_\min]=7-E[X_\max]=7-\frac{119}{24}=\boxed{\frac{49}{24}}.$$

bof
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    This should be the top answer – Laksh Jan 27 '19 at 21:04
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    Nice generalization! – Idonknow Jan 03 '20 at 01:47
  • If the values are symmetrically distributed then why is it always the case that $E[X_{\text {max}}] + E[X_{\text {min}}] = 2\ E [X]\ $? Could you please explain it? Thanks in advance. – Anil Bagchi. Aug 25 '21 at 05:28
  • @AntonioClaire Don't recall what I was thinking when I wrote this answer 5 years ago. Seems intuitive. Let's see. Do you agree that it would be true if the values were symmetrically distributed about zero? Say the sides of the cube were labeled $-3,-2,-1,+1,+2,+3$? Is it clear in that case that $E[X_\max]+E[X_\min]=0$ ? Maybe the general case could be derived from that by some simple algebra. – bof Aug 25 '21 at 06:20
  • @bof yeah I agree with you when $X$ is symmetrically distributed about zero. Now if $X$ is symmetrically distributed about it's mean then $X - E[X]$ is symmetrically distributed about zero and then using the above formula for $X - E[X]$ and noting that $(X - E[X]){\text {max}} = X{\text {max}} - E[X]$ and $(X - E[X[){\text {min}} = X{\text {min}} - E[X]$ we are through. – Anil Bagchi. Aug 25 '21 at 07:18
12

Let $X_{i}$ be the value from individual die

Let Z be the max value: $Z = max(X_1, X_2, X_3)$

The CDF of Z is $F_Z(x) = P(Z\le x) = P(X_1\le x)P(X_2\le x) P(X_3\le x) = F_{X_1}(x)^3 = (x/6)^3 $

Therefore PMF of Z: $f_Z(x) = F_Z(x) - F_Z(x-1) = (x^3-(x-1)^3)/6^3 $

$ E(Z) = \sum_{x=1}^6 xf_Z(x) = 119/24 $

5

Let X denote the largest value, then:

  • $P(X=1)=\frac{ 1}{216}$
  • $P(X=2)=\frac{ 7}{216}$
  • $P(X=3)=\frac{19}{216}$
  • $P(X=4)=\frac{37}{216}$
  • $P(X=5)=\frac{61}{216}$
  • $P(X=6)=\frac{91}{216}$

Hence the expected value is:

$$1\cdot\frac{1}{216}+2\cdot\frac{7}{216}+3\cdot\frac{19}{216}+4\cdot\frac{37}{216}+5\cdot\frac{61}{216}+6\cdot\frac{91}{216}=\frac{119}{24}$$

Details:

Let $C_n$ denote the number of combinations with largest value $n$.

Observe that $C_n=n^3-\sum\limits_{k=1}^{n-1}C_{n-1}$, therefore:

  • $C_1=1^3=1$
  • $C_2=2^3-1=7$
  • $C_3=3^3-1-7=19$
  • $C_4=4^3-1-7-19=37$
  • $C_5=5^3-1-7-19-37=61$
  • $C_6=6^3-1-7-19-37-61=91$
barak manos
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1

Picture the cube of possible outcomes. The cells that represent a maximum of $6$ lie in a greedy half of the outer layer, which has $6^3-5^3=216-125=91$ cells in it.

The next layer represents max $5$, and has $5^3-4^3=125-64=61$ cells in it.

We can proceed in a similar manner and arrive at the sum of the whole cube:

$$6\cdot(6^3-5^3)+5\cdot(5^3-4^3)+4\cdot(4^3-3^3)+3\cdot(3^3-2^3)+2\cdot(2^3-1^3)+1\cdot(1^3-0^3)$$

$$=6^3(6)-5^3(6-5)-4^3(5-4)-3^3(4-3)-2^3(3-2)-1^3(2-1)-0^3(1)$$

$$=6\cdot6^3-5^3-4^3-3^3-2^3-1^3-0^3$$

$$=1296-225$$

$$=1071$$

Divide by the number of cells in the cube $6^3=216$, and the answer is:

$$\frac{1071}{216}=\frac{119}{24}\approx4.96$$

JMP
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