5

My question is related to this question, with the exception that I'm looking for the average when taking only the lower of two dice rolls.

My question is:

What is the average of rolling two dice and only taking the value of the lower dice roll?

This formula is used for the "take higher roll" case:

$$ E[X] = \sum_{x=1}^6\frac{2(x-1)+1}{36}x = \frac{1}{36}\sum_{x=1}^6(2x^2 - x) = \frac{161}{36} \approx 4.47 $$

...and includes the term "$2(x−1)+1$". Could someone please explain how that term would have to be changed so that it applies for "take lower roll"? From the original example, I'm somewhat confused where "$x-1$" comes from, although it is mentioned in the comments.

PS: I would have written this question as a comment to the original question, but I lack the points to be allowed to write comments.

N. F. Taussig
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DerpMeister
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  • If $R$ represents a die roll, it has the same distriburion as $7-R$. Since $\min[R_1,R_2]=7-\max[7-R_1,7-R_2]$, you can use the "the higher of two rolls" result to find the "lowest of two rolls" result. – eyeballfrog Jul 04 '22 at 13:29
  • Take this comment: "$\require {cancel}$To roll an x with this scheme, you either (i) roll an x with die 1, and a number $\cancel{less}$ more than x with die 2, or (ii) roll an x with die 2, and a number $\cancel{less}$ more than x with die 1, or (iii) roll an x with both dies. The possibilities for (i) and (ii) are $\cancel{x-1}$ $6-x$, so the total is $\cancel{2(x-1)+1}$ $2(6-x)+1$" – fleablood Jul 04 '22 at 15:29

5 Answers5

3

In your question you link to a question on the expected maximum of two die rolls. Based on the solution to that question (namely the expected maximum is $\frac{161}{36}$), we have:

(1) For any roll of two dice: Minimum = sum - maximum. So...

(2) $E$(Minimum) = $E$(sum) - $E$(maximum). So...

(3) So $E$(minimum) = $7 - \frac{161}{36}=\frac{91}{36}$

paw88789
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The number of total dice rolls is $36.$ The number of rolls with a minimum die value of $x$ is $2(6-x)+1=13-2x$. Therefore $P(X=x)=\frac{13-2x}{36}.$ Then $$ E[X]=\sum_{x=1}^6 P(X=x)\cdot x=\sum_{x=1}^6\frac{13x-2x^2}{36}= \frac{13}{36}\sum_{x=1}^6x-\frac{1}{18}\sum_{x=1}^6 x^2. $$

subrosar
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  • Thank you for your answer. Could you please elaborate on why 2(6-x)+1 is the number of rolls with a minimum die value of x? – DerpMeister Jul 04 '22 at 11:47
  • @DerpMeister There are $6-x$ ways to get $x$ on the first die and $>x$ on the second, $6-x$ ways to get $x$ on the second and $>x$ on the first and $1$ way to get $x$ on both dice. – subrosar Jul 04 '22 at 17:33
  • Thank you, @subrosar. – DerpMeister Jul 05 '22 at 15:00
1

Let the random variable $X$ be defined as the maximum value when two fair six-sided dice are rolled. We wish to find $E[X]$.

Since there are $k$ possible values which are at most $k$, there are $k^2$ ordered pairs of values in which both dice display a number which is at most $k$. To ensure that the maximum value is $k$, we must subtract the $(k - 1)^2$ cases in which both values are at most $k - 1$. Since there are $6^2 = 36$ possible ordered pairs of values, the probability that the maximum value is $k$ is $$\Pr(\max = k) = \frac{k^2 - (k - 1)^2}{36} = \frac{k^2 - (k^2 - 2k + 1)}{36} = \frac{2k - 1}{36}$$ Since the maximum value can assume any integer value from $1$ to $6$, the expected value for the maximum value is \begin{align*} E[X] & = \sum_{k = 1}^{6} k\Pr(X = k)\\ & = \sum_{k = 1}^{6} \frac{k(2k - 1)}{36}\\ & = \sum_{k = 1}^{6} \frac{2k^2 - k}{36}\\ & = \frac{1}{18} \sum_{k = 1}^{6} k^2 - \frac{1}{36}\sum_{k = 1}^{6} k\\ \end{align*} Use the formulas \begin{align*} \sum_{k = 1}^{n} k^2 & = \frac{n(n + 1)(2n + 1)}{6}\\ \sum_{k = 1}^{n} k & = \frac{n(n + 1)}{2} \end{align*} with $n = 6$ to finish the calculation.

Let the random variable $Y$ be defined as the minimum value when two fair six-sided dice are rolled. We wish to find $E[Y]$.

On a six-sided die, there are $7 - k$ values which are at least $k$. Hence, there are $(7 - k)^2$ ordered pairs of values in which both dice display a number which is at least $k$. To ensure that the minimum value is $k$, we must subtract the $[7 - (k + 1)]^2 = (6 - k)^2$ cases in which both dice display a number which is at least $k + 1$. Since there are $6^2 = 36$ possible ordered pairs of values, the probability that the minimum value is $k$ is $$\Pr(\min = k) = \frac{(7 - k)^2 - (6 - k)^2}{36} = \frac{49 - 14k + k^2 - (36 - 12k + k^2)}{36} = \frac{13 - 2k}{36}$$ Since the minimum value can assume any integer value from $1$ to $6$, the expected value for the minimum value is $$E[Y] = \sum_{k = 1}^{6} k\Pr(Y = k) = \sum_{k = 1}^{6} \frac{k(13 - 2k)}{36}$$ I will leave it to you to finish the calculations.

N. F. Taussig
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There are several ways to work out the expected value. But since you already know the expected value of the higher of the two dice, the easiest way is to use that knowledge.

Let $X$ be the lower value of the two dice, so $E(X)$ is what you want to compute.

Let $Y$ be the higher value of the two dice. Then the answer to the other question tells you that $$E(Y)=\frac{161}{36}.$$

Let $Z$ be the sum of the two dice. It is well-known that $E(Z)=7.$

But the sum of two dice is always the sum of the lesser value and the greater value, even when both dice roll the same number (because in that case the “lower” value and the “higher” value are the same).

That is, $Z=X+Y.$

By linearity of expectation, that implies that $E(Z)=E(X)+E(Y).$

Solve for $E(X)$:

$$ E(X) = E(Z) - E(Y). $$

The rest is just plugging in the known values and doing a little arithmetic to simplify the answer.

David K
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Here's your payoff matrix:

{{1, 1, 1, 1, 1, 1}, 
 {1, 2, 2, 2, 2, 2}, 
 {1, 2, 3, 3, 3, 3}, 
 {1, 2, 3, 4, 4, 4}, 
 {1, 2, 3, 4, 5, 5}, 
 {1, 2, 3, 4, 5, 6}}

Sum these elements and divide by $36$.

David G. Stork
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