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# Matching values of Die throws pt. 3

## Question

Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die?

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7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n

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I'll be a smart a** and say it's seven minus the expected value of the lowest valued die.

Seriously, I'll work on it.

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We seek the expected value of the highest individual score when n dice are thrown. We first find pn(k), the probability that the highest score is k. There are kn ways in which n dice can each show k or less.

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Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die?

Let PN( i ) be the chance of having i as the highest score on simultaneous rolls of N dice. Obviously, PN( 0 ) = 0. The general expression for PN( i ) is

Now that we have PN( i ), the expected value is simply

E = sum PN( i )* i for i from 1 to 6.

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7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n

Agree.

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7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n

Nice compact form!

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7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n

For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.

So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.

In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

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The table below shows E(n), correct to four decimal places, for various values of n, from 1 to 50. As expected intuitively, E(n) approaches a number, as n increases.

Expected values

n E(n)

1 3.5

2 4.4722

3 4.9583

4 5.2446

5 5.4309

6 5.5603

7 5.6541

8 5.7244

9 5.7782

10 5.8202

20 5.9736

30 5.9958

40 5.9993

50 5.9999

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7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n

For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.

So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.

In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

That's the obvious and natural approach. And that's exactly how I got the formula.

When you say "close but", you mean that the formula is wrong?

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it wouldn't be the first time I made a mistake on one of my own problems , but yes, i derived a slightly different answer.

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it wouldn't be the first time I made a mistake on one of my own problems , but yes, i derived a slightly different answer.

I applied witzar's formula to various different N, and I reproduced the same exact table that you gave in here

The table below shows E(n), correct to four decimal places, for various values of n, from 1 to 50. As expected intuitively, E(n) approaches a number, as n increases.

Expected values

n E(n)

1 3.5

2 4.4722

3 4.9583

4 5.2446

5 5.4309

6 5.5603

7 5.6541

8 5.7244

9 5.7782

10 5.8202

20 5.9736

30 5.9958

40 5.9993

50 5.9999

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There are kn ways in which n dice can each show k or less.
For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.
So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.
In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

Hence E(n) = [6(6nâˆ’5n) + 5(5nâˆ’4n) + 4(4nâˆ’3n) + 3(3nâˆ’2n) + 2(2nâˆ’1n) + 1(1nâˆ’0n)]/6n. = 6 âˆ’ (1n + 2n + 3n + 4n + 5n)/6n.

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There are kn ways in which n dice can each show k or less.

For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.

So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.

In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

Hence E(n) = [6(6nâˆ’5n) + 5(5nâˆ’4n) + 4(4nâˆ’3n) + 3(3nâˆ’2n) + 2(2nâˆ’1n) + 1(1nâˆ’0n)]/6n. = 6 âˆ’ (1n + 2n + 3n + 4n + 5n)/6n.

Nevermind, they are equivalent. See witzar's

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I concede to Witzar.

There are kn ways in which n dice can each show k or less.
For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.
So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.
In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

Hence E(n) = [6(6nâˆ’5n) + 5(5nâˆ’4n) + 4(4nâˆ’3n) + 3(3nâˆ’2n) + 2(2nâˆ’1n) + 1(1nâˆ’0n)]/6n. = 6 âˆ’ (1n + 2n + 3n + 4n + 5n)/6n.

Let's take a look at the highlighted equation above

Let's play with only 1 die so n = 1. So the expected value of the highest number is simply (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5

But if you plug it in the above formula, you'll get a different value for the expected face number

6 - (1 + 2 + 3 + 4 + 5 + 6)/6 = 6 - 3.5 = 2.5

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There are kn ways in which n dice can each show k or less.

For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.

So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.

In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

Hence E(n) = [6(6nâˆ’5n) + 5(5nâˆ’4n) + 4(4nâˆ’3n) + 3(3nâˆ’2n) + 2(2nâˆ’1n) + 1(1nâˆ’0n)]/6n. = 6 âˆ’ (1n + 2n + 3n + 4n + 5n)/6n.

7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n =

= 6 + 6^n/6^n - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n =

= 6 - (1^n + 2^n + 3^n + 4^n + 5^n)/6^n

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There are kn ways in which n dice can each show k or less.

For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.

So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.

In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

Hence E(n) = [6(6nâˆ’5n) + 5(5nâˆ’4n) + 4(4nâˆ’3n) + 3(3nâˆ’2n) + 2(2nâˆ’1n) + 1(1nâˆ’0n)]/6n. = 6 âˆ’ (1n + 2n + 3n + 4n + 5n)/6n.

7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n =

= 6 + 6^n/6^n - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n =

= 6 - (1^n + 2^n + 3^n + 4^n + 5^n)/6^n

if they are equivalent then why don't they produce the same expected values as Bushindo pointed out.

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There are kn ways in which n dice can each show k or less.

For the highest score to equal k, we must subtract those cases for which each die shows less than k; these number (k âˆ’ 1)n.

So, k is the highest score in kn âˆ’ (k âˆ’ 1)n cases out of 6n.

In other words, pn(k), the probability that the highest individual score is k, is (kn âˆ’ (k âˆ’ 1)n)/6n.

The expected value, E(n), of the highest score is the sum, from k = 1 to 6, of k Â· pn(k).

Hence E(n) = [6(6nâˆ’5n) + 5(5nâˆ’4n) + 4(4nâˆ’3n) + 3(3nâˆ’2n) + 2(2nâˆ’1n) + 1(1nâˆ’0n)]/6n. = 6 âˆ’ (1n + 2n + 3n + 4n + 5n)/6n.

7 - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n =

= 6 + 6^n/6^n - (1^n + 2^n + 3^n + 4^n + 5^n + 6^n)/6^n =

= 6 - (1^n + 2^n + 3^n + 4^n + 5^n)/6^n

if they are equivalent then why don't they produce the same expected values as Bushindo pointed out.

I misread the equation. Yours was

6 âˆ’ (1n + 2n + 3n + 4n + 5n)/6n.

6 âˆ’ (1n + 2n + 3n + 4n + 5n + 6n) /6n.

Mea culpa =)

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if they are equivalent then why don't they produce the same expected values as Bushindo pointed out.

They do of course. Bushindo made a silly error evaluating your formula.
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oh good, i can stop pulling my hair out now. Well done witzar.

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I have a question. What will be the expected value if only 1 dice is rolled?

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I have a question. What will be the expected value if only 1 dice is rolled?

i believe 3.5

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Ok thanks. I didnt realise expected value may be a fraction.

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