Matching values of Die throws pt. 3

[BMAD]

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

[bonanova]

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.

[BMAD]

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

[bushindo]

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

Here's my answer

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

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

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

[witzar]

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

[bonanova]

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

Agree.

[bushindo]

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

Nice compact form!

[BMAD]

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 − 1)ⁿ.

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

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

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

[BMAD]

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

Edited April 30, 2013 by BMAD

[witzar]

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 − 1)ⁿ.

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

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

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

[BMAD]

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.

[bushindo]

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

[BMAD]

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 − 1)ⁿ.
So, k is the highest score in kⁿ − (k − 1)ⁿ cases out of 6ⁿ.
In other words, p_n(k), the probability that the highest individual score is k, is (kⁿ − (k − 1)ⁿ)/6ⁿ.

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

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

[bushindo]

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 − 1)ⁿ.

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

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

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

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

Nevermind, they are equivalent. See witzar's

[BMAD]

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 − 1)ⁿ.
So, k is the highest score in kⁿ − (k − 1)ⁿ cases out of 6ⁿ.
In other words, p_n(k), the probability that the highest individual score is k, is (kⁿ − (k − 1)ⁿ)/6ⁿ.

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

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

Edited April 30, 2013 by BMAD

[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 − 1)ⁿ.

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

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

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

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

[BMAD]

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 − 1)ⁿ.

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

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

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

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

[bushindo]

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 − 1)ⁿ.

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

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

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

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

But I misread it as

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

Mea culpa =)

[witzar]

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.

[BMAD]

oh good, i can stop pulling my hair out now. Well done witzar.

[dark_magician_92]

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

[BMAD]

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

i believe 3.5

[dark_magician_92]

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