 # Probability - Dice

Hi,

Thank you!!

If the six-face dice is numbered by 1, 2, 3, 4, 5, and 6, but the eight-face dice is numbered by 1, 2, 3, 4, 5, 6, 6, and 6, what is the expected number if you roll each type of dice many times, respectively?

i did my calculations and i get number 7 as the expected number, however, i am not very sure

Draw yourself a 6x8 table. List out the sum of each combo, add them up and divide by 48. I get an answer of 7.625.

Are they fair dice?

Are you looking for the expected sum? Or the expected value on each?

you can think of expected number as the “average” number you would get if you repeated the experiment many, many times. as s2k pointed out, you need to know if the die is fair (i.e. all outcomes are equally likely) or not (in which case, at least one of the outcomes will be more likely than the others).
in general, expected value for discrete events (such as rolling of a die) is defined as follows:
E(X) = sum from x = min(k) to x = max(k) of x*p(x = k)
(mathjax doesnt seem to work for me)
E(X) = \sum_{x=min(k)}^{max(k)} x * p(x = k)

for a fair, 6-sided die numbered 1 through 6, the expeted value is, by definition,
1 * (1/6) + 2 * (1/6) + … + 6 * (1/6) = 21/6 = 3.5
that is, if you roll a die many times, on average you would get 3.5
the answer will be different if the die is not fair. for example, if 1 had 0 probability of occurring and 2 had twice the probability, then the expected value would be
1 * (0/6) + 2 * (2/6) + … + 6 * (1/6) = 3.67

use the same steps to find the expected value of the 8-sided die.

edit: if the question is asking for the expected value of the sum, there is a neat formula from probability theory:
E (X + Y) = E(X) + E (Y)
that is, the expected value of the sum of two random variables is the sum of the expected values of the individual random variables.
the expected value for a fair, 6-sided die is 3.5
the expected value for a fair, 8-sided die with the extra 2 numbers being 6 is 4.125
thus, the expected value of the sum is 3.5 + 4.125 = 7.625, which agrees with bread.

edit: mathjax works now, thanks magician It didn’t occur to me to treat each die as independent!!! That makes things a little easier! yes, both dice have an equal probability.
To be honest, the question is very unclear to me; meaning that I do not understand If we have to find the expected sum or the expected value on each.
I do believe that is asking for the expected sum.

Thank you

Thanks a lot!!!

you’re very welcome 1 Like

I did my research and I should calculate separately the mean of each die.

I also have another question in the same problem and again I am not that sure about the results. The question is: if you randomly pick one die from a black box with two six-face dice and one eight-face die, what is the expected number you can roll?

Assume that p(select 2 six dice)=p( select 8 side die)
2/3*(3.5)+1/3(4.125)

Does it seem right?

yea, that seems right to me. the problem wants you to know the law of total probability.
for this problem,
E(X) = E(X|A) * P(A) + E(X|B) * P(B)
where A is the event of selecting the 6-sided die
and B is the event of selecting the 8-sided die
(and with the implied constraint P(A) + P(B) = 1)

in general,
E(X) = \sum_{i=0}^n E(X_i|A_i) * P(A_i)

with the constraint
\sum_{i=0}^n P(A_i) = 1, A_i all mutually exclusive

edit: fixed

1 Like

I think what you meant was that each die has a 1/3 chance of being picked, but since there are 2 six-sided dice, then the probability of picking one of them is 2/3, as you correctly identified. Mutually exclusive, not independent.

Two events cannot be both mutually exclusive and independent.

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thanks for the correction. it seems that my memory was rusty.

This question has confused me so much. Every person I talked to has different results. @not_a_CFA @breadmaker

what other answers did you get? it is instructive to understand why the other answers are wrong.

At least you’re young enough to have a memory. 1 Like

@not_a_CFA in this question for example, the other person claims that there is not an expected value for the Six faced die, and on the other hand, he thinks that the expected number to be rolled for the 8-sided die will be 6 as it has higher a higher probability. And it doesn’t make any sense to me, because we haven’t been asked to find the mode

@not_a_CFA in this question he said that we have a 66.66 % chances of blindly selecting a six faced die, and 33.34% of randomly selecting the 8sided die from the black box.
Then he calculated the probabilities
P1=1/62/3+1/81/3= 15.28%
P2= 15.28%
P3=15.28%
P4=15.28%
P5=15.28
P6= 1/62/3+3/81/3=23.61%
Therefore the expected number should be 6 as it has higher chances to be rolled

He started with the law of total probability first:
P(X = 1) =
P(X = 1 | 6-sided die) * P(6-sided die) + P(X = 1 | 8-sided die) * P(8-sided die) = (1/6) * (2/3) + (1/8) * (1/3) = 11/72
repeating this calculation, you have
P(X = 1) = … P(X = 5) = 11/72
P(X = 6) = 1 - 5* 11/72 = 17/72
now you just need to use the definition of expected value (see my previous post above)

E(X) = 1* 11/72 + 2* 11/72 + … + 6 * 17/72 = 3.708333…