Please help me start this question.


New Member
Consider an experiment where we roll a single die once and let X represent the number of dots that appear on the top face of the die. Suppose that the die is not fair and that P(X=x)=x/k for x= 1,2,3,4,5,6.

All I would like to know is what does the k in x/k mean in this question? How would I go about determining the probability for x at a given number when the dice is unfair by an unknown value? Thank you in advance for any help. :)


New Member
When a die is unfair by a unknown amount, many tosses are required. One method I've
used is chi square. If the "unfairness" is large enough, two effects in the chi square
analysis make it certain the die is unfair.

Let me give you a numerical example that's easy to detect. Here, five faces are
slightly lower in probability that 1/6 and one face is slightly higher:

.98/6 + .98/6 + .98/6 +.98/6 + .98/6 + 1.1/6 = 1

The sum of the dice "face up" probabilities equal one.

I won't go into the details of the chi square analysis.

Hope this helps.

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New Member
Thank you for the responses ArtK and JesperHp. I understand the concept of an unfair dice but I'm just a little confused about how to determine the constant of unfairness without any information given. The question that follows is what value of k would yield a valid distribution for X. So am I simply just making up my own unfair probability and filling in my chart? I think I'm really over thinking this question and as the only way to make sense of it is to make up the unfairness of the dice rolled myself such like you did in your answer ArtK. Clarification would be greatly appreciated. Thank you.:)


Ambassador to the humans
Let me ask you this. Would this be a valid distribution:

P(X = 1) = 1/2
P(X = 2) = 2/2
P(X = 3) = 3/2
P(X = 4) = 4/2
P(X = 5) = 5/2
P(X = 6) = 6/2

if the answer is no then why not? What needs to be true to make something a valid probability distribution?


New Member
No this would not be a valid distribution as the probability must add up to = 1. In this case the distribution = 21/2 which is not a valid probability.


Super Moderator
So am I simply just making up my own unfair probability and filling in my chart? I think I'm really over thinking this question...
Yes, you are over thinking the problem (as you originally posted).....That is to say, the constant \( k \) can take on any value - other than 1/6 - such that it creates a discrete probability mass function, which has already been suggested in a specific form above (i.e. ArtK).


TS Contributor
While \( k \) is yet to be determined, the values for \( x \) can be taken is known. So \( k \) is not arbitrary.