# What is the sample space of a dice labelled with $1,2,2,3,3,3$ for the standard dice?

When we roll a dice labeled with $1,2,2,3,3,3$ for the standard dice. What is the sample space of this activity?

If someone argues the probability of getting $1$ is $\frac{1}{3}$. Because the person argues the sample space $S=\{1,2,3\}$ and the event of getting $1$ is $E=\{1\}$. So $$P(1)=\frac{n(E)}{n(S)}=\frac{1}{3}.$$ How can we correct this arguing?

Similar question:

If we draw a ball in a bag which contains $99$ black balls and $1$ white ball. If someone says the sample space $S$ should be $\{black, white\}$ and the event of getting a black ball $E=\{black\}$. So the probability of getting a black ball is $$P(black)=\frac{n(E)}{n(S)}=\frac{1}{2},$$ what's is the wrong with this reasoning?

• Question for you: must all outcomes in a sample space be equally likely? – kccu May 24 '16 at 2:41
• Don't argue with him. Invite him to gamble with you. – bof May 24 '16 at 2:47
• @kccu All outcomes are not equally likely. My question is why the classical formula of the probability $P(E)=n(E)/n(S)$ leads a wrong answer to this problem. – MS.Kim May 24 '16 at 3:06
• It should be clear in the examples you've given. Although the sample space in the first problem is $\{1,2,3\}$, the die has six sides and all six are equally likely to be rolled. In this case $P(1)$ is the number of ways to roll a $1$ divided by the total number of possible rolls, or $1/6$. Similarly, $P(black)$ is the number of ways to draw a black ball divided by the total number of balls, or $99/100$. – kccu May 24 '16 at 3:28
• We're still exploiting equally likely outcomes, but taking into account the fact that there are $99$ distinct black balls, not just one. Similarly there are $6$ distinct sides on the die, though some are labeled the same. – kccu May 24 '16 at 3:28

To correct your first argument, although the sample space is $\{1, 2, 3\}$, the key observation is that the probability masses assigned to each of these points are no longer equal. So the "classical probability" model no longer applies to this case and it's now not legitimate to use the formula $P(E) = n(E)/n(S)$.
Specifically, we have $P(\{1\}) = 1/6, P(\{2\}) = 1/3$ and $P(\{3\}) = 1/2$.