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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?

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    $\begingroup$ Question for you: must all outcomes in a sample space be equally likely? $\endgroup$ – kccu May 24 '16 at 2:41
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    $\begingroup$ Don't argue with him. Invite him to gamble with you. $\endgroup$ – bof May 24 '16 at 2:47
  • $\begingroup$ @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. $\endgroup$ – MS.Kim May 24 '16 at 3:06
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    $\begingroup$ 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 $99/100$. $\endgroup$ – kccu May 24 '16 at 3:28
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    $\begingroup$ 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. $\endgroup$ – kccu May 24 '16 at 3:28
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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 $P(\{3\}) = 1/2$.

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  • $\begingroup$ When can we apply the classical formula of probability? When all outcomes are equally likely? What's the definition of probability when all outcomes are not equally likely? $\endgroup$ – MS.Kim May 24 '16 at 3:13
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    $\begingroup$ For your first question, the classical formula applies when "all outcomes are equally likely", which is your second question. Given you are solving a probability problem (instead of constructing a statistical model), whether all outcomes are equally likely can be read from the context, which is usually quite obvious and intuitive. Your third question is more complicated, which you might want to consult the axiomatic definition of probability, for example, see this link. $\endgroup$ – Zhanxiong May 24 '16 at 3:20
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Possibly a return to the definitions is needed. The sample space is simply the set of possible events - absolutely nothing about the probability masses is implied by a sample space, other than if the set is complete, the sum of the probabilities is one. The formula you state refers to a property of a fair die - by definition each side is equally likely to appear. There is nothing 'classical' about this, it's a set of words used to establish a situation that we can reason about . There is no reason to not choose any the three probabilities that sum to one as the three probability masses assigned to the sample space.

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