New to probability - Is this true?

Solution 1:

Let's find the probability of, say, rolling a 5. You could roll a 5 right away with probability $\frac{1}{20}$, roll a number over 12 and then roll again and get a 5 with probability $\frac{8}{20} \frac{1}{20}$, roll two numbers over 12 and then roll a 5 with probability $\frac{8}{20} \frac{8}{20} \frac{1}{20}$, etc. All these ways of rolling a 5 are mutually exclusive so their probabilities sum up. In total the probability of eventually rolling a 5 is given by the geometric series below, which I evaluated using the usual geometric series sum formula.

$$S = \frac{1}{20} + \frac{8}{20} \frac{1}{20} + \frac{8}{20} \frac{8}{20} \frac{1}{20} + \ ... \ = \sum_{i=0}^{\infty} \frac{1}{20} \left(\frac{8}{20}\right)^i = \frac{\frac{1}{20}}{1-\frac{8}{20}} = \frac{1}{12}$$

The fact that when you reduce the sample space there are only 12 of the 20 equally likely possible outcomes remaining, one of which must occur eventually, is a nice informal way to find the sum of this infinite series.

Solution 2:

What you are looking for is the notion of "conditional probability". The process which you are describing can be thought of "sampling the dice" conditioned on the event that the result is between $1$ to $12$.

Let $A$ be the event that the dice falls between $1$ to $12$. This event has probability $12/20$. Now, denote by $B_i$ the event that the dice falls on the number $i$, where $i$ is between $1$ and $12$. Each $B_i$ has probability $1/20$. Then, you have that:

$P(B_i \mid A) = P(B_i\cap A)/P(A) = \frac{1/20}{12/20} = \frac{1}{12}$,

which is indeed uniform over the numbers $1$ to $12$.

Solution 3:

Yes. The probabilities of rolling any given number are still equal, so when you discard the extras you are left with twelve numbers with equal probability. Therefore the probability of any single number must be $\frac{1}{12}$.