A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. (For example, a six is three times as probable as a two.) What is the probability of getting an even number in one throw?
If we set \(P(1)\) equal to \(\frac{1}{n}\) then the probabilities of each possible outcome are:
\[ m(1) = \frac{1}{n} \\ m(2) = \frac{2}{n} \\ m(3) = \frac{3}{n} \\ m(4) = \frac{4}{n} \\ m(5) = \frac{5}{n} \\ m(6) = \frac{6}{n} \\ \]
Since:
\[ m(1) + m(2) + m(3) + m(4) + m(5) + m(6) = 1 \]
Then:
\[ \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \frac{4}{n} + \frac{5}{n} + \frac{6}{n} = 1 \]
Solving for \(n\) we get:
\[ \begin{align} \frac{1+2+3+4+5+6}{n} &= 1 \\ n &= 21 \end{align} \]
So the probability of each possible outcome is:
\[ m(1) = \frac{1}{21} \\ m(2) = \frac{2}{21} \\ m(3) = \frac{3}{21} \\ m(4) = \frac{4}{21} \\ m(5) = \frac{5}{21} \\ m(6) = \frac{6}{21} \\ \]
If \(E\) is the event that the result of a roll is an even numbner, then:
\[ E = \{2, 4, 6\} \]
The probability of getting an even number in one throw therefore would be:
\[ \begin{align} P(E) &= m(2) + m(4) + m(6) \\ P(E) &= \frac{2}{21} + \frac{4}{21} + \frac{6}{21} \\ P(E) &= \frac{12}{21} \\ P(E) &= \frac{4}{7} \end{align} \]
If we simulate a small number or throws there can be a large discrepancy between what we would expect and what we get:
set.seed(150)
throws <- sample(1:6, 50, replace=TRUE, prob=c(1,2,3,4,5,6)/21 )
counts = table(throws)
counts## throws
## 1 2 3 4 5 6
## 3 4 6 9 15 13p_expect = 4/7
p_expect## [1] 0.5714286p_even = sum(counts[c(2,4,6)])/50
p_even## [1] 0.52But if we increase the sample size to a very a large number (in this example we set it to 100,000) we get a result that is almost exactly what we would expect…
set.seed(150)
throws <- sample(1:6, 100000, replace=TRUE, prob=c(1,2,3,4,5,6)/21 )
counts = table(throws)
counts## throws
## 1 2 3 4 5 6
## 4804 9603 14221 19038 23601 28733p_expect = 4/7
p_expect## [1] 0.5714286p_even = sum(counts[c(2,4,6)])/100000
p_even## [1] 0.57374