Let's imagine a die roll.
You roll the die n times.
Example: I have a $6$ sided die. Assuming the distribution of the die is perfect, the chances of getting any single number are $1/6$. The chances of getting two consecutive same numbers (by rule of product) are $1/36$. The chances of getting any two specific numbers (let's assume a random pair like $(2,3)$) are $1/36$ also.
Let's assume that pairs of type $(n,m)$ are equivalent to pairs of type $(m,n)$ - we count them as equal. So when we roll the die n times, is the chance of some pair $(m,n)$ appearing greater than the chance of a pair $(n,n)$ appearing? That is to say is the chance of getting $(2,3)$ & $(3,2)$ in a row bigger than just the chance of getting $(4,4)$ & $(4,4)$ in a row?
Answer
The probabilities of outcomes of the die do not depend on which outcomes you consider
to be equivalent. The die rolls and stops rolling as it always does.
When we were paying attention to the sequence in which the numbers appeared, there
was just a $\frac{1}{36}$ chance of getting $(4,4),$
a $\frac{1}{36}$ chance of getting $(2,3),$
and a $\frac{1}{36}$ chance of getting $(3,2).$
When we only consider events distinct based on the number of times each face of
the die has occurred, there is still only a $\frac{1}{36}$ chance
that face $4$ occurred twice in two rolls,
and there is still a $\frac{1}{18}$ chance that $2$ and $3$ each occurred once.
Compare this to what happens if we only write down the sum of the numbers that occur,
not the actual numbers themselves. The probability that the total is $5$ is $\frac 19,$
but the probability that the total is $8$ is $\frac{5}{36},$ which is greater.
No comments:
Post a Comment