Outcome probability in Laplace experiments

Finding the exact probability of an outcome in a random experiment is difficult and involves lot's of repetitions and counting. However, if we make one further assumption about the experiment, we can determine the outcome probabilities on theoretical grounds - no repetition is required.

This assumption is that no outcome is preferred, or in probability speech, that all outcomes have the same probability to occur. Clearly, this is quite a restrictive assumption and is not true for most experiments. Still, for a few popular experiments this is quite a reasonable assumption to make, as for example

A fair coin. The word fair implies that there is no preference for a side to occur, that is, after many repetitions we have $p(H)=p(T)$ .
A fair die. Again, it is implied that after many repetitions each side of the die occurs with the same frequency, that is, $p(1)=p(2)=...=p(6)$ .
Selecting a ball at random from of box of balls. If we assume that each ball has the same chance to be selected, this is a Laplace Experiment. This is, for example, the case, if all balls have the same size, and selection is made blindly. If there are $m$ balls in the box, we have $p(1)=p(2)=... = p(m)$ .

Here is the exact definition, and an important implication: A random experiment where all $m$ possible outcomes $o_1, o_2, ..., o_m$ have the same probability to occur,

p=p(o_1)=p(o_2)=...=p(o_m)

is called a Laplace experiment.

Theorem 1

The probability for an outcome in a Laplace experiment is

p=\frac{1}{m}=\frac{1}{\vert S\vert}

Exercise 1

Prove the above assertion that in a Laplace experiment with $m$ outcomes, the outcome probability is $p=\frac{1}{m}$ .

Solution

With $p(o_1)+p(o_2)+...+p(o_m)=1$ follows $m\cdot p=1$ , and thus $p=\frac{1}{m}$ .

Exercise 2

Argue if the random experiment below are Laplace experiments. Always calculate the probability of the indicated outcome.

You roll a fair die once, p(6)=?
You roll a fair die twice, p(66)=?
You flip a coin $3$ times, p(HHH)=?
You select at random a letter from the word MARK, p('A')=?
You select at random a letter from the word HALLO, p('L')=?

Solution

Yes, $S=\{1,2,3,4,5,6\}$ , thus $m=6$ and $p=1/6$
Yes, because a die has no memory - if a 6 was thrown in the first roll, this has no effect on the second roll. $S=\{11,12,13,...,64,65,66\}$ , thus $m=36$ and $p=1/36$
Yes, because a coin has no memory. $S=\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ , thus $m=8$ and $p=1/8$ .
Yes, because random selection implies that each position will in the long run be selected with the same percentage. So each letter will be selected with the same percentage. Thus $S=\{M,A, R, K\}$ and $p=1/4$ .
No. Although each position is selected with the same percentage in the long-run, the $L$ will be selected more often than the other letters (as there are two $L$ 's in the word). Actually, we can still calculate the probability for selecting an $L$ : if each of the $5$ positions is selected $n$ times out of the $N$ repetitions, we have

N=n+n+n+n+n=5n

Because there are two letters L, it is

p(L)=\frac{2n}{N}=\frac{2n}{5n}=\frac{2}{5}=0.4

Exercise 3

A biased coin with $p(H)=0.7$ is flipped twice. Is this a Laplace experiment? Argue.

Solution

No, if the experiment is repeated many times, the percentage of observed $HH$ is much higher than the percentage of observing $TT$ (because $H$ occurs more often than $T$ ). So not all outcome have the same probability to occur, so not a Laplace experiment.