The theory of probability

  Introduction
  Foundation
  Calculating
  Presize definition
  Binomial distribution
  Hypergeometric distribution
  Combined events
  Inseparable events
  'Ion Saliu's Paradox'
  Combinatorics
  Markov Chains
  Random Walks
  Contacts

The calculations are based on what I call exponential sets. The Birthday Paradox is a particular case of exponential sets. Such sets consist of unique elements and also duplicates. The unique part of an exponent is equal to the arrangements type of the set. The part with at least two elements equal to one another is the difference between exponents and arrangements. In the probability problem of four dice, the Birthday Paradox parameters are: lower bound = 1, upper bound = 6, total elements (number of dice) = 4. The probability to get at least two dice showing the same point face when throwing four dice is: 0.7222 or 1 in 1.385. Easy to verify. Throw four dice. In almost three out of four rolls, at least two dice show the same face.

Also, the probability to get the four dice show the same point face is precisely calculated by using the exponential sets. A die has six faces — always! To get exactly 1-1-1-1 = 1/1296; the probability to get exactly 6-6-6-6 = 1/1296; the probability to get exactly 1-2-3-4 = 1/1296.

The pick 3, 4 lottery games should be considered forms of dice rolling — therefore inseparable phenomena. A drawing machine is a 10-face die. The pick 3 game is like casting three 10-faceted dice. The slot machines, by extension, are the equivalent of casting multi-faceted dice (usually three dice).

~ Another probability problem that pops up in forums and newsgroups and emails. A jar contains 7 red balls, 6 black ball, 5 green balls, and 3 white balls. We can construct a huge variety of probability problems with the 21 balls. For example, the probability to draw exactly 5 balls with this exact composition: 2 red, 2 black, 1 white. We must apply the hypergeometric distribution of each color.
- exactly 2 red of 5 drawn in 7 red from a total of 21 balls: 0.375645 (1 in 2.662)
- exactly 2 black of 5 drawn in 6 red from a total of 21 balls: 0.335397 (1 in 2.982)
- exactly 1 white of 5 drawn in 3 white from a total of 21 balls: 0.4511278 (1 in 2.217)
The combined probability is the product of the three: 0.056838 or 1 in 17.6.

How about the probability to draw 5 balls and get at least one ball of each color? Applying now the 'W' option of SuperFormula.EXE ('Win at least Lotto & Powerball'):
- at least 1 red of 5 drawn in 7 red from a total of 21 balls: 0.9016 (1 in 1.109)
- at least 1 black of 5 drawn in 6 red from a total of 21 balls: 0.8524 (1 in 1.173)
- at least 1 green of 5 drawn in 5 green from a total of 21 balls: 0.7853 (1 in 1.273)
- at least 1 white of 5 drawn in 3 white from a total of 21 balls: 0.5789 (1 in 1.727)
The combined probability is the product of the four: 0.3494 or 1 in 2.86.

The Birthday Problem

Converse probabilities are used to work out the infamous birthday problem.  Many people find the answer puzzling but it can be proved by either asking your personal manger for birthday dates or flicking through a the who’s who in your reference library.

The question is:-

"How many people should be gathered in a room together before it is more likely than not that two of them share the same birthday?"

Ignoring the issues of leap years the problem is solved as follows:-

When the first person enters the room and announces their birthday, the probability of the second person sharing the same birthday is 1/365. Conversely, the probability of the second birthday being different is the opposite of the first calculation, 364/365.  When two birthdays are known, the probability of the third being different is 363/365, as there are now two 'favourable' outcomes among 365.  The compound probability of birthday 2 being different from birthday 1, and of birthday 3 being different from the other two, these being independent outcomes, is:-

(364/365)*(363/365) = 0.991796 or 99.2% chance that two people will not share the same birthday.

Note the start of the sequence is (365/365). We have removed this as it does not affect the result of the calculation.

All that is necessary now is to continue adding terms to the fraction until it equals less than 1/2 or 50%, since as soon as the probability is less than 1/2 that all birthdays are different, the probability is clearly more than 1/2 that any two are the same. In other words it is more likely than not that two people in the room share the same birthday. The following chart shows the number of the people in the room and the probability that they DO NOT share the same birthday.
  

The fraction drops to less than 1/2 with 23 iterations, so it is more likely than not that in any gathering of 23 or more persons, two of them will share a birthday.

Only 50 people need be present for the 'coincidence' of two of them having the same birthday to become, roughly, a 30-1 on chance.

In a company of 100 employees the odds are more than three million to one on that two share a birthday.

The birthdays proposition is one where a gambler who can estimate probabilities can make money from unsuspecting punters.