http://saliu.com/theory-of-probability.html#Trials

Quote:

8. The Probability (Odds) of Inseparable Events, Single Trial
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.



Idk, that seems wrong to me for some reason. Anyone else have a take on it? It just looks a bit too simple. Then again, no one said it had to be hard right? lol

Last edited by PrenceOfDarkness; 02/15/12 17:23.

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