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- Statistics depends on the law of large numbers. Given enough opportunities for specific events to happen, despite how unlikely to be for each opportunity, it will eventually happen.
-. · Another way to put it: the number of combinations of interacting elements increases exponentially with the number of elements.
-. A “combination” of “n” different taken “r” at a time with no attention to order increases exponentially. The formula uses “!” which means “factorial”, which is 5! = 5*4*3*2*1 = 120. “Factorial” is multiplication in a descending sequence of numbers down to the number 1.
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-------------------------------- Combination = n! / r! (n - r)!
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-. To illustrate this formula let’s use a third-grade class of 30 students. If the teacher chooses to have them work individually than the number of combinations is 30.
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------------------------ Combination = 30! / 1! (30 - 1)! = 30*29 / 29 = 30
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-. But if the teacher asked the class to work in pairs how many commendations are there?
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------------------------- Combinations = 30! / 2! (30 - 2)! = 30*29 / 2 = 435
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-. There are 435 different pairs that could work together.
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-. How about if the teacher asked students to work in groups of 3 that would be 10 groups. How many different combinations are there?
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------------------------- Combinations = 30! / 3! (30 - 3)! = 30*29* 28 / 6 = 4040
-.. There are 4060 combinations of these student groups.
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- The number of combinations is growing exponentially from 30 ,to 435, to 4060, to 27,405 to………………….
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-. What would happen if the teacher said to the class okay you can work together in any combinations you want?
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-. Number of possible groups of students working together in a class of 30 students is a set of “n” elements with (2^n - 1) possible subsets that can be formed
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------------------- Combinations = 1,073,74,823
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----------------- Combinations = 2^30 - 1 = 1,073,74,824 -1
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-. Lastly let's put the class of students on the Internet. Now the class is growing to 2,500,000,000, the number of www.users.
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------------------------ 3 * 10^18 pairs = 10^750,000,000 possible pairs
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-. Probability that any and all of these can interact with any of the others is 10 followed by 750,000,000 zeros .
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-. Another example of how combinations can work is the famous problem of “how many people need to be in a room before there is a probability that two have the same back birthday“.
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-. The answer is 23 people in a room and the probability is better than 50 % that two have the same birthday. But how do we get to that number?
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- A single birthday has 1 chance in 365 days of occurring that 0.27 % So, any one person has one chance in 365 of having my same birthday.
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-. And, to take the reverse, there 364 / 365, or 99.7 % probability that any particular person will have a different birthday.
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-. If “n” is the number of people in the room then (n-1) is probability of 364 / 365 of having a different birthday.
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- If you combine these probabilities of being different it becomes 364/365 * 364/365 * 364/365 * 364/365 ………………….. (364/365)^22 = 0.94
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-. When “n” is 23 there is 94 % probability that none of them will share the same birthday as you.
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-. The probability that at least one of them has the same birthday as you is ( 1 - 0.94). or 0.06. A 6% probability that one of the 23 people shares your birthday.
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-. Six percent is a very small number, If there are 23 people in a room the likelihood is high that no one has the same birthday as me but they may have the same birthday matching someone else.
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-. Either someone has a same birthday as me or no one has me. These two likelihoods add up to one, 0.94 + .06 = 1
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-. Now if the question is different that any two people have the same birthday as each other. Not only is (n -1) people sharing the same birthday but (n-1)*2 / 2 pairs of people in a room share the same birthday. When n = 23:
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-------------------------------- (n-1) * n /2 = 23 * 22 / 2 = 253
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------------------------------ 23! / 2! * ( 21!) = 253
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-. That is more than 10 times larger than (n -1) which is 22
-. There are 253 possible pairs of people but only 22 pairs include me.
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-. Will that get us the answer that the odds of two people having the same birthday begins to occur when 23 people are in the room.
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-. Let’s look at the contrary way what that none of the 23 people have the same birthday. We've already determined that for two people the answer is 364/365 which is by 99.7 %.
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-. Now at a third person in the room and the probability becomes a 99.4 %.
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-. At a fourth person and it is 94.1%.
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-. When you get to 23 people the probability is 49 % that don't have the same birthday and - - 0.49 = 51% probability that two people will have the same birthday
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- The law of combinations grows at an exponential rate. The probability with 10 people is 12%. The probability with 20 people is 40%. The probability with 23 people is 50.7%. the probability with 100 people is 99.99996 %
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(1) The probability that all the birthdays are
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----------------------different = (366-n ) / 365^n
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--------------------- same = 1 - 365! / 365^n ( 365-n)!
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- Each person in the room has a 365 chance to match a birthday. For 23 people the number of chances is 365^n = 365^23.
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RSVP, with comments, suggestions, corrections. Index of reviews available ---
--- Some reviews are at: -------------------- http://jdetrick.blogspot.com -----
---- email request for copies to: ------- jamesdetrick@comcast.net ---------
---- https://plus.google.com/u/0/ , “Jim Detrick” ----- www.facebook.com ---
---- www.twitter.com , --- 707-536-3272 ---- Monday, February 17, 2014 ---
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