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---------------------------- 2222 - The power of combinations?
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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. If it can happen it will happen!
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- Another way to put it: the number of combinations of interacting elements increases exponentially with the number of elements.
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- 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 combinations 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 = 4060
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- 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 exact 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 the same birthday as 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 is the likelihood 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 add 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 :
(1 - 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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- The probability that all the birthdays are the same:
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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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- Pass it on to anyone who wants to learn the power of combinations. The power of teamwork, networking, and the world-wide-web are demonstrated in this math.
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- December 31, 2018. 1650
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----- Comments appreciated and Pass it on to whomever is interested. ----
--- Some reviews are at: -------------- http://jdetrick.blogspot.com -----
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--------------------- Monday, December 31, 2018 -------------------------
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