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What are these strange terms, and what do they mean? Like other topics in mathematics, most people understand and work with these ideas at some time in their life, but they may not understand the common lingo of mathematicians.
Both terms work with the idea of creating the total number of different lists. First think of the total number of outcomes after rolling 2 die. Let’s say that the first die rolled gives a 1. How many different outcomes exist given that the first die shows a 1? There are 6 different numbers on the second die, so the answer is 6. Now fix the first die to a 2, and repeat the process. Once again, you get 6 more outcomes to add to the first 6 outcomes. The total number of outcomes from rolling 2 die? 6x6 = 36. You simply take the total number of outcomes for each die and then multiply them (thus the name “multiplication rule”). Pretty simple.
Now think of a baseball team. There are 9 hitters on the team (not including the Yankee’s pitchers in the dugout who start fights. That’s a different type of hitter). How many ways can you arrange those hitters in the line up. For the 1st position, you can choose from 9 players. For the 2nd position you can choose 8 players, etc., etc. Continue this process until you reach the 9th batting position, for which you can only choose 1 person. What’s total number of ways that you can arrange 9 hitters in a batting order? Following the example with the die we can use the multiplication rule which gives
9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880.
This multiplication is known as factorial, and the shorthand for it is
9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.
If we shorten the number of batters to 5, then we follow the same ideas given above, except that we only list the first 5 numbers in sequence. With 9 total hitters and only 5 batters we have 9 x 8 x 7 x 6 x 5 = 15,120. This sequence is called a permutation of 9 objects choosing 5. It is written as 9P5.
Combinations are best described at a later date in another post, so I’ll just focus on the problem at hand: the “Birthday” Problem.
Here is the proposed question: how many people must be in a room until it is likely (greater than 50%) that two or more people share a birthday? The initial assumption might be 183, as 183/365 is greater than 50%. However, the required number is much lower than that.
To solve this problem it is easier to consider how many people will not share a birthday. To start off, if there is one person in the room then—duh!—that person won’t share a birthday with anybody else. Simple enough. If there are two people in the room, then the second person must have a birthday that is not the same as the 1st persons birthday; hence, that person’s birthday must be on one of the 364 remaining days. For the 3rd person, his birthday must be on one the remaining 363 days of the year. The pattern is similar to the baseball lineup mentioned before. When there are 5 people in the room, if nobody shares a birthday, then the number of different arrangements of birthdays is 365 x 364 x 363 x 362 x 361 = a really big number.
But how many total combinations of birthdays exist? To answer that question we look to the die rolling example. The 1st person’s b-day can be on one of 365 days. The 2nd persons b-day can be on one of any 365 days. Etc. For 5 people, the total permutations of birthdays is 365 x 365 x365 x 365x 365 = another big number.
A probability is given by taking the total subset of events (nobody shares a birthday) and dividing it by the total number of possible events (total outcomes). Given that there are 5 people in the room, the probability that 5 people do not share a birthday is
6.3 trillion
------------- = .97
6.47 trillion
and the probability that at least two people do share a b-day is 1 - .97 = .03, or 3%. This probability is pretty small, but as there are more people in the room the probability gets larger very quickly. Look at the following chart. Note that the horizontal axis only contains 60 days. As the number of days approaches 60, the probability of at least 2 people sharing a birthday is very close to 1. The initial guess of 183 was way off. If 183 people are in a room, then it is almost certain that at least two people share the same birthday. For our problem of when it will be more likely, the answer is 22, which is where the two dotted lines cross.
It's the sort solution that a person might at first find incredulous, but there is the answer. I hope you enjoyed this little aside into the world of probability. I'll hopefully include more interesting topics from that world at future dates.

| The Prioress You scored 5% Cardinal, 74% Monk, 58% Lady, and 35% Knight! |
You are a moral person and are also highly intellectual. You like your solitude but are also kind and helpful to those around you. Guided by a belief in the goodness of mankind you will likely be christened a saint after your life is over. |
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| Link: The Who Would You Be in 1400 AD Test |