### Math, Byatches

So for the record, my knowledge base extends slightly beyond cartoon trivia, bong resin, and fashion commentary...it doesnt go far, but i can hold my own in the math knowledge arena as well. So with that in mind, I figure I may as well write about two cool math related things that could be good for conversation about poker, cards, general proposition betting, etc.

**Shuffling cards**

Lets say you take a deck of cards. If you could perfectly shuffle the cards by splitting them exactly in the middle and then shuffle in right, left, right, left, etc.... all the way through, there are two types of shuffles. An in shuffle is where you start shuffling left to right. An out shuffle is where you start shuffling right to left.

In Shuffle: 1 2 3 4 5 6 7 8 ==> top to bottom

1 2 3 4 5 6 7 8 ==> split into L and R

5 1 6 2 7 3 8 4 (top to bottom)

Out Shuffle: 1 2 3 4 5 6 7 8 ==> top to bottom

1 2 3 4 5 6 7 8 ==> split into L and R

1 5 2 6 3 7 4 8 (top to bottom)

The trick or trivia or whatever you wanna do to impress people is to ask: if you could perfectly shuffle a deck of cards by in-shuffling and shuffle another deck by out-shuffling, how many times does it take to get the deck back to its original order? The crazy part is it takes 52 perfect in-shuffles to return it to its original order. It only takes 8 out shuffles.

A followup to this shuffling issue is, lets say youre in a home game and some jackass is shuffling the cards over and over and over. its like he thinks that by shuffling them a ridiculous amount of times it will spread the high cards out. It has been proven that 3/2 log2

*n*(n = # of cards in the deck) shuffles is sufficient, so for a deck of 52 thats about 8 or 9 shuffles.

**Birthday Proposition**

This math problem is basically to try to figure out in a room full of people, what the odds are that two people in the room have exactly the same birthday. When you ask someone to try and think about it, usually they will come up with 1/365 * 1/365 is the odds two people have the same birthday. Unfortunately, this is one of those crazy math problems.

Instead of going at it from same day, calculate the odds that two people will NOT have the same birthday. so in a group of

*n*people with

*d*being the number of possible days (365), if n=2, the odds that they will not have the same birthday is (d-1)/d. the odds that three people will not is (d-1)/d * (d-2)/d. with n people, it works out from:

or simply put:

*exponentially*higher that 2 will have the same day.

## 1 Comments:

oh, internet boyfriend! How you are melting my loins with your mind!

i heart you.

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