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.

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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:

This number is really hard to figure out if you dont have a ridiculous scientific calculator, but basically that Q1 is the odds that n people do NOT have the same birthday. since the denominator is increasingly bigger, Q1 is heading towards zero as the number of people increases.  if you calculate 1-Q1 then you are calculating the odds of two people with the SAME birthday. Q1 gets smaller as n gets bigger, which means as more and more people are added your odds become exponentially higher that 2 will have the same day.

 

If you are really fucking confused right now and you have no idea whats going on, i think you should just download this little program i made and youll see what i mean. (you need to have microsoft access 2000 or better to view)

 

If you are ever at a lame party or ya know, you just need to impress a medium sized group of people, you can definitely lay odds on the possibility that two people in the group have the same birthday. So even though there are 365 days in a year,

 

just as an example:

30 people, the odds are 70.63% that two people have the same bday.

35 people, the odds are 81.43% that two people have the same bday.

40 people, the odds are 89.12% that two people have the same bday.

46 people, the odds are 94.82% that two people have the same bday.

 

i will take 10% collectors fee for any of you who find a really big whale who wants his whole bankroll taken.