# Combinatorics : Which side is heavier?

$\begingroup$ I ‘m not bang-up at statistics ( which is why I try answering these questions sometimes ), but here it goes and I ‘ll count on the community to tell me if I ‘m incorrectly .
Because I ‘m not great, I have to break these problems down into little quite elementary chunks .
first gear, I started with the probability that each slope is equal. I believe that that could be expressed as  \mathbb { P } ( \text { Both Sides Equal } ) = \mathbb { P } ( \text { 0 Heavy } ) + \mathbb { P } ( \text { 2 Heavy } ) \cdot \mathbb { P } ( \text { even Split } ) 

I decided that that ‘s the probability that you do n’t choose any heavy coins ( thus they must be equal ) or that you choose 2 heavies and one is on the left .
I started out by saying that : $\mathbb { P } ( \text { 0 Heavy } ) = \frac { \binom { n – 3 } { k } } { \binom { nitrogen } { k } }$
After trying to figure out $\mathbb { P } ( \text { 2 Heavy } )$ I think that I stumbled on this : $\mathbb { P } ( i\text { Heavy } ) = \frac { \binom { 3 } { iodine } \binom { n – 3 } { kelvin – i } } { \binom { north } { k } }$
I besides decided that most of the time I wanted to look at selecting both sides at once and then splitting them evenly so I started thinking of $k$ as $2k$ .
OK, so that covered $\mathbb { P } ( \text { 0 Heavy } )$ and $\mathbb { P } ( \text { 2 Heavy } )$, but I had to figure out the probability that given 2 heavies

Thanks to Gerry Myerson, I ‘m convinced this section was incorrect .

I ‘d randomly split them evenly between the sides of the scale. I decided I ‘d flip the heavy coins and if they were heads I ‘d put them on the leave and tails I ‘d put on the right. That ‘s metaphorically of course. But, basically, there are $2^2$ outcomes of flipping 2 coins and 2 of them result in an tied burst so $\mathbb { P } ( \text { evening Split } ) = 2 / 2^2 = 2 / 4 = .5$ .

My newfangled think is that there will always be alone 2 ways to have the 2 grave coins together : both on the impart or both on the right. There are $\binom { k } { \frac { k } { 2 } }$ ways to split the coins. frankincense, I think this is correct :  \mathbb { P } ( \text { even Split } ) = 1 – \frac { 2 } { \binom { k } { \frac { k } { 2 } } } 
If that ‘s decline, I think, the probability of both sides weighing the lapp is :  \mathbb { P } ( \text { Both Sides Equal } ) =\frac { \binom { north – 3 } { k } } { \binom { north } { k } } +\mathbb { P } ( \text { even Split } ) \cdot\frac { \binom { 3 } { 2 } \binom { normality – 3 } { k – 2 } } { \binom { north } { k } } 

OK. indeed, the conditions for the left side being heavier than the right side are the lapp as the conditions for the right side being heaver than the left so $\mathbb { P } ( \text { Left Heavier } ) = \mathbb { P } ( \text { Right Heavier } )$ .
thus, I only tried to calculate $\mathbb { P } ( \text { 1 Side Heavier } )$. I spent more time than I ‘d like to admit before I realized that $\mathbb { P } ( \text { 1 Side Heavier } ) = \mathbb { P } ( \neg\text { Both Sides Equal } )$ .
nervously, I submit this answer as I run some tests to make sure I ‘m at least on the right track .

source : https://ontopwiki.com
Category : Finance