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

Read more: How to Organize a Coin Collection?

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 .

## Leave a Comment