GingerAle

You seem to have the right idea; you just need to multiply each combination by the available number of suits.  As each suit is selected, there is one less suit to consider. 

\(\binom{13}{6} \cdot (4) \cdot \binom{13}{5} \cdot (3)\ \cdot \binom{13}{2} \cdot (2) \\\\ \text { } \\ = 4,134,297,024 \text { Combinations}\)

GA

Rusty armor, dull sword, and Bad manners. 

He probably doesn’t even know the multiplication tables. 

Caution!!!  This is an impostor!   

Not a very good one, either!

Get recttttt

OH! You really are a child! Sorry, I thought you were a teen acting like one.

You spelled Piperson’s name wrong.

I noticed, but I was too lazy to give enough of a dámn to correct it. The disease has a fast incubation rate.

GA

Hi Miss Rosala,

Of course, we want your post(s). Sometimes your audience may be small, but we are loyal as puppies (and chimps).

 P.S.  Re-visit your birthday post. More have added their greetings. 

GA

Right Piperton! Melody is going to solve a million-dollar math problem for someone too lazy or stupid to present it properly.  What you have is worse than CDD.  You have Contagious Don’t Give a Dámn Disease (CDGDD). 

This is the “Beal Conjecture”

IF

\(A^{x}+B^{y}=C^{z}, \)

\(\text { where A, B, C, x, y, and z are positive integers with x, y, z > 2,}\\ \text { then A, B, and C have a common prime factor. } \)

GA

HI Miss Melody

Happy Birthday Miss Rosala!

I thought this was your 16th birthday, too.

This is why I thought so:

Here’s my logic.

This is a weighted average of probability.  The weights are proportional to the probability of each event.

The first event is the probability of either a black ball or a white ball.  Four of the five bálls are black and one is white. The black ball has a (4/5) 0.80 probability and the white ball has a (1/5) 0.20 probability of being drawn

IF a black ball (0.80 probability) is drawn, then bin “b” probabilities are in play.

The weights are calculated the same as above. There are four b***s – three are $1 bálls and one is a $7 ball. The probability of drawing a $1 ball is (3/4) 0.75 and the probability of drawing a $7 is (1/4) 0.25

0.75 * $1 = 0.75

0.25 * 7$ = 1.75

Total       =  2.50 average per event

$2.50* 0.80 (Black ball Probability) = $2.00

The logic for the white ball event is the same.

---------------------

Ugh!  CDD strikes again!!

I’m not posting it because I see from Sir Alan’s post that my numbers are wrong. I was using one $500-ball and eight $8-bálls. (I think I added 1 to the 8 instead of the five).

 I added Sir Alan’s solution to Naus’ Tortoise and Hair AI formula for blonds, so it now works for red-heads.

-----------------------

Here's the fully repaired answer:

Black Ball probability 0.80

                                $1 Ball drawn probability 0.75    Expected Value = 0.80 * 0.75 * 1 =  $0.60

                                $7 Ball drawn probability 0.25    Expected Value = 0.80 *0.25 * 7 =   $1.40

                                                                                                                          Sum         $2.00

White Ball probability 0.20

                           $8 Ball drawn probability (5/6) 0.8333 Expected Value = 0.20 * 0.8333* 8 = $1.33 

                       $500 Ball drawn probability (1/6) 0.1667 Expected Value = 0.20 * 0.1667 * 500 = 16.67

                                                                                                                                    Sum         $18.00

18.00+2.00 = $20.00 Expected Value