Question

I have a big question to ask about probabilities and estimated value.
The problem is there are 4 cups. In each cup there is 4 jellybeans. One green,one red, one blue, one orange. You have to pick one jellybean out of each cup. If you get 2 of the same, you get 50 dollars. If you get 3, 100 dollars. If you get all 4, you get 150 dollars.

So can someone help me with finding the probabilities?

Answer 1

1/16, 1/64 and 1/256

Answer 2

No i mean in order to get het estimated value

Answer 3

of the whole thing i need all the probailities and they need to equal one

Answer 4

sorry it is 1/4, 1/16 and 1/64

Answer 5

I want to get hte estimated value, that requires all the probabilities.

Answer 6

I don't get it. I thought you wanted the probability of picking two beans of the same color, 3 beans of the same color and 4 beans of the same color

Answer 7

estimated value of what?

Answer 8

well here's teh thing, itsays i get points for getting 2 of the same, 3, 4, so on

Answer 9

right.

Answer 10

But its liek a game, and I have to price it, and to price it correctly i need to find the estimated value

Answer 11

Its like a carnvial game and im trying to find a good price for people to pay to play

Answer 12

YOUR QUESTION IS NOT DIRECT WILSON

Answer 13

Its not clear enough?

Answer 14

oh okay.

Answer 15

i have a long list of what i currently have, but they dont add up to one like they should

Answer 16

EMPHASIZE ON THIS QUESTION

Answer 17

That's why i came on here :P.
The probabilites to get all the same would be 1/256. Just multiply 4 times.
To get all three, 3/256 multiply by 4 times

Answer 18

see, the probability of picking two jelly beans of the same color is 1/4
the probability of picking 3 jellybeans of the same color is 1/16
and the probability of picking 4 jellybeans of the same color is 1/64
right?

Answer 19

Well lets focus on pickign 2 jelly beans of the same color.
It would be 1/4 x 1/4 x 3/4 x 1/2.

Correct?

Answer 20

how did you get that?
i think it is 1/4

Answer 21

There are 4 cans. Each has 4 jelly beans.

Green, Blue, red, Orange.

Its a 1/4 chance to get a green on the first, 1/4 chance to get green again on the second can, 3/4 to get non green jelly bean, 1/2 to make sure i get a different one frmo the one i get from the 3rd can.

Answer 22

the first time you pick a jelly, it doesnt matter what color you pick. so the probability of picking from the first can is 1. the next time you pick a jelly bean, it has to be of the same color.
so the probability of that is 1 times 1/4 = 1/4

Answer 23

the next time you pick a jelly bean, it shouldn't be of the first color. so the probability of that is 3/4
and the 4th time you pick a jelly bean, it shouldn't be of the 1st and 3rd color, so that probability is 1/2
so the probability of picking exactly two beans of the same color is 1/4 x 3/4 x 1/2

Answer 24

make sense?

Answer 25

actually it depends on what order you pick the beans too.

Answer 26

It does?!?!?!?!

Answer 27

So if i do it like you said, make it
1 x 1/4 x 3/4 x 1/2.

I wouldnt have to account for each color? Like thats my probability for twos?

Answer 28

Like i wouldnt have to do that 4 times for each color?

Answer 29

suppose you picked green on the first time. suppose you picked orange on the second try.
suppose you picked blue the next time. to make exactly two beans of the same color on the fourth try, you can pick either green or blue or orange.
so in this case the probability is 1 x 3/4 x 1/2 x 3/4

Answer 30

So if i imultipl ythat out i would get 1/32. Would i have to multiply it by 4 so that i would get the probability for each color?

Answer 31

hmmm.. this is a head scratcher. let me get back to you on this one.
meanwhile, try asking amistre64 or dumbcow or INewton for the answer.

Answer 32

i think you have to use permutations in your probability.
http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

Answer 33

Alright :P, i've been tryign to solve this problem the whole day.

Answer 34

Unfortunately i dont hink any of them are online

Answer 35

well, I guess you'll have to wait for them. Both dumbcow and amistre64 are online most of the time.

Answer 36

This is due tomorrow, am i screwed? :P

Answer 37

lol, maybe

Answer 38

I think polpak gave you the correct answer when you posted this question the last time.

Answer 39

No it wouldn't work, because to get the estimated value, i still need the probabilities to add to 1.

Answer 40

Here what if i made it 3 cups and 3 colors of jelly beans.

Answer 41

Blue, Green, and Red.
For a set of two blues would it be like, 1/3 x 1/3 x 2/3

Answer 42

no no no! you are not getting used to the idea that it doesn't matter which color you pick the first time.

Answer 43

Hmm, okay so then 1 x 1/3 x 2/3?

Answer 44

stil lthere?

Answer 45

okay. so the first time it doesn't matter which color you picked. you will pick one bean. so the probability of picking a bean is 1, since you are picking one bean out of the first can.
suppose in your second attempt, you did not pick the bean of the same color. the probability of that is 2/3. so now you have picked two beans of two different colors. now, to pick 2 beans of the same color in your third attempt, you need to pick a bean of the same color as the first attempt, or the same color as the second attempt. so the probability of that is 2/3
so in this case, your probability is 2/3 * 2/3
if you picked the same color as the first attempt on your second attempt, the probability of that is 1/3. the probability of NOT picking the same color as the first time on the third attempt is 2/3
so in this case the probability is 1/3 * 2/3. so, the order in which you pick the beans is important. Do you get it?

Answer 46

Omg, thats amazing. I didn't see that angle on the 2/3.
That makes sense

Answer 47

So because we start with the one, we do not have to replicate or multiply otu probabilities by 3 to show the different probabiltiies for the colors?

Answer 48

exactly!

Answer 49

If so, then 1 x 2/3 x 2/3
and
1 x 1/3 x 2/3
is all i neeed? o.o

Answer 50

it doesn't matter which color you pick. so you don't have to replicate it for the different colors

Answer 51

This seems to short to be true :0

Answer 52

yup, those are the two probabilities for picking 2 beans of the same color depending on the order.

Answer 53

but it is true. if it makes sense, then it is true.

Answer 54

Hmm, okay, how about the chance of all the beans being different colors.
That would be 1 x 2/3 x 1/3?

Answer 55

yup

Answer 56

okay, so tehse would be depended events right?

Answer 57

yes.

Answer 58

The formula would then be P (A and B and C) = P(A) x P(AlB) x P(AlC and B) ?

Answer 59

I think so, but I'm not sure. it's been a very long time since I did probability. Don't remember all the formulae. try googling it.

Answer 60

haha i did xD I think ima use this path for now, thanks soooo much!

Answer 61

Much appreciation for the help, sorry to bother :P

Answer 62

hey, you posted one of the best questions to hit this forum in a long time. thank you!