## Wilson<3Pho 4 years ago 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?

1. Owlfred

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

2. polpak

Ok, so it doesn't matter which one you pick from the first cup, but for all other cups they have to match it. So whichever one you pick in the first cup, what are the odds that you'll get the same from the second?

3. polpak

err the probability rather. odds is something different.

4. animalsavior94

i was thinking since a cup has 4 jelly beans and 1 each of a color so to get like example red u have a 1/4 chance to get it out of a cup...but idk..lol if it makes sense?

5. Wilson<3Pho

So to calculate the amount of probabilities, I have to find the chances of getting 1 1 0 0 then 1010 then so on and so on?

6. polpak

No no no. You just have to find the probability for one cup. For example, if I pick a red bean first, what is the probability I'll get a red bean second?

7. Wilson<3Pho

but then i have to do it with each color :P so after wards calculating teh amount of getting 2s, i have to find 3s, 1110 1011 1101 and so on?

8. Wilson<3Pho

they are independent of each other the color of bean you get. The amount of money is not

9. polpak

Yes, but the probability will tell you what the expected earning is.

10. polpak

So start with the first case. No matter which color you pick from the first cup, what is the probability you will pick that same color in the second cup?

11. Wilson<3Pho

1/4 Chance to get it from the first cup, but to get from second cup would i use a P(A and B) formula?

12. polpak

Yes.

13. polpak

Well wait.

14. polpak

What do you mean by a 1/4 chance for the first cup?

15. polpak

The first cup doesn't matter. You can pick any color.

16. Wilson<3Pho

Oh.

17. polpak

You have a 1/4 chance for the second cup to match the first.

18. polpak

and a 1/4 chance for the 3rd cup.

19. polpak

and a 1/4 chance for the 4th cup.

20. polpak

So P(A and B and C)

21. Wilson<3Pho

so that means to get 4 straight of the same color its a 1/64 chance?

22. polpak

Right.

23. Wilson<3Pho

Mhm. so if i wanted to find three of the same color, it would be P(A and B) and to find two just P(A) ?

24. polpak

yep.

25. Wilson<3Pho

So in an estimated value write out... 1/4(50) + 1/16(100) + 1/64(150) ?Or must I do this for every color

26. polpak

Nope, that's it. It doesn't matter which color you get first.

27. Wilson<3Pho

:0 no way, wouldnt i have to do 1/4(50) + 1/4(50) + 1/4(50) + 1/4(50) to represent the 4 different colors i might get?

28. Wilson<3Pho

And the nrepeat with the others?

29. polpak

No. Because you'll only pick 1 of those 4 colors.

30. polpak

And whichever one you pick you'll only have a 1/4 chance to get the same color on your next pick.

31. Wilson<3Pho

It seems too simple to be true :0

32. polpak

The expected earning is how much you will make on average per attempt.

33. polpak

Each attempt you're only dealing with one color (the one you pick in the first cup).

34. polpak

Mathematics is often easier than you think.

35. Wilson<3Pho

Hmm, so if im trying to price this game like how much people have to pay to play

36. Wilson<3Pho

I just do the estimated value out

37. Wilson<3Pho

and just price on that number? So in this case its 21.1, so i might price it at 25?

38. polpak

39. polpak

yeah.

40. Wilson<3Pho

Haha wow, in that case i think im good =D Thanks so much for your help! Ill post here if another problem arises :D

41. Wilson<3Pho

Will fan you! :D

42. polpak

be sure to click good answer to mark this question as solved.

43. Wilson<3Pho

Ohh mmkay no problem

44. Wilson<3Pho

Wait a minute, doesn't the probabilities have to add up to 1?

45. Wilson<3Pho

!!!

46. Wilson<3Pho

!

47. polpak

No they don't. The probability that you will pick a red, plus the probability to pick a green, plus the probability you will pick a blue, plus the probability you will pick an orange must add to one. But you don't care about those. You just want to know the probability you will pick the same color you picked from the first cup. If it added to 1 that would mean that you would always get the same color.