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Wilson<3Pho

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?

  • 2 years ago
  • 2 years ago

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  1. Owlfred
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    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 years ago
  2. polpak
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    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?

    • 2 years ago
  3. polpak
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    err the probability rather. odds is something different.

    • 2 years ago
  4. animalsavior94
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    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?

    • 2 years ago
  5. Wilson<3Pho
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    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?

    • 2 years ago
  6. polpak
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    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?

    • 2 years ago
  7. Wilson<3Pho
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    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?

    • 2 years ago
  8. Wilson<3Pho
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    they are independent of each other the color of bean you get. The amount of money is not

    • 2 years ago
  9. polpak
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    Yes, but the probability will tell you what the expected earning is.

    • 2 years ago
  10. polpak
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    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?

    • 2 years ago
  11. Wilson<3Pho
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    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?

    • 2 years ago
  12. polpak
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    Yes.

    • 2 years ago
  13. polpak
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    Well wait.

    • 2 years ago
  14. polpak
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    What do you mean by a 1/4 chance for the first cup?

    • 2 years ago
  15. polpak
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    The first cup doesn't matter. You can pick any color.

    • 2 years ago
  16. Wilson<3Pho
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    Oh.

    • 2 years ago
  17. polpak
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    You have a 1/4 chance for the second cup to match the first.

    • 2 years ago
  18. polpak
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    and a 1/4 chance for the 3rd cup.

    • 2 years ago
  19. polpak
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    and a 1/4 chance for the 4th cup.

    • 2 years ago
  20. polpak
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    So P(A and B and C)

    • 2 years ago
  21. Wilson<3Pho
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    so that means to get 4 straight of the same color its a 1/64 chance?

    • 2 years ago
  22. polpak
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    Right.

    • 2 years ago
  23. Wilson<3Pho
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    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) ?

    • 2 years ago
  24. polpak
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    yep.

    • 2 years ago
  25. Wilson<3Pho
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    So in an estimated value write out... 1/4(50) + 1/16(100) + 1/64(150) ?Or must I do this for every color

    • 2 years ago
  26. polpak
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    Nope, that's it. It doesn't matter which color you get first.

    • 2 years ago
  27. Wilson<3Pho
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    :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?

    • 2 years ago
  28. Wilson<3Pho
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    And the nrepeat with the others?

    • 2 years ago
  29. polpak
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    No. Because you'll only pick 1 of those 4 colors.

    • 2 years ago
  30. polpak
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    And whichever one you pick you'll only have a 1/4 chance to get the same color on your next pick.

    • 2 years ago
  31. Wilson<3Pho
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    It seems too simple to be true :0

    • 2 years ago
  32. polpak
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    The expected earning is how much you will make on average per attempt.

    • 2 years ago
  33. polpak
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    Each attempt you're only dealing with one color (the one you pick in the first cup).

    • 2 years ago
  34. polpak
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    Mathematics is often easier than you think.

    • 2 years ago
  35. Wilson<3Pho
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    Hmm, so if im trying to price this game like how much people have to pay to play

    • 2 years ago
  36. Wilson<3Pho
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    I just do the estimated value out

    • 2 years ago
  37. Wilson<3Pho
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    and just price on that number? So in this case its 21.1, so i might price it at 25?

    • 2 years ago
  38. polpak
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    And add your profit margin.

    • 2 years ago
  39. polpak
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    yeah.

    • 2 years ago
  40. Wilson<3Pho
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    Haha wow, in that case i think im good =D Thanks so much for your help! Ill post here if another problem arises :D

    • 2 years ago
  41. Wilson<3Pho
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    Will fan you! :D

    • 2 years ago
  42. polpak
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    be sure to click good answer to mark this question as solved.

    • 2 years ago
  43. Wilson<3Pho
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    Ohh mmkay no problem

    • 2 years ago
  44. Wilson<3Pho
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    Wait a minute, doesn't the probabilities have to add up to 1?

    • 2 years ago
  45. Wilson<3Pho
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    !!!

    • 2 years ago
  46. Wilson<3Pho
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    !

    • 2 years ago
  47. polpak
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    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.

    • 2 years ago
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