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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?
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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?
err the probability rather. odds is something different.
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?
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?
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?
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?
they are independent of each other the color of bean you get. The amount of money is not
Yes, but the probability will tell you what the expected earning is.
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?
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?
What do you mean by a 1/4 chance for the first cup?
The first cup doesn't matter. You can pick any color.
You have a 1/4 chance for the second cup to match the first.
and a 1/4 chance for the 3rd cup.
and a 1/4 chance for the 4th cup.
so that means to get 4 straight of the same color its a 1/64 chance?
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) ?
So in an estimated value write out... 1/4(50) + 1/16(100) + 1/64(150) ?Or must I do this for every color
Nope, that's it. It doesn't matter which color you get first.
: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?
And the nrepeat with the others?
No. Because you'll only pick 1 of those 4 colors.
And whichever one you pick you'll only have a 1/4 chance to get the same color on your next pick.
It seems too simple to be true :0
The expected earning is how much you will make on average per attempt.
Each attempt you're only dealing with one color (the one you pick in the first cup).
Mathematics is often easier than you think.
Hmm, so if im trying to price this game like how much people have to pay to play
I just do the estimated value out
and just price on that number? So in this case its 21.1, so i might price it at 25?
And add your profit margin.
Haha wow, in that case i think im good =D Thanks so much for your help! Ill post here if another problem arises :D
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Ohh mmkay no problem
Wait a minute, doesn't the probabilities have to add up to 1?
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.