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Mathematics 28 Online
OpenStudy (anonymous):

Probability dice help

OpenStudy (anonymous):

OpenStudy (adamaero):

Do you just add them up and pick the largest? or is there something more to do?

OpenStudy (anonymous):

I think that each individual die has different numbers and let's say three people have one die each. Person A has die A person B has die B and person C has die C. We roll the dice and whoever has the highest number wins

OpenStudy (anonymous):

Die A has numbers 3,3,5,5,7,7 Die B : 2,2,4,4,9,9 Die C: 1,1,6 ,6,8,8

OpenStudy (anonymous):

I've been making a number of every possible probability that the numbers will roll. It it's too confusing for me

OpenStudy (adamaero):

@Data_LG2 Just adding up numbers and choosing the largest quantity?

OpenStudy (anonymous):

I'm not sure about this one. What do you mean by adding up the numbers? **I'm still trying to figure out how to do this

OpenStudy (anonymous):

I don't think you have to add any numbers. It's more like um...if you roll the three dice, which dice will have the highest number based on the difference set of numbered faces

OpenStudy (anonymous):

@esmer can you take the whole screenshot of the sheet? because I don't think it's right to assume that there will be three persons that will be involved here unless additional information is provided.

OpenStudy (anonymous):

Like for example for one roll one possibility could be 3 for die A 2 for B and 1 for C which makes die A with the highest number therefore die A wins

OpenStudy (anonymous):

oh nvm...

OpenStudy (anonymous):

The bottom part is just for me to write a journal on which die is best

OpenStudy (anonymous):

I'm thinking die B would be the best one

OpenStudy (anonymous):

For Die A to win this would be some possible rolls 3=A 2=B 1=C 5,2,1 5,4,1 7,2,1 7,4,1 7,4,6 7,2,6 For Die B to win 4=B 3=A 1=C 9,3,1 9,5,1 9,3,6 9,3,8 9,5,6 9,5,8 9,7,6 9,7,8 9,7,1 For Die C to win 6=C 3=A 2=B 6,3,4 8,3,2 8,3,4 8,5,2 8,5,4

OpenStudy (anonymous):

I think you have to compare the values of P(A>B), P(B>C), P(C>A) i'm still trying to find out how to do this so this may take a while.. i guess creating a tree diagram helps

OpenStudy (anonymous):

|dw:1423437169610:dw| do the same thing for P(B>C) and P(C>A) and let's see what happens.. i'll try to do it

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