Dice probability question. Need to know if I am right / wrong

blin

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Suppose that we have 2 dice, one Red and one Black. The Red die is defective, the probability to get 6 is 1/3. Now if we throw the 2 dice, the die with the biggest value wins and if we get equal values the Black one wins. Which one of the dice has the biggest probability to win on throwing them both at the same time?

My solution:

P(red die for not rolling six) = 1 - 1/3 = 2/3Since there are 5 other possible values we have P(red die for each value) = 2/3 * 1/5 = 2/15
Now for the Red Die to win we have 15 possibilities: (1,2) (1,3) (1,4) (1,5) (1,6*) (2,3) (2,4) (2,5) (2,6*) (3,4) (3,5) (3,6*) (4,5) (4,6*) (5,6*) => Of which 5 have 6 in it. While for the Black one 21 of which 1 have 6 in it.

P(red to win) = 10 * (1/6 * 2/15) + 5 * (1/6 * 1/3) = .5
P(black to win) = 20 * (1/6 * 2/15) + 1 * (1/6 * 1/3) = .5

Both have equal possibilities to win, however i am not really sure about this.

Please help!
 
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Suppose that we have 2 dice, one Red and one Black. The Red die is defective, the probability to get 6 is 1/3. Now if we throw the 2 dice, the die with the biggest value wins and if we get equal values the Black one wins. Which one of the dice has the biggest probability to win on throwing them both at the same time?

My solution:

P(red die for not rolling six) = 1 - 1/3 = 2/3Since there are 5 other possible values we have P(red die for each value) = 2/3 * 1/5 = 2/15
Now for the Red Die to win we have 15 possibilities: (1,2) (1,3) (1,4) (1,5) (1,6*) (2,3) (2,4) (2,5) (2,6*) (3,4) (3,5) (3,6*) (4,5) (4,6*) (5,6*) => Of which 5 have 6 in it. While for the Black one 21 of which 1 have 6 in it.

P(red to win) = 10 * (1/6 * 2/15) + 5 * (1/6 * 1/3) = .5
P(black to win) = 20 * (1/6 * 2/15) + 1 * (1/6 * 1/3) = .5

Both have equal possibilities to win, however i am not really sure about this.

Please help!

I would say that's a pretty good analysis and you should have more faith in yourself.
 
I would say that's a pretty good analysis and you should have more faith in yourself.

Thank you Ishuda, it's just that i had the teacher answer on my notes and it says the Black die is favorite to win.
 
Suppose that we have 2 dice, one Red and one Black. The Red die is defective, the probability to get 6 is 1/3. Now if we throw the 2 dice, the die with the biggest value wins and if we get equal values the Black one wins. Which one of the dice has the biggest probability to win on throwing them both at the same time?
Here is a table: \(\displaystyle \begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{(1,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2,4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3,4)}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4)}&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}&{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&{(6,5)}&{(6,6)}\end{array}\)
Lets say the first of each pair is red and the second is black.
 
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Thank you Ishuda, it's just that i had the teacher answer on my notes and it says the Black die is favorite to win.

What is the complete problem then? Is the probability of each pip other than six on the one die being throw the same, i.e. 2/15? Is the other die a fair die, i.e. the probability of getting any one particular pip equal to 1/6?

If that is the case, did the teacher tell you where you were incorrect and, if so, where was it?

Unless something changes from the problem as presented (odds, who wins what, etc.), I am very inclined to believe your teacher is incorrect.
 
What is the complete problem then? Is the probability of each pip other than six on the one die being throw the same, i.e. 2/15? Is the other die a fair die, i.e. the probability of getting any one particular pip equal to 1/6?

If that is the case, did the teacher tell you where you were incorrect and, if so, where was it?

Unless something changes from the problem as presented (odds, who wins what, etc.), I am very inclined to believe your teacher is incorrect.

The problem is just as stated above. The other die is a regular one. I don't know, i got some notes from a friend and it has just the answer without calculating. It might be she is incorrect. Thanks for your help! :)
 
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