If I choose four cards from a standard -card deck, with replacement, what is the probability that I will end up with one card from each suit?
Four people sit around a circular table, and each person will roll a standard six-sided die. What is the probability that no two people sitting next to each other will roll the same number after they each roll the die once? Express your answer as a common fraction.
There are 6 possibilities for the first person, 6 for the second, 6 for the third, and 6 for the fourth.
6^4 = 1296 total die roll possibilities.
Out of these 1296,
there are 6 possibilities for the first person but only 5 for the second. Again 5 for the third, and 4 for the fourth.
6*5*5*4 = 600
600/1296 = 75/162
You are very welcome!
Circular tables always complicate things.
When maths questions talk about round tables they assume no external points of reference. In other words rotations are the same thing.
6^4 would certainly be the number of possible combinations if the people were in a nice straight row but it is too high for a round table.
1234 is the same as 2341, you cannot count that combination twice....
1234 = 2341 = 3412 = 4123
Maybe 6^4 should be divided by 4 but I am not sure if this is over simplified.
I'll assume that, even though it could easily be wrong. 6^4/4 = 324
Now now if I chose a number and make the others all different, I get 6*5*4*3 /4 = 120 posibilities.
Now I chose a number make the opposite one the same and the other 2 different I get 6*1*5*4/4 =30possibilities
Now if I have 2 pairs of oposite numbers the same I get 6*1*5*1 /2= 15 posibilities
Makes a total of 165 possibilities where no adjacent ones are the same.
So that could be a prob of 165/324
Please do not believe me. I would be extremely surprised if my answer was not total rubbish.
But I do not think that Coolstuff's answer is correct either.