This probably isn't that difficult of a problem for most of you, but I am stuck on it. Can anyone help?
Question: What is the minimum number of times you must roll a number cube to have a 50% chance of getting a 6.
Thanks in advance!
This probably isn't that difficult of a problem for most of you, but I am stuck on it. Can anyone help?
Question: What is the minimum number of times you must roll a number cube to have a 50% chance of getting a 6.
Thanks in advance!
The probability that you get no six on first roll is 5/6. The probability that you get no sixes on two rolls is (5/6)(5/6). And in general the probability that you get no sixes on n rolls is (5/6)^n. Now if n=3 then that number is .578 (thus it is likely you get no sixes on 3 rolls), and if n=4 then that number is .482. Thus, it is unlikely to get no sixes after 4 rolls. That number is 4.
Hello, november7!
Same solution, different spin . . .
What is the minimum number of times you must roll a number cube
to have a 50% chance of getting a 6. ?
Consider rolling the die times and not getting a 6.
The probability of getting a number other than 6 is:
Then: .
And we want this probability to be less than 50%.
So we have .
Take logs: .
Divide by . . . a negative quantity.
. .
Therefore, we must roll at least 4 times.