Does anybody know the odds on rolling five of a kind with one roll of five dice? I thought it was 6 to the fifth power(7776 to one) but I am being told it is actually much less. How do you calculate it? Thanks.
Question on dice odds?
- Thread starter BreakaLeg
- Start date
6^4 power
first roll out of 5 is arbitrary.....(any number can be made into 5 of a kind with 4 more same rolls -- outcomes only depend on the next 4 throws)
for 5 5's, 5 6's, etc. etc. you are correct -- odds are 6^5 for each possible combination....
hope that explains it
first roll out of 5 is arbitrary.....(any number can be made into 5 of a kind with 4 more same rolls -- outcomes only depend on the next 4 throws)
for 5 5's, 5 6's, etc. etc. you are correct -- odds are 6^5 for each possible combination....
hope that explains it
Each dice has six sides with numbers one through to six. To roll anynumber is a 5/1 chance.
To roll five dice for one roll each and obtain the same number is as follows;
6 x 6 x 6 x 6 x 6 = 7775/1
So yes your first thought was correct.
To roll five dice for one roll each and obtain the same number is as follows;
6 x 6 x 6 x 6 x 6 = 7775/1
So yes your first thought was correct.
no it isn't.......first roll is arbitrary.....
if you don't know, then don't give out wrong information.....
correct answer is 6*6*6*6:1
if you don't know, then don't give out wrong information.....
correct answer is 6*6*6*6:1
freeze is right, not bulldog.
Think about the odds of rolling a pair. there are 6 ways to do it, and 36 possible rolls, so 1/6 chance of rolling a pair. Its a 1/36, that is 1/(6*6) probability of rolling a SPECIFIC pair. If you have N dice, it's 6^(N-1) OR (6^N)/6 (divide by 6 because it can be any one of the pairs).
So, the probability is 1/1296, or 1295-1 against it happening.
Think about the odds of rolling a pair. there are 6 ways to do it, and 36 possible rolls, so 1/6 chance of rolling a pair. Its a 1/36, that is 1/(6*6) probability of rolling a SPECIFIC pair. If you have N dice, it's 6^(N-1) OR (6^N)/6 (divide by 6 because it can be any one of the pairs).
So, the probability is 1/1296, or 1295-1 against it happening.
