1 in (6*6*6)
or
1 in 216 or about 0.46%
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if you are talking about getting a roll that totals 6 - there is only one way to roll that - get all ones. as such there are 6^6 -1 = 46656 -1 = 46655 other ways to roll the dice. so the odds of not rolling a total of 6 with 6 dice is 46655/46656 = .~99.997857%. If you are talking about not rolling a 6 on ANY of the dice, there are 5 ways to roll each die that will give you something other than 6 so the number of ways to not roll any 6's is 5^6 = 15625. That means the odds of not rolling any 6's on 6 dice is 15625/46656 = ~38.489798%
-4
The possibilities are 2, 4 and 6 which are all multiples of two. So the odds are 50:50
Approximately 1/1120 * You have a 27.777% chance of rolling either a six or an eight on the first roll. * You have a 13.888% chance of rolling the same thing as the first roll. * You have a 13.888% chance of doing it again. * You have a 16.666% chance of rolling a seven. To find total probability in this situation, you multiply all of your chances together. .27777 x .13888 x .13888 x .16666 = 0.000892886 = .0892886% =1/1119.964 You may just be finding this out now, but casinos have known it for a long time. The casinos simply make their payouts less than the probability and they can't lose. While I recognize that your question deals with the probablity of an event recurring, for the average gambler, there is another way to look at this. Dice have no memory. Therefore, each time you roll them, the probability of rolling a given number remains the same. Each time you roll the dice, the chances of rolling a 6 before a 7 is 6 to 5 against you. The same odds apply to rolling an 8 before a 7.
If we roll 2 dice simultanosly the sample space consists of 6 rows and 6 col so the answer is 6*6 i.e 36 elements.If we roll 6 dice simultanosly the sample space consists of 36 rows and 36 col so the answer is 36*36 i.e 1296 elements.