If you roll 6 times, what are the odds of getting a given number? It can’t be 100%, since there’s a good chance you won’t get the number on any roll, so what percent is it?
Edit: A lot of you are misinterpreting the question. I am asking what the combined odds are to get a given number at least once, not just the 6th roll.
In: Mathematics
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1/6 * 6 =100. Say you want to roll a 4. Stastically speaking, its near a 100% chance that a 4 will turn up at least once in 6 rolls.
66.5%.
To find the answer, you multiply out the odds of **NOT** getting a desired result. So, for each roll you have 5/6 chance of **NOT** getting the result you want, and you multiply that out by the number of rolls.
5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 33.5%, which you then subtract from 100% to get 66.5%.
It’s a 1/6 chance of rolling any number, no matter what you’ve rolled before. Previous rolls have no effect on future rolls.
Note that this is known as a ‘binomial distribution’ if you want to google it.
As noted elsewhere, the question you’re asking can be reduced to “what is the chance of getting no matches out of 6 rolls?” – which is trivial to calculate.
To see why this is so, consider a simpler question involving three rolls. You’ve got a 1/6th chance of success (S) and a 5/6th chance of failure (F). With three rolls, there are the following possibilities: SSS, SSF, SFS, SFF, FSS, FSF, FFS, FFF.
To calculate the chance of any of those sequences, we just need to multiply the chances of each element of the sequence. So the chance of a sequence like FSF would be (5/6) * (1/6) * (5/6).
For the question “are there any successes at all?”, the answer would be the sum of all the possible sequences except for FFF (the only sequence with no success). Since those sequences represent every possibility, they must add to 100% – and thus 100% – chance(FFF) is equal to the answer you seek.
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