Math CentralQuandaries & Queries


Question from Howard, a parent:

My son is desperately trying to work out pecentage addition.
If he has a 16.66% chance of rolling a particular number on a dice, and he rolls for it 6 times, he knows (from experience) that his odds of getting at least 1 right are not 100% (16.66% + 16.66% + 16.66% etc).
He asked me if there was an equation that would give him the correct percentage chance of success when trying X times for a Y% chance of something.
I didnt know, but I figure there must be one. Can you help?

Thanks, my son is 13.


Adding is not the thing to do.  Multiplication is the thing to do.  But you need to express the chance of success (probability of success, or fraction of times that the number comes up) as a decimal between 0 and 1 rather than as a percentage. The best way to convince yourself not to add is to notice that if you add 16.66% together seven times, it eventually exceeds 100%.  We know that's not possible.  We also know that there are sequences of 7 rolls where the number doesn't turn up, so the fraction of time it happens is less than 1 (that is, it happens less than 100% of the time).

It is easier to calculate the chance that the particular number never turns up in the 5 rolls.  If all numbers are equally likely to turn up, then the probability that your number doesn't turn up on any one roll is .8333 (83.33%).  The probability that it doesn't turn up on neither roll 1 nor roll 2 is .83333 x .83333 (83.33% of 83.33%).  The probability that it doesn't turn up on any of three rolls is (.8333)3, and so on.

The probability that it turns up at least once is 1 minus the probability that it doesn't turn up at all.

To compute it directly (i.e. without the subtraction), you would need to consider the cases where the number turned up exactly once among the 6 rolls, exactly twice, and so on.  This is more complicated.

I hope this answers your question.


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