I swear I leaned this in school at some point but its been bothering me .
If you throw a six sided die, the odds of rolling a six are 1/6.
But what are the odds of rolling at least one six if you throw two dice at the same time? I thought it was 1/6 + 1/6 or 2/6, but suspect I might be wrong.
This is the idea that has me stuck. If you throw 6 dice at the same time what are the odds of rolling at least one 6 ? It can’t be 6/6 or 100%. The odds of rolling at least one 6 are certainly high but absolutely not 100%, so the logic I used for the two dice can’t be correct. There must be a formula for this but I’m having trouble searching for it . Thanks !
In: 12
The odds of rolling a six on two die is 11/36. This is slightly less than 1/3 you would get by adding 1/6+1/6.
Why? Because sometimes when you roll a six on the second die you have already rolled a six on the first die, so it doesn’t add it’s full value
To calculate this, you add the chance of rolling a six on the first die (1/6) to the chance of rolling a six on the second die when the first die was *not* a six (which is 5/6*1/6 or 5/36)
To add a third die you add your two die probability (11/36) to the chance of getting a six on the third die only (which is 25/216), which is 91/216.
You can calculate this easier by calculating the inverse: what is the chance of rolling no 6’s?. With one die it is 5/6, the inverse of 1/6. With two die it is (5/6)^2, or 25/36, the inverse of 11/36. With three it is 125/216, the inverse of 91/216.
Thus with 6 die the chance of missing all is 5^6 / 6^6, or 15625/46656, or 33.4%. The inverse (chance of getting at least 1 six) is 31031/46656, or 66.6%.
The way I find helpful to think about these kinds of questions is to work out all the ways you can’t get a six. Each die has a 5/6 probability of landing on a number other than 6 or roughly an 82% chance. If you multiply the probability for all six dice and subtract that product from one you will be left with the odds of rolling at least one 6. 1-(5/6)^6 =0.6651020233 or approximately a 66.5% chance
This is not “ELI5” but try thinking of it this way:
To get AT LEAST 1 6, you must throw 6 dice and none can come out 6, right?
So on the 1st dice you need 1 to 5 out of 6, on the 2nd as well and so on.
So:
(5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6) – multiply this and its 0,335 aprox or roughly 1 in 3 of getting no 6. So the odds of getting at least 1 6 is the reverse -> 100% – 33,5% or aprox 66,5%
Rule 2 forbids straightforward questions.
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