Probability total number on dice is 11 or more
- Thread starter rad6210
- Start date
I know the probability is 1/2, but can't see a way to work it out except by listing the possibilities;
but this "listing" can be greatly shortened, like if 1st two are 6,6 or 6,5 or 5,6 or 5,5
then automatically you've got 24 cases at 11 or more.
On average, after 6 rolls of a dice, you get 3.5; so 3 dice means 3.5 * 3 = 10.5;
perhaps that's rounded to 11, hence the 1/2 probability...
We'll see what our probability expert Galactus has to say to that :wink:
but this "listing" can be greatly shortened, like if 1st two are 6,6 or 6,5 or 5,6 or 5,5
then automatically you've got 24 cases at 11 or more.
On average, after 6 rolls of a dice, you get 3.5; so 3 dice means 3.5 * 3 = 10.5;
perhaps that's rounded to 11, hence the 1/2 probability...
We'll see what our probability expert Galactus has to say to that :wink:
Hello, rad6210!
A variation of Denis' approach . . .
\(\displaystyle \text{The possible sums are: }\;\underbrace{3,4,5,6,7,8,9,10}_{\text{10 or less}},\underbrace{11,12,13,14,15,16,17,18}_{\text{11 or more}}\)
\(\displaystyle \text{In }hal\!f\text{ of the outcomes, the sum is 11 or more.}\)
\(\displaystyle \text{Therefore: }\;P(\text{11 or more}) \;=\;\frac{1}{2}\)
A variation of Denis' approach . . .
\(\displaystyle \text{The possible sums are: }\;\underbrace{3,4,5,6,7,8,9,10}_{\text{10 or less}},\underbrace{11,12,13,14,15,16,17,18}_{\text{11 or more}}\)
\(\displaystyle \text{In }hal\!f\text{ of the outcomes, the sum is 11 or more.}\)
\(\displaystyle \text{Therefore: }\;P(\text{11 or more}) \;=\;\frac{1}{2}\)
Thanks!