Seemingly simple problem...

[AdmiralWen]

Perhaps I'm just missing the obvious, but I seem to need some help with this particular probability problem.

4 fair die are tossed. The largest result is discarded. What is the probability that the remaining 3 die adds up to 11 or greater?

I know it's possible to count the total possibilities, but what would be the "mathematical" way of doing this problem?

Thanks.

[Gandalf]

counting the possibilities

[AdmiralWen]

Originally Posted by Gandalf

counting the possibilities

So... is that the only way? Counting the possibilities?

[joni308]

here's my thought:

for the smallest 3 to add up to 11 then among those 3 there should be at least a 4 (as 3X3 = only 9)

Therefore the one discarded HAS TO be a 4,5 or 6.

So, all the favorable outcomes, if I were to rank them from the smallest to the biggest will look like this:

X X X 4
X X X 5
X X X 6

Now, fill in the X's such that they are in an increasing order and they add up to 11. (Remember, the last number is the one discarded)

So I have

3 4 4 4 in (4 choose 3) = 4 ways

1 5 5 5 in 4 ways
2 4 5 5 in 12 ways
3 3 5 5 in 6 ways
3 4 4 5 in 12 ways

1 4 6 6 in 12 ways
2 3 6 6 in 12 ways
1 5 5 6 in 12 ways
2 4 5 6 in 24 ways
3 3 5 6 in 12 ways
3 4 4 6 in 12 ways


So 122/1296 = 0.094 prob. of getting exactly 11?


Can you guys go over this really quick for any errors? I wrote down very quick whatever came to mind after first reading the problem so I didn't even take a moment to double think it.

Any sleeker- or simply different- methods would be appreciated.

[joni308]

OK, I just realized (after reading again the question) that whatever solution I have provide is for "add up 11 (exactly 11)"

But I see I can extend the same idea to 'GREATER than 11".

[Gandalf]

.094 for exactly 11 looks right, as does the method. I don't know if there is a more efficient way, and suspect it would not be efficient to spend time looking for a better way.