probability n dice [resolved]

If three fair standard dice are tossed, what is the probability that the results will be three consecutive integers?

About 32% if this is correct :

VB Code:

Private Sub Form_Load()
Dim i, j, k, Yes, No As Integer
 
For i = 1 To 6
    For j = 1 To 6
        For k = 1 To 6
            Select Case i + j + k
                Case 6, 9, 12, 15
                    Yes = Yes + 1
                Case Else
                    No = No + 1
            End Select
        Next
    Next
Next
 
MsgBox ((Yes) / (Yes + No)) * 100 & " %"
End Sub

sorry but thats wrong, :S
why ??!
9 = 2+3+4
but also
9 = 2+2+5
see ??!

lets not use code,
use math ;) :P

the answer will be
(1/6) ^ 2,
only in one case:
if its loop like
meaning:
5,6,1
or
6,1,2
is alright,
thats the only exception that we should exclude

so its: (1/6)^2 - (1/6)^3 * 2
=1/54
this might seem complex but think about it,
its not.

in other words:
the first dice is is free to be anything from 1 to 4
ie. chance 4/6
AND the second one should be the one next to it
ie. 1/6 chance
AND (hence we should multiply the chances)
the thrid should be next to the second which has chance
1/6

multiply all chances:

4/6 * 1/6 * 1/6
=4/216
=1/54


a sure answer !! ;)

But the order of the dice probably doesn't matter. So 3,1,2 would be consecutive integers. So you would have 3! or 6 possibles for each of the 4 integer sets or 6 * 4 = 24 possible combos. 24/216 = 1/9. :)

well, lets see this case ............
mmmm............


ok, a little change to our tree,

adding the other possibilities:

x , x+1 , x+2
x , x+2 , x+1

x+1 , x , x+2
x+1 , x+2 , x

x+2 , x , x+1
x+2 , x+1 , x

where x ranges from 1-4,
each line has probability 1/54 (as calculated previously)

so the answer on this new assumption is
6/54
=1/9

yeah, work horse is right

Possibility of getting a certain event with 3 dice 6*6*6 = 1/216

There are 6 ways to get 1,2,3; 6 ways to get the next, until we hit 6 (4 different series in total) so 6*4/216.
6*4/216 = 24/216 = 1/9.

I didn't read workhorse's post before doing this, but we just did probabilities in math and I had to test it out.

6,1,2 isn't consecutive thuogh, so there are only 4 permutations out of 216 that are valid.

Oh well, it was just an idea... Even if it was wrong :(

Kedaman said - 4 permutations are possible our of 216.

No 4 combinations are possible and each can have 3! permutations.

In short the correct answer is already there 1/9

Thus easy thread resolved.

fundu: depends how you interpret the question