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Jan 24th, 2004, 01:44 PM
#1
Thread Starter
Addicted Member
probability n dice [resolved]
If three fair standard dice are tossed, what is the probability that the results will be three consecutive integers?
Last edited by dolor; Jan 26th, 2004 at 01:39 PM.
- you've been privileged to read a post by Miz
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Jan 24th, 2004, 05:35 PM
#2
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
Has someone helped you? Then you can Rate their helpful post. 
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Jan 24th, 2004, 08:58 PM
#3
Addicted Member
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 !!
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Jan 24th, 2004, 10:54 PM
#4
Fanatic Member
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.
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Jan 25th, 2004, 09:42 AM
#5
Addicted Member
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
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Jan 25th, 2004, 11:26 AM
#6
Fanatic Member
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.
Don't pay attention to this signature, it's contradictory.
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Jan 25th, 2004, 12:33 PM
#7
transcendental analytic
6,1,2 isn't consecutive thuogh, so there are only 4 permutations out of 216 that are valid.
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Jan 26th, 2004, 08:03 AM
#8
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Jan 26th, 2004, 12:02 PM
#9
Member
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.
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Jan 26th, 2004, 04:32 PM
#10
transcendental analytic
fundu: depends how you interpret the question
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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