@Betlist: That is a neat diagram you posted. What tool did you use to
generate it?
On 4 May 2013 23:33, Joseph DeVincentis dev...@gmail.com wrote:
In your cases with 10 diamonds, you have probabilities of 1/16, 4/16,
6/16, 4/16, and 1/16 of being in the respective cases. To get to the first
But that would say there are 32 possibilities, but as drawn, there are only 30.
From first case for N=10, you go to the first case in N=11 in all cases,
because the diamond have to slide to the right...
Dňa nedeľa, 5. mája 2013 5:33:09 UTC+2 /dev/joe napísal(-a):
In your cases with 10
@Jugesh: I have to admit, it was created in mspaint O:-)
Dňa nedeľa, 5. mája 2013 8:11:46 UTC+2 Jugesh Sundram napísal(-a):
@Betlist: That is a neat diagram you posted. What tool did you use to
generate it?
On 4 May 2013 23:33, Joseph DeVincentis dev...@gmail.com wrote:
In your
There are 30 possibilities, but they are not all equally likely, because of
the result that after 4 diamonds are stacked on one side, the next one is
forced to the other side.
On Sun, May 5, 2013 at 3:48 AM, Betlista betli...@gmail.com wrote:
But that would say there are 32 possibilities, but
Thank you, I think I understood it ;-)
Dňa nedeľa, 5. mája 2013 15:40:37 UTC+2 /dev/joe napísal(-a):
There are 30 possibilities, but they are not all equally likely, because of
the result that after 4 diamonds are stacked on one side, the next one is
forced to the other side.
On
In your cases with 10 diamonds, you have probabilities of 1/16, 4/16, 6/16,
4/16, and 1/16 of being in the respective cases. To get to the first case
with 11 diamonds, if you are in the 1/16 case, you will always go there,
and in the 4/16 case, you will go there half the time, so the correct
