Re: No result is likely!!!

Van, thanks for your response, and in part you wrote:

"Assuming fair dice and no shooter influence, on average the
individual faces of each die are expected to show one time every six
rolls. Of course the sums of the pips on the faces that show are not
expected to be evenly distributed, but I would hardly call that a
destruction of equilibrium."

I guess I used the wrong word -- equilibrium-- but you got the gist of
my post. Mr Shank is using one die and when one die is used each
result of 1 through 6 has an equal chance of showing.

But when two dice are used, each result DOES NOT have an equal chance
of showing.

Yet, he is using ONE DIE as the basis to extrapolate the results of TWO
DICE.

What I am questioning is that the results of ONE DIE cannot be
extrapolated to show the results of TWO DICE.

Here is how we can see this:

with one die each of the numbers one through 6 in a perfect world shows
up one time each:

1 1
2 1
3 1
4 1
5 1
6 1

but with two dice each of the results in a perfect world would show up
this way:

2 1
3 2
4 3
5 4
6 5
7 6
8 5
9 4
10 3
11 2
12 1

Now, what has happened? How is it that with one die all six numbers
should have an equal distribution of once for each face or number-- but
for two dice the perfect distribution changes?

It changes because with two dice there is a new factor affecting the
outcomes. Im not a math guy so I can't explain but I would hope that
the math guys can come up with it.

Why is this important? Because Mr Shank used the "one die results" to
point out my questioning the perfect distribution of the point of 4.
What I question is that it wins one out of three times. I suggest that
having two dice in the game adds an extra element so that Mr Shank's
"one die model" cannot be used to explain that the 4 will win one out
of three times.

I suggest that with two dice, the point of 4 might in fact win fewer
than one out of three times. I said that because on each and every
roll, the 4 is an underdog of 2 to 1. I asked before and I'll say it
again: what power in the cosmos will instruct the dice to overpower the
2 to 1 edge so that the point of 4 will win?

I will concede that if the game of craps were played with ONE DIE, and
the goal of the game was to repeat on the hop the number you just
rolled, then the point of 4 would win one out of six times. But craps
is not played with one die with each face having a one in six chance of
showing; craps is played with two dice and the distribution of numbers
favors the number 7 showing six times while the distribution shows the
number 4 showing three times.

What I am asking for, please, is an explanation. Please don't call me
an idiot, a moron, a jerk, or a troll. Just educate me.

Mr Shank says he no longer reads what I post. Perhaps someone else
would like to copy it so Mr Shank will see my question.

thanks.

.

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