No result is likely

MoneyLA's legendarily stupid question:

"With the chance of a win stacked against the 4 by "2 to 1" would it
not make more sense and be more realistic that the 4 would win fewer
than 1 in 3 times?"

and another classic:
[QUOTE]
"Mark, you wrote: "The math predicts outcome probabilities for a
SINGLE ROLL just as well as it does for millions of rolls."

This is TRUE. The SINGLE ROLL probabilities tell us that the house
will win any "odds bet."
[END QUOTE]

and, finally:
[QUOTE]
Yes, Van, that is what I am asking. If on each independent roll the 4
is a 2 to 1 underdog (true odds)
shouldn't the 4 win fewer than one in three times?
[END QUOTE]

So, MoneyLA basically thinks that an event that has, each roll, only a
..33 probability can NEVER occur, doesn't he? Of course, the logical
conclusion to be drawn from this is that he should make only lay bets
against the 4 and 10, which could NEVER lose, could they?

I'm not posting these JUST to make fun of MoneyLA's comically low IQ;
there is an issue here that might be a useful review for those of us
with normal-or-better intelligence.

When you roll one fair die, making sure it travels down the table and
bounces off the pyramids, which face will show? Well, it doesn't
depend in any way on that die's "history", as math quack DiMauro would
have us believe. Nor does it depend on some tangible or intangible
characteristic of the table, as 777 and our latest troll, usalistings,
would have us believe. On what does it depend? Nothing. That's why we
cannot predict it. But one face or another must come up, right? How
can that happen, when each face is an underdog? After all, each face
has only a 1/6 chance of coming up, right? This is far more of an
"underdog" situation than the 4 or 10 vs. the 7. So, in MoneyLA's
confused world, no face could ever show, so the die would just hang in
mid-air forever. But apparently that's not what happens; the die
lands, bounces off the pyramids in some direction, bounces and rolls
for a bit and then comes to rest, with one face showing.

DI aside, I think we all believe that each face has an equal chance to
show, and since there are six faces, the probability of each showing
up is 1/6, or .166666..... There are no "favorite" faces, and at this
point, no face is better or worse for the casinos or the players.
There are no bets on the result of a single die.

If this is true for one fair die, then, presumably, it's true for
another fair die, or any fair die. We could decide to use any number
of dice and make up any rules we wanted to to create a game with dice.

Suppose we choose to use two dice and combine the number of pips on
the two faces in some way? We are going to assume that the two dice do
not affect each other, i.e. that each retains its property of
"equiprobability" when rolled along with another die. The fact that
one die stops with the one-pip face showing does not make it any more
or less likely that the other will show one pip, or any other number
of pips. IOW, the two results are INDEPENDENT of one another.

So, we now believe that each face of each die, when two are rolled at
the same time, has an equal chance of showing. We can conclude, then,
that for each die:

p(1) = 1/6
p(2) = 1/6
p(3) = 1/6
p(4) = 1/6
p(5) = 1/6
p(6) = 1/6

and that this is true for each and every roll, whether we are talking
about one roll, never to be repeated, or 800 million billion rolls.

Given this assumption, a casino could combine these two independent
results any way it wanted to, and offer any bets it wanted to, which
the players could take or leave. How would the casino figure out how
much to pay on a winning bet in order to 1) entice players to make the
bet AND 2) make a profit on it? They'd have to be able to figure out
the probability of the player winning or losing the bet, wouldn't
they? How would they do that?

Well, the "multiplication theorem" of probability says that the
probability of two independent events occurring is the product of
their individual probabilities. So:

p(1 on 1st die,1 on 2nd die) = P(1 on 1st die) * p(1 on 2nd die)

>From this, we can calculate the probability of every possible
combination of dice results for throwing two dice.

p(1 on 1st die) = 1/6
p(1 on 2nd die) = 1/6
SO
p(1 on 1st die AND 1 on 2nd die) = 1/6 * 1/6 = 1/36

p(1 on 1st die) = 1/6
p(2 on 2nd die) = 1/6
SO
p(1 on 1st die AND 2 on 2nd die) = 1/6 * 1/6 = 1/36

Presumably, everyone can see that every possible combination is going
to have a probability of 1/36.

What about the situations where we have (1 on 1st die/2 on 2nd die)
and (2 on 1st die/1 on 2nd die)? Can we distinguish these? Not unless
we have dice of different colors or we throw them separately. So, we
now have reduced the number of "distinguishable" results from 36 to
how many?

1:1
1:2, 2:1
1:3,3:1
1:4,4:1
1:5,5:1
1:6:6:1
2:2
2:3,3:2
2:4,4:2
2:5:5:2
2:6:6:2
3:3
3:4,4:3
3:5,5:3
3:6,6:3
4:4
4:5,5:4
4:6,6:4
5:5
5:6,6:5
6:6

Count 'em up, you get 21. So, even though there are 36 possible
outcomes, we only distinguish 21. So, what this this do to our
probability distribution? Well, the "additive theorem" of probability
says that, if two events are mutually exclusive, the probability of
one OR the other occurring is the SUM of their individual
probabilities. So:

p(1 on 1st die/2 on 2nd die OR 2 on 1st die/1 on 2nd die) =
p(1 on 1st die/2 on 2nd die) + p(2 on 1st die/1 on 2nd die) =
1/36 + 1/36 = 2/36 = 1/18

Looking at the above list of distinguishable outcomes, we get a set of
6 outcomes with 1/36 probability and 15 with 1/18. Another probability
theorem states that the probability of a certain event is 1. Since
it's certain that both dice will land and eventually stop with a face
showing, the sum of all these probabilities must be 1, right?

6(1/36) + 15(1/18) = 6/36 + 15/18 = 6/36 + 30/36 = 36/36 = 1

So, we seem to be on solid ground so far.

In Part II, tomorrow, we will use this information to develop a bet
and figure out how to
1)get people to make it and
2) make money on it
Cheers,
Alan Shank
.

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