--- In mepbmlist@y..., "Draugnar" <Draugnar@o...> wrote:
The key to the probability is that EACH person still has a 75%
chance of
succeeding. But the chances of BOTH succeeding are significantly
less. So,
by this logic, IF the other downgrade happens first, then your
downgrade has
a statistically smaller chance of succeeding. Think if it this way.
If you
flip a coin ten times, 5 of those should be heads and five tails.
Any given
flip of a coin has a 50/50 shot for heads and tails. But if the
first 4
coin tosses come out heads, then the probability is (5 to 1) that
the next
coin toss will be tails.
Ack, I'm confused by this argument and I'm the one whose argument is
being paraphrased.
Each coin toss is an independent event, with no "memory" of what
happened previously. If the first 4 coin tosses come out heads, then
the probability is still even that the next toss will be tails. To
believe otherwise is the Gambler's Fallacy - "There's been a run of
heads, so there probably won't be another." It doesn't work that way.
The point of my original post was not that the downgrade attempts
affect each other - the point was that given two events which might or
might not happen with probability p, the probability that _both_ will
happen is p^2. Or, to walk through it in more detail:
Suppose Arthedain and Cardolan are both trying to downgrade Rhudaur,
with a 75% chance each of succeeding.
3 times in 4, Arthedain succeeds. 1 time in 4, Arthedain fails. You
could model it by saying that you roll a d4, and a 4 represents a
failure.
You can write the cases out and record the number hypothetically
rolled:
Ar. succeeds (1)
Ar. succeeds (2)
Ar. succeeds (3)
Ar. fails (4)
Arthedain's success or failure does not affect Cardolan's chances to
succeed. So in 3 cases out of 4 of Arthedain's success, Cardolan
succeeds; in one case out of 4, Cardolan fails even after Arthedain
has succeeded. You can write those cases out too:
Ar. succeeds (1), C. succeeds (1)
Ar. succeeds (1), C. succeeds (2)
Ar. succeeds (1), C. succeeds (3)
Ar. succeeds (1), C. fails (4)
Ar. succeeds (2), C. succeeds (1)
Ar. succeeds (2), C. succeeds (2)
Ar. succeeds (2), C. succeeds (3)
Ar. succeeds (2), C. fails (4)
Ar. succeeds (3), C. succeeds (1)
Ar. succeeds (3), C. succeeds (2)
Ar. succeeds (3), C. succeeds (3)
Ar. succeeds (3), C. fails (4)
Ar. fails (4), C. succeeds (1)
Ar. fails (4), C. succeeds (2)
Ar. fails (4), C. succeeds (3)
Ar. fails (4), C. fails (4)
As you can see from this list, there are 16 possible outcomes. In 9
of them, both nations succeed at the downgrade. In 3 of them,
Arthedain succeeds but Cardolan fails. In 3 of them, Cardolan
succeeds but Arthedain fails. And in 1 of them, neither nation
succeeds. This rule - the probability of n nations all successfully
downgrading, where the probability of each nation downgrading is p, is
p^n - holds regardless of the number of nations. Or, more
intuitively, the chances of Arthedain and Cardolan both successfully
downgrading must be less than the chances of just Arthedain
downgrading, because it doesn't help if Arthedain succeeds but
Cardolan fails, and there's a chance for Cardolan to fail.
If you restrict yourself to considering only cases where Arthedain has
already succeeded, Cardolan's 3/4 chance is the only factor. That's
the situation described above with the 4 heads in a row case. But
it's not what you're interested in when planning strategy.
His statistical analysis was just demonstrating this. When using a
random
number generator, unless the seed is changed between each random
number, you
clearly see the laws of probability in action as the computer is
never truly
random.
This doesn't really have anything to do with seeding; the laws of
probability best describe with truly random events. Bad random number
generators produce patterns which break the independence of the
events.
From: Wilson Reis [mailto:will@i…]
As my math teacher would say : That looks like "rabbit's and
wands".
From which sleeve did that came ? I never heard that simultaneous
downgrades
would be harder.
Please, tell us a bit more about that 
Hope the above helped. It's the same logic as guessing the
probability of one nation succeeding in three risky downgrades in a
row, or why it's less likely to throw heads three times in a row with
join than it is to throw heads once.
I think some study of probability is a good idea for any gamer who
plays games in which random elements are used.
-Peter
···
-----Original Message-----
So we have to decide whether to attack now or wait until we make sure the downgrade has succeeded...
