[lit-ideas] Re: Worst Case Scenarios
- From: Teemu Pyyluoma <teme17@xxxxxxxxx>
- To: lit-ideas@xxxxxxxxxxxxx
- Date: Tue, 25 Apr 2006 21:57:42 -0700 (PDT)
--- Omar Kusturica <omarkusto@xxxxxxxxx> wrote:
> *I disagree. The only countries that have recently
> talked about the use of nuclear weapons are the US,
> Britain and France.
I don't get your reference to Britain, but no nuclear
state has ruled out using the weapons if attacked by
nuclear weapons. This is a necessary condition for
MAD.
> Again, I cannot help but note the
> inconsistency in your position. Just a few posts
> ago,you were predicting a nuclear attack by the US
on
> Iran.
>
If Iran has nukes, it is possible. As long as they
don't, not a snowballs chance in hell as the Americans
say. Even Seymour Hersh agrees on this.
> > And I've completely ignored the chance that
> nuclear
> > weapons could be used against non-nuclear states,
> > which makes the above calculation much worse.
>
> *Indeed you have. This shows that the probability of
> *use of nuclear weapons* (as distinct from conflict,
> which takes two sides presumably) does not
> necessarily rise in proportion to the number of
> buttons.
I didn't even try to construct such a model, but
fine... We're talking probabilities, so as Leibnitz
would put it, let's calculate.
We'll assume 30 players in this merry little game of
death
B is the button nations, and B=9
N is the number of wannabe nuclear nation, N = 21
X is the number wannabees getting nukes.
p is the probability of Bs bombing one another
q is the probability of Bs bombing Ns
P is the initial total probability
Px is the total probability after x new nuke states
P = B*( (B-1)*p + Nq) = 72p + 189q
Px = (B+x) ( (B+x-1)*p + (N-x)q)
Let's try x=6, six more nuke states
P6 = 15*(14p + 15q) = 160p + 175q
The only way you could argue that the total
probability of nuclear war is now lower is by assuming
some weird values for p and q, So we would like to
know how much bigger q than p should be in order for
there to be at least as low probability after
wannabees got nukes.
q = xp
P = 72p + 189xp
P6 = 160p + 175xp
P = P6
72p + 189xp = 160p + 175xp
14xp = 88p
x= 88/14 = 6.29
That is you would need more than six times the
probability of nukes being used by those that have
them against every single nations that hasn't. I find
it hard to believe that q > p nevermind q > 6p.
Cheers,
Teemu
Helsinki, Finland
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