Re: Dishonest Forks? | Was - Re: [Namedroppers-honest] Brian Smith asks: ?Who is Dean Anderson?

On Fri, 12 Mar 2010, Jeff King wrote:

> On Fri, Mar 12, 2010 at 12:30:03AM -0500, Dean Anderson wrote:
> 
> > > OK, I tried to follow your math. But it really doesn't make any sense
> > > to me.
> > 
> > Well, I hope you figure it out before they give you a PH.D. in... what?  
> > computer science?
> 
> Why are you even mentioning this except to be inflammatory? 

Ok. That wasn't fair.  I'm just getting tired of some of the BS going
around.  But you are playing into it. "One guy wrote the code, One wrote
the doc, had nothing to do with the code but post, one had nothing
whatsoever to do with it, and a sockpuppet distributes it".  That
clearly wasn't what happened.

> An attacker does not succeed by guessing your port. He succeeds by
> guessing your port and qid pair.

Yes. Thats a reasonable way to analyze it, IF you are really only going
send only a 200 of packets. We aren't. We are going to brute force the
qid, so they aren't random.  Just like the source and dest IP addresses
aren't random.  To match, you have to have the right set of IP addresses
too.

> > Possibly, you should consult the reference I cited. CRC handbook of
> > Applied Cryptography.
> 
> I did.  It gives the formula. I never said the formula was wrong. I said
> you misapplied it.  Strangely, section 2.15 of the CRC handbook does not
> mention DNS spoofing at all, so the application of their balls and urn
> formula to the problem at hand is left up to us.

Yes. I explained which formulas and why:  2 Urn's no replacement, 
because for one qname, there will be two hundred ports.

> 
> > > you calculate the probability based on a particular number of packets.
> > 
> > ???? If you know one, you can calculate the other.
> 
> A probability does not have an expectation. A _random variable_ with a
> distribution has an expectation. The random variable here is the number
> of packets the attacker sends. That is the third value in the birthday
> attack formula (the number of balls in the second urn).  

There is no expectation value here.  (Sum of tries * value of try)  
would give an expected port number, which is nonsense.  There is a pure 
probability.

> The probability you get from the formula is the chance of collision
> given a certain number of packets the attacker sends. But in the math
> you posted, you arbitrarily set the third parameter to 200. Why?

Because both urns are estimated to be the same size: 200. The attacker 
knows there are 200 ports on the server. He is looking for that 200.  
The attacker can't send all the qids for all the ports (~4billion 
packets). So they will pull 200 ports, exercise the qids, and start 
over.  So: 2 urns of 64510, each picks 200 ports, no replacement.

If we do it your way:

2 urns: size 65536 * 64510, server picks 200, attacker picks X million
packets, no replacement.  Except we know there is a pattern that isn't
represented here: In this case, you aren't hitting all the qids of a
given port.  This is a complicated summation that take a lot of work to
compute.  It might be a better representation of the problem for the
attacker, since they know how many packets they are going to send on a
particular qname.

		--Dean


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