How does the probabilty for universal hashing work out?

Question

Hey for universal hashing we say the following:

Definition:
A randomized algorithm $H$ for constructing hash functions $h\colon U \to \{1,\ldots,M\}$ is universal if for all $x \neq y$ in $U$, we have $\Pr_{h\gets H} [h(x) = h(y)] \leq 1/M$

I actually just can´t understand how the probability can be $1/M$ in any case. I alway try to think about any random function mod M as example and compare it with rolling a dice, but with a dice we have for every instance a probabilty of 1/6 as we have 1/M for mod M. But when I roll a dice two times after I have at least $\frac{1}{6}^2$ but saying we have binomial probabilty it looks even worse thinking that the dice would be thrown y times.

Can you help me understand how 1/m could be reached within this context?

That is a definition to achieve. Did you look at the Wegman-Carter — kelalaka, Dec 12 '20 at 18:23
Actually it's the follow up here that answers your question. — Modal Nest, Dec 12 '20 at 21:34

score 4 · Accepted Answer · answered Dec 12 '20 at 22:17

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Actualy chance of 2 dice throws to have the same value is 1\6, not 1\36. And definition also compares 2 values.

What you are asking is called https://en.wikipedia.org/wiki/Birthday_problem and it is still present even with universal hashing. Or dice throws. If more than 2 are compared.

answered Dec 12 '20 at 22:17

Surprised Seagull

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Thank you! This link made it even more clear regarding birthday problem and rolling a dice. – OuttaSpaceTime Dec 14 '20 at 07:40

How does the probabilty for universal hashing work out?

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