So now we can quickly write down the following.

Just look at the first two people. We know that the probability of no cascade

just like what we discussed before is that if x1 is zero, x2 is one and you flip a

coin to maintain one. Or if x1 is one, x2 is zero and you flip a

coin to maintain zero, this will be the probability 1-P times P times half.

This is P times 1-P times half and you add up the two together you get P times 1-P

same as before. Now the probability of an up cascade,

however, is different because we now explicitly assume the correct number is

fixed at one. So the up cascade is probability that x1

is one, x2 is one or x1 is one, x2 is zero but you flip a coin and decide to write

down one. The probabilities are respectively P P +

P times one - P times half, and that equals to P times one + P over half.

And a probability of a down cascade after two user.

You can write a similar expression. It is one - P^2 + one - P times P / two,

That equals one - P times 1,2 - P / two.

So that's the expressions for these three possible events.for P times one - P for no

cascade, P times one + P / two for up cascade and P, one - P times two for a

down cascade. Now this is also the probability of a

correct cascade. This underline number is, is soon to be

fixes, one and this is an incorrect. Cascade.

Now this is just for the first two users. We can now write down the expression for a

general two end users after even number of users.

What would happen? The derivation is a slightly involved but

you can either do that as home exercise or look at a textbook derivation.

Basically follows the same principle as before with Cnet.

First of all, no cascade that probability is simply P1-p, one - P the whole thing

n times. Now up or correct cascade turns out to be Pp+1.

P + one one - P - P^2 n times / two one - P + P^2 And the probablity of down

and incorrect cascade turns out to be one - P two - P one - P - P^2 n times /

two minus, two one - P + P^2. This is just basically, a denominator that

we use to add up the terms. This factor of n is really a factor

expressing, the probability, no cascade. Okay?

But it's, the other part of the numerators are different because we already assumed

the underlying number is one. Therefore, that breaks the symmetry.

And therefore, up and down probabilities are not the same anymore, after one or n

pairs of users. Now we can plot this on a graph, or

actually four different, cases, okay? One pair, so two user.

Two pair, four user. Five pair, ten user.

100 pairs, that's 200 users. Now, in fact, you can only see, visually,

three lines for one to five pairs. Because beyond five pairs, it pretty much

stays as the same. Visually, you cannot use naked eye to tell

the difference anymore. And I am plotting here the probability of

the correct cascade. As a function of P, the common probability

of getting the correct part of signal across all the users, ranging from half to

one, has to be bigger than half and one is the largest you can get.