0:25

What this is really saying is that we need an odd number person, followed by

an even numbered person, to show the same action in order for a cascade to start.

So, given that it doesn't start after the first

two pair, after the first pair that it would have

to start then it could have to start here after

the second pair if they both show the same action.

ANd if it didn't start here, it could start after the next pair and so forth.

What can happen is you can't trigger a cascade by

having an even person followed by an odd numbered person.

Because, remember, that if this even person is showing a different value than

what the first person did, then, the second person must've just flipped a coin.

Right, and if the second person flipped a coin, that means that

the first persons back in the same shoes as the first, as, then.

That means the third person is back in the same shoes as the first person was.

So, you would need then this third person to show their,

to just show their public or their private signal and the

next person to be showing their, the same value as what

the first person showed, so, that would trigger a cascade then.

1:32

And in the case of no cascade, the next odd-numbered person

as we said is just going to start over from the beginning, right?

So, after the 100th person goes, if there's no cascade then the person 101

is going to be in the same exact position as the first person was.

And, but the idea here in the main take away is that

there's many opportunities for cascade to start, as the number guessing continues.

Let's run through a few calculations to see how

these factors interplay for the first pair of people.

2:03

So, let's make two assumptions, first is

that, let's assume the correct number is one.

So, the moderator has decided that one is the correct number.

And let's assume that the probability that he shows one to any person is 80%.

And that implies also the probability that he chose

zero, to a person is going to be 20%.

So 80% of the time, a person is going to see the correct number, which is one, and

20% of the time, a person's going to see

the incorrect number, it's their private signal, of zero.

2:32

So let's try to figure out what's the

probability that there's going to be no cascade

after the first pair of people go, so,

after the first two people go in this case.

And order to do that, let's trace out all the possible

sequence of events that could occur in this tree diagram here.

So, just to illustrate what this is saying, this is basically showing what the

different, what the public actions will look

like for different sequences of private signals.

So if PRVI, the first private signal, is one then we go

down this path, over here, and then if PRVII also turns out

to be one, then, as we said, well those are two matching

private signals therefore the first two public signals are going to be one.

If on the other hand the second private signal

is zero then as we said these two are

not matched so therefore the second person's going to

flip a coin, to determine what their public action is.

So, if the flip turns out to be one then they're going to be the same of one.

And if it turns out to be zero then they're

going to be different, which which will be one and zero.

On the other hand, if we go back to PRVI and we go on top of the tree over here,

so we'll go up rather than down, if PRVI is

zero than it's really just the mirror of the logic.

Then, if PRVII is also zero they're both going to be,

both the public actions are going to turn out to be zero.

Which then is triggering incorrect cascade obviously, because

the correct probability should be, be a credit.

The correct number is one, and if PRVII is one then the two public, the two private

signals are different and then therefore, the second person's going to flip a coin.

And if the flip turns out to be one then I'm going to

get two different public action 0 and 1, which means no cascade, and

if it turns out to be zero then like in this case,

you're going to get both of them be the same meaning there's a cascade.

5:07

And in that case we see that, public action

one is zero and public action two was one.

So, that's the, that's the result that we're getting here.

So, the question is what's the chance that this is going to happen?

So, in order to find this total probability, we have

to find the probability of each of these individual events occurring.

So the first case what's the probability

that the first private signal's equal to 0?

Well, it's simple, we know that one is shown with a chance of 80% and

0 is shown with a chance of 20%, so this is just a 20% chance.

5:38

Probability of the second private signal is shown as a one, again, what's

the probability that the person's yet to show a one now, which is 80%.

And then, the flip coin toss, what's the probability that that, is heads in the

case that heads goes to one or tails in the case tails goes to one.

And that's going to be a 50% chance, this is always, 50/50.

So in order to find the total probability

of each of these happening, it's an AND operation.

And for those AND operations, we're assuming of course that, these are

independent events, and they are independent

events, we just multiply the probabilities.

So, we have 0.2 times 0.8 times 0.5, and that comes out to be 0.08.

So, we have an 80% chance of this sequence happening, given the assumptions.

Now, the second case is if the first

private signal is a one, the second private

signal is a zero and the flip result is a zero, so it's the exact mirror image.

In this chart rather than going up to the top, we're going to

the bottom for the two pa for the two public actions to be different.

And, so in order for that to happen, the probability

for the first private signals one is going to be 80%.

6:52

The second private signal being a zero is

the probability again of any private signal being

a zero, which is 20% and the probability that the flip result is a zero is 50%.

So, it's the exact same as above except these, the 80

and the 20 are now switched from where they were before.

So, this evaluates the same thing 0.2 times 0.8 times 0.5, which is 0.08.

So, now the total probability of having no

cascade, since it could have come from either one

of these events, we just add the two probabilities,

so we have 0.08 plus 0.08 which equals 0.16.

So, there's a 16% chance of having no cascade, after the first two people go.

So now let's look at the case in which a cascade is going to occur.

7:55

So, the first two public actions on the board have to be the same.

So, we can trace out the four cases of this happening, the first one is

shown here, the second one here, the third one here, and the fourth one here.

So, for each of these to happen we

can have, the private signal of one being zero.

And the private signal two also being zero, in which case

the first two public actions are both going to be zero.

If the second private signal, when the first private signal is zero, is a one,

then if the flip result is a zero, we're still going to get two 0's.

On the other hand, if the first private signal is a one, then the

second one is also a one, this is just going to be a one.

And it even if the second private mode was a zero is

the flip results one, then we're going to see two public actions of one.

9:03

So, there's two cases of that that were just drawn

out, the first one is that both private signals are one.

So, the probability of that happening is simple to compute.

We know that the probability of a private signal being shown, is

one is 80%, so this is just 80%, and this is 80%.

Since both of these have to happen at the same time this means that we

have to multiply 0.8 times 0.8, which is 0.64, or 64%.

So, there's a 64% chance of that happening,

which is high but, you, that should make sense

because the, the chance that people are getting

shown the, the correct private signal is very high.

The second case is when the first private signal is equal to one

and the second private signal's a zero but the flip result is a one.

And in that case, for each of these down here, we know this is 80%,

this for someone to be shown the incorrect

private signal, that's going to be a 20% chance.

For the flip result is always going to be 50%.

So, if it flips [UNKNOWN], one is going to be 50%.

We know it, we multiply these again, 0.8 x 0.2

x 0.5 and that is equal to 0.08 which is 8%.

So this total chance here, of having a correct cascade is going to be 72%.

Which is high, but again it's expected because the probability

of any person getting shown the correct private signal is high.

So we expect that if there's going to be a

cascade, at least that it would be a correct cascade.

The other hand, incorrect cascade has two cases as well, the other two cases.

In each one of these cases we need both the public actions to be zero, because

that's the incorrect value, and then it's going to trigger a cascade of 0's instead.

So, from these two cases the first one is that the first private

signal's a zero, the second one's a one and the flip result's a zero.

So, this means that the first private signal's

a zero, the second one's a one and

the flip result turns out to be zero, and which means that we're in this case.

And for that to happen, we can just say what's the probability of a private

signal being 0, 20%, this one's going to be 80%, and the flip result is always 50%.

So, we multiply that out, we have 0.2 times 0.8

times 0.5, which is 0.08, translates to 8%.

And the second case is that the first two private signals are 0s.

So, the probability that the first two people

get unfortunately showing the incorrect number is 20%,

for the first one and also 20% for

the second one because of they're the same probability.

So, we multiply this out 0.2 times 0.2 we get 0.04, which is just 4%.

So, if we add these two together, we get

12% for being the probability of an incorrect cascade.