So, we just saw how a cascade could start after the first two people. But, note that it could just as well start after the next two people, and the next two, and so on. All it takes is two people in a row showing the same public action, such that the first person is an odd-numbered person, so a person 1, 3, 5 and so forth. And the second person is an even-person number, I mean person 2, 4, 6 so on. 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. 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. 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. 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. So, there's two cases here where we do not have a cascade. We either have to come down this way, all the way to the bottom, or we have to come up this way all the way up to the top, for the two, first two public axes to be different. So, as we said, the no cascade is going to occur if if pub one and pub two are different. So let's look at each of those two cases right now and see what is the chance of them occurring, given the assumptions that we made. So, the first case is that PRVI is zero, and PRVII is one, so we're going PRVII being zero. PRVII being one and the flip result being one, which means we're coming up the top up here. 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. 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%. 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. So, in order for us to have a cascade there's going to be four different cases. But basically for a cascade to happen we know that we need the first two public actions, pub one to be equal to pub two. 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. So, correct cascade is only two of these cases, right? And the two of these cases for correct cascade, or when the public actions are both one because the public action of one then represents the true value. And, so this happens when pub one equals pub two which is equal to one. 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. So, the probability of having a cascade at all, is adding these two together. So this is the probability of it being correct is 72%, probability of it being incorrect, is 12%. And if we add these two together, probably halving any cascade, at all after the first two people, it's going to be 84%. Now, notice how probabilities have to add up to 100%. So we have three different cases, or two cases depending on how you divide them up, after the first two people. We can have no cascade, we can have a correct cascade, or it can have an incorrect cascade. So, probability of no cascade we had last time, was 16%, the probability of having a correct cascade is 72%. The probability of having the incorrect cascade is 12%. Now, if you add these up, you're going to get a 100%, as expected because probabilities need to add up to a 100%. That's a way to check your calculations to make sure that they worked out correctly.