 So, let us discuss the implications of there is the formula that we derived for the success ultimate success probability of a gambler. So, that was equation 6 here equation 6. So, it says basically it is a ratio of 2 powers one taken with respect to the starting amount and the other one taken with respect to the goal amount. So, let us consider various cases for example, the first case 1 is unbiased random walk unbiased case you can still call it as it is a random walk in the money domain. So, it is unbiased. So, that is p equal to q equal to half. So, if you look at our formula it says f k of reaching l we will write in this case as 1 minus q by p to the power k your by 1 minus q by p to the power l you would write like this also. Now, we have a small problem if q equal to p this is 1. So, it 1 minus 1 is 0 and denominator is also 0. So, when q equal to p equal to half expression becomes 0 by 0 type. But we know how to evaluate those 0 by 0 cases there is something called use L Lopitha's rule which says in the limit if you have two quantities let us say y x divided by z x and you have to take the limit at the some point limit at some point I am not going to specify it is y prime x by z prime x you go on doing it until you get a not a 0 by 0 term. So, this is just an illustration, but in this case specifically you will come. So, in our case therefore, limit p equal to q of f k l will be it is a derivation derivative with. So, we supposing we write it as 1 minus 1 or or limit x equal tends to 1 of a quantity called x to the power k divided by 1 minus x to the power l and this will be basically d by dx of x to the power k minus k x to the power k minus 1 divided by minus l x to the power l minus 1 evaluated at x equal to 1 which is simply k by l. Hence to summarize for an unbiased random worker or unbiased gambler for a unbiased gambler that is p equal to q equal to half is neither winning streak nor losing misfortune then f k l is k by l simply says that if he has low amount his chance of winning is small closer he is to the final amount the chance of winning is high. So, you can optimize of course, depending on some assessment of the winning probability and how much money he has to start with that could be another interesting problem, but this only mentions a value neutral probability that the probability of winning or losing winning depend is k by l or probability of losing obviously, implies probability of getting ruined f k 0 is 1 minus that. So, it is l minus k by l. So, how far he has to stay from how close he has to start to the winning amount can be decided on his his assessment of what risk he can take to the amount he is putting. Now, let us consider case 2 where the person has a tendency to lose basically every time he bats for some reason. So, let us say q q is greater than p probability of losing is more than the probability of winning then we can write f k l equal to q by p to the power k minus 1 as we derived q by p to the power l minus 1. So, this of course, gives you the probability for any l and k even though he is a basically a kind of a loser there is still a probability that he could win for a finite amount. However, is the probability of winning becomes weaker and less and less as the amount becomes larger and larger in a fashion which is not linear even here in the previous case p equal to q case equal to half case the probability reduced as l increased starting from a given point he had no chance of winning of course, but the probability came down linearly here the probability will come now down exponentially because q by p is more than 1. As you can see if q by p is more than 1 which is some quantity which is let us say 1.1 or something. So, an l is large it is going to be exponentially large eventually a small 0.1 is sufficient to make it grow to a large value as l tends to infinity. So, we can now consider case 3 which is this will be more clear when we look at case 3 that is p greater than q that is a winning streak case. So, then for the case 3 we can write since p is greater than q we can write for convenience sake it is slightly differently I mean same way, but we can now write it as q by p to the power k divided by 1 minus q by p to the power l. So, q is less than p winning streak now let us see what happens if l tends to infinity person says I have just let us say I have to only 10 dollars and I want to know whether I have some chance of winning if I bet a million dollars or for that matter even higher amount what is my chance of winning that and I he has a feeling that he has a winning streak 5 percent 10 percent whatever. So, then this this formula says that as l tends to infinity f k l tends to be a finite nonzero finite value nonzero finite is of course, finite, but nonzero value given by 1 because when l tends to infinity q by p will tend to 0 why because p is larger than q. So, it will be a smaller number less than 1 and any small number to the power infinity is going to be 0. So, it is going to be therefore, only numerator left which it does not depend on l which is q by p to the power k. So, now of course, if it starts with k very small this probability is very small ultimate probability, but if it starts with the reasonable amount of k knowing his own assessment of q divided by p this advantage of this success probability could be significant to win a very large amount. So, we illustrate these 3 cases by a graphical form with respect to the starting site k say let us say up to some l value here. Then we first saw that for the unbiased winner that is p equal to q equal to half it is a straight line let us terminate here. So, this is a straight line and for a player with higher p value there is a probability that he could win starting from k equal to l definitely he will win. So, the probabilities this is let us say l here then, but then the probability is always above than an unbiased case. So, this is for p greater than q in the same fashion for a person p less than q the probability will go as a sub linear curve. So, this is a supra linear curve and a sub linear curve when you reach l of course, the probability is always 1 because you are almost at the winning point and interestingly the problem now therefore, is the ultimate willing probability when the betting amount is very high that probability is 0. So, the so finally, as l tends to infinity f k l tends to 0 if p equal to q again tends to 0 in a more faster manner if p is a less than q tends to a value 1 minus q by p to the power k if p greater than q. So, this case is of quite some interest and if you now look back at the derivations we made for the problem of an absorber just single absorber case which defined the point 0 of a semi infinite lattice over the positive side of the x axis then l tending to infinity is a very similar to that problem. Now, the point l is now going to infinity. So, that success point actually does not have much role to play, but the point 0 which is for any finite k k is closer to 0 than l has an important point a important role in deciding on the final success probability. We showed that for an unbiased random walk in the presence of a absorber also had exactly the similar results for an unbiased random walker is the probability of ultimate escape from the point 0 was 0. For a biased random walker if his forward jump probability was weaker than his backward jump probability then also is the probability of escaping the absorber was 0. In contrast the probability that the random walker will escape the absorber was a finite for the case when p was greater than q and we gave an example of a positive and a negative ion being driven by an electric field from a plane parallel plate. So, as we mentioned this is a plane electrode and apply a positive potential with respect to a far of point then positive ion would escape and negative ion would deposit on the absorber and there is no chance for the negative ion to escape the absorber. Whereas, there was a finite probability that the positive ion would escape the absorber and in fact, this is a method that is used for collecting judge particles. So, there is a close similarity between an apparently entertainment problem like the gamblers ruin with respect to the real physical system behavior. Most excitingly and interestingly the analogy does not stop here more and more research has been progressing in this area of finding out the probability of either going one way or the other by of course, generalizing the p by q ratios to handle problems on the one hand connected with the nucleation theory on the other hand connected with the phase separation in granular media and all that. How systems are separate from one phase to another in a binary system for example, there is a certain connection to this gamblers ruin problem and that is why it is of utmost interest. In the course of one study and research one will I am not surprised if somewhere one finds that your problem can be formulated in some measure in the lines analogous to the concept of ultimate probability that we defined and in the solution that we obtained correspondingly. Just to give taste of how it is connected to a problem like a nucleation for example, analogy with the nucleation problem. Those of you who have some idea of nucleation or would have heard of something called a critical nucleus would know that just because water vapor has reached a super saturation you do not get a rain or you do not get large drops. It has to pass through a process of forming a small particles first small droplets and these droplets have to grow. They have to grow to a certain size beyond which other processes take over and large particles are formed large droplets are formed. So, in the early stages even though thermodynamics favours the formation of a liquid phase actual kinetics dis favours the formation because small particles have a large surface area and the surface free energy offers resistance. All this basically implies for asking a question that if this is the droplet size space or droplet mass space or monomer number of monomer number of atoms molecules in a droplet and we say 0 0 mass then each monomer adds a certain mass and let us say there is a critical size which has to grow and if that size is let us say L then you can ask a question that that under a certain condition if one starts with the droplet of n mon k monomers what is the probability that it could execute a kind of random walk with the certain probabilities of p and q's and eventually reach L. It is a very genuine question in nucleation theory. However, to formulate in the way we have done we cannot assume p equal to q or even constant p and q. This problem so, what is the question is what is the probability a molecule that a cluster or a droplet starting with the k monomers. Monomers means its molecules eventually reaches a critical size L of course, exact nucleation theory tells that reaching a critical size alone is not sufficient, but it is a precondition. It is much more nucleation theory is much more than just reaching a critical size, but at the moment let us go to the minimum requirement condition. So, this is the question. So, then we have to now postulate that if it is already a droplet somewhere let us say you have put a seed some seed particle of some size k then you have to have the probability of its going to k plus 1 and a probability of it becoming k minus 1. What are these called? Yes. So, p and q, p's and q's represent condensation of a monomer and evaporation processes equivalently. So, the winning streak and a losing tendency is equivalently represented by the condensation and evaporation processes in a case of a droplet. So, they can therefore postulate we do not need absolute values of condensation rates or evaporation rates. All that one needs is condensation rate divided by evaporation rate because if you see it will be always p by q ratio. So, once you know the formula all many terms drop out, tendency terms only remain and p by q can be formulated. And let us say that we have information let us assume p by q or q by p either way or to begin with p's and q's are site dependent or site when I say site here it is actually the volume dependent. So, but to be very specific it is more convenient to call it a size dependent that is p has to be now replaced with p i where i is the site index i varying from 1 to l and i is the molecular monomer size and similarly q be replacing q i. Now, this is where one requires a an engineering mapping of a mathematical situation what we identify as variables how does it connect to the reality of the process. So, this is where the ingenuity lies. So, we can see that when we do this identification of p i being the condensation rate condensation probability that is probability that a droplet acquires 1 molecule equivalent to 1 dollar bed 1 molecule and becomes i plus 1 a droplet of size i we should say a droplet of i molecules becomes or becomes a droplet of i plus 1 molecule. Analogously we can define the evaporation rate probability as a q i is again the probability that i the droplet and i droplet let us write it as i droplet evaporates evaporates 1 molecule and becomes i minus 1 droplet. So, we have exactly the same formulation and we ask now a question same way ask the question about ultimate question ultimate probability for the droplet for the system here system is the droplet to reach the critical required for onward growth. So, if you follow the steps it will be we go back to the steps it will have the same formulation like for example, as we seen equation yes as we saw in our equation 1. So, exactly, but we must then redefine the q's and a p's in terms of the site itself they are no longer constants, but the methodology runs throughout. So, just to keep the continuity we just write that here also I could write f k l probability that it will reach l that is the critical cluster starting from a droplet of k molecules is now q k that is important point f of k minus 1 l the probability that it would shrink by 1 and then from there it will become a critical cluster eventually any number of time is allowed one has enough time let us say. So, it shrinks by 1 molecule and then eventually becomes a critical cluster or it grows by acquiring one molecule through a probability p k and then it goes over to become a critical cluster l. Hence, we see a exactly the same scenario here and the same boundary conditions also work if there is virtually no molecule no droplet there is no chance of it ever reaching l. So, you have the same issue f not equal to 0 and if it is already a critical cluster then it exists. Hence, the probability if k equal to l is 1 and further we have we can formulate our p's and q's in such a way that p k plus q k is always 1. So, so actual rates will depend on times. So, fundamentally how we manage to get rid of the actual dynamics and kinetics is because the concept of step is a far reaching thought. The real time there is no there is a constraint of having to do it in a real time. In steps all that matters you you are free to take your next step the system will wait. So, as a result of which actual rates do not matter the relative probabilities only matter. So, p k plus q k equal to 1 simply implies that in the next step it has to either shrink or to evaporate. So, that is all. So, in a computer simulation it is very easy to execute these things. So, once this assumptions are made formulations are set you just have to go through every step that we have done virtually runs through like for example, the we defined j k here it will be the same way our j k definition is going to be as earlier j k is going to be f k minus 1. Now, l we are just omitting it is going to be there at the end. So, you will arrive at a general formula now j k define and 1 arrives j k is going to be again with an unknown j 1, but it is not a closed solution you have to write it as a product now k equal to 1 to let us say i the dummy index is i equal to 1 to k minus 1 of the ratio q i by p i. So, you need now here you need physics you cannot do it formally, but nevertheless it has been reduced to a product for the j k and then from this j k we can easily go over to f k evaluation because the same steps go on putting the values etcetera and we get with this we get. So, f n for example, is going to be sigma i equal to 1 to n j k some of all these fluxes is going to decide the value of f n at any point up to a point some up to a point n. So, again use the boundary conditions etcetera that is use f l equal to 1. So, this leads to the solution of implies j 1 for example, you can get it as 1 by 1 plus sigma k equal to 2 to l we just want to be careful about not taking the very first monomer. So, we that is why this that is why we separated it out. So, we evaluate j 1 and from this point the final answer is very clear now because if you follow the same steps f n l is going to be you can actually write it as some product evaluated at n divided by 1 plus some product evaluated at l where sp is just a definition sp of any value k is defined as k equal to 2 to k equal to 2 to not k let us keep it dummy i equal to 2 to k of product the sum and product together. So, i equal to 2 to k and let us say j equal to 1 to i minus 1 it will be. So, that is sp k that is of u i by p i. So, that is the definition. So, if you can evaluate f sp k then you say assign values of k equal to n and k equal to l and you have the probability ratio. This is basically the aim of introducing this example is not to actually solve a nucleation problem we do not intend to do that. Want to give you a taste of how stochastic techniques that we are learning in this paper modeling techniques is in this study is useful and can be rewarding to attempt at real world physical engineering problems. So, with this basic spirit in mind we will explore phenomena several phenomena and also give examples of what is the physical implication of this phenomena in engineering problems. Thank you.