 Hi, I'm Zor. Welcome to Unizor education. I would like to present to you one problem which is actually subdivided into sub problems, if you wish. I put it in the category of advanced problems in theory of probabilities. Well, it's not really a difficult problem, but it's an important one. I consider, if we will go through all the details for this problem, it will be very beneficial for us. Okay, this is part of advanced math course presented on Unizor.com, which is actually intended for teenagers who are willing to spend time and efforts to get deeper understanding of mathematical concepts, which basically aimed at just development of their minds. That's the most important goal of mine. Alright, so let's go to this particular problem. Here is the general description. We're playing a game, a game of roulette. According to, I think it's called American rules or Las Vegas rules, whatever it is, there are numbers from 1 to 36 and then there is zero and double zero on a spinning wheel and numbers can be red or black. When the dice falls on zero, double zero, you're losing. Now, if it's one of these, it depends on your bet. Now, you can bet on a particular number or you can bet on a color. For instance, there are two colors, red and black. And depending on what exactly your betting, your payoff is different, obviously. We will consider only one way of betting. Let's say you're betting on a color red. So half of these 18 numbers are red and the other half is white, no, black, sorry. So if it's one of these, it means it's 18 red numbers. Altogether, we have 38 numbers, right? 38 cells on the wheel. So basically, if you bet on the red, your probability of winning is 18 38s, right? And probability of losing correspondingly is 20 38s. So this is wind and this is loose. That's probability on a single spin. Now, let's talk about strategy. The person comes and thinks about this way. I will bet one unit of currency. Let's say, let's call it Bitcoin. So I'm betting one Bitcoin on the first spin of the wheel. If I lose with this probability, I will double the amount. I will put two coins, two Bitcoins next time. If I will lose, I will double the game. I will put four coins, etc. But if I win, on the first win I get, I will finish the game. So what happens is the following. Let's just discuss it in more details. Let's say you're spinning wheel just once. There are two possibilities. Other you win, in which case you bet one Bitcoin and the payoff if you are betting on the color is double your bet. So it will be minus one you bet and then you get two if you win. So you're actually winning one, right? Or you can lose. In which case you are basically losing your one bet. And then you start the next game and you're putting double the amount, which is two. Now you can always either win or lose. Now you are betting two now, but now if you win, you get double. Which means you end your game again with one Bitcoin in plus. And if you're losing, well, you're losing your minus two Bitcoins and you're playing the third time, doubling again. It can be win or lose. And now you're betting four. Now if you're betting four, you are getting back eight if you win. So you're losing this, this, this, and then you're winning this. Minus one, minus two, minus four, again one. So it looks like on every time if you win, you are gaining one Bitcoin. And that's the end of the game. Or if you're losing, you continue the game by doubling the amount. So I would like to basically research this particular strategy, how good it is or how bad it is. So we know the game and we know the strategy. So let's go to the problems. Okay, let's consider that you would like to be able to stay in the game for end spins. That's your goal. Let's say spin takes a minute. And you would like to be able to be in the casino for 10 minutes for sure. Well, but you're doubling the amount all the time, right? Which means you have to have enough money to double the amount for end times in a row. Now, if you would like to be able to stay in the game for end spins, you have to be prepared to bet end times. So the first time you bet one and you lose. The second time you bet two and you lose, et cetera, et cetera. And on the nth spin, this is two to the power of zero to the power of one. So that should be n minus one. So that's your bets. And you're losing all the time because you're continuing the game. You have to be able to stay in the game for end times, for end spins. Which means you have to have this sum. That's the amount of capital which you need. So that's my first problem. What is the amount of capital which you need to be able to assure yourself to stay in the game for end spins, for end times of spinning? Well, obviously this is equal to what? Two to the nth degree minus one. Now this is a geometric progression. And obviously I don't remember the formulas, but I do remember that to summarize the geometric progression, you have to multiply the sum by the factor, factor is two and subtract. If you subtract from this this, you will get s is equal to two to the power of n minus one. Now I will write this out and that's my answer to the first problem. You need two to the power of n minus one capital, bitcoins, if you wish. If your initial bet is one bitcoin, you need two to the power of n minus one bitcoins. If your initial bet is ten dollars, well you need two to the power of n minus one times ten dollars. Alright, so that's my first problem. Next problem. Next problem is the following. I would like to list all the elementary events, mutually exclusive events, which can happen during these n games for which you have enough capital. So you have, let's assume you have the capital two to the power of n minus one, which means you can play up to ten, up to n games. And my question is what are mutually exclusive results of these experiments? Well, first you spin once and you win and this is the end of the game, right? So the event number one is win the first spin. Event number two. You lose the first spin and win second spin. And that's the end of the game because you have strategy to end the game as soon as you win, right? Event number three. You lose first and second game, but you win the third. Event number n, you lose first, second, etc. n minus first game and then you win nth. And finally, event number n plus one, you lose first, second, etc. n minus first and nth game, the very last game you end. And that's also the end of the whole game because you don't have any more money. You're running out of money, which means this is the end of the game. So these are completely independent, mutually exclusive events. You can call them elementary events which comprise the whole game. The whole game can be divided into these outcomes. There are no more outcomes if you have two to the power of n minus one capital and the first bet is one. So that's my second problem to list all the different outcomes. Next. Well, obviously the next one is to attach the probability to each one of them. Well, let's just think about it. What's the probability of winning the first spin? Well, the probability of winning once is 1838, right? So this is 1838. Now, this, you lose the first one, the probability is 2038 and then you win the second. You lose two times in a row. Now, by the way, the events are completely independent. That's why I'm multiplying the probability to signify that both should be actually happening. Now, these are three different spins independent. The probability of first and the second is 2038, so it's square. And the probability of this one is 1838, winning for the third one. How about this one? You lose n minus time, spins, right? So it's 2038 to the power of n minus one times 1838, the very last one. And finally, it's 2038 to the power of n minus one, n, sorry, n, n times in a row you lose. That's the probability of each event, each outcome of this particular game. Now, just to be sure that this is correct, I always recommend to add these together and check if you get one as a result of this, because the sum of all the probabilities must be equal to one. Let's just check it out. Okay, so we have to summarize these guys. Well, these all have 1838 as a factor, so it would be 1838 times one plus 2038 plus 2038 square, etc. plus 2038 to the power of n minus one, right? And plus this one equals... Okay, what is this? Well, again, this is geometric progression. And again, if you remember, to summarize it, you have to multiply it by the factor and subtract. So, let's just count it separately. If you have s equals one plus two plus, etc., plus q to the power of n minus one, then qs is equal to... oh, sorry, not two. This is a q. Then q square, etc. So, the q times s is q times q square plus, etc., plus qn minus one plus q to the power of n, right? I'm multiplying each member by n, subtract. Now, these guys are all cancelling out. And I have s one minus q equals one minus q to the power of n, right? So, s is equal to one minus q to the power of n divided by one minus q. So, let's do it here. Now, q is 2038, right? So, I have to replace this with this s, which is one minus 2038 to the power of n divided by one minus 2038. All right? And this one. One minus 20 over 38 is 1838. And this is 1838. So, these two are cancelling out. Now, you have one minus 2038 to the power of n and plus 2038 to the power of n, which means it's equal to one. So, my check is fine. Sum is equal to one. That's good. So, this is my next problem. So, I have proven that. Well, not proven. I didn't prove that all these numbers are correct. But I just checked some elementary procedure that it satisfies certain obvious rule. Now, my next problem is to find the expectation of the winning or losing, whatever it is. I mean, if it's negative, it will be losing. If it's positive, it will be winning, right? So, the expectation of the results of these games. Now, let's think about what is the expectation. If you have an outcome, outcome is basically the value of the random variable, right? Now, what is our value of the random variable, which is the winning? In this case, we win one. In this case, we win again one. You remember, we were basically calculating this. In the very beginning of this lecture, I was just talking about what we are winning if we are ending the game on the win. Summarize all the previous losses and the last winning. And I always have one as a result, right? So, I'm basically referring to this beginning of the lecture. And in all these cases, I have one as a result. That's my winning, except the last one. In the last case, I'm losing completely all my capital. Remember what capital was? My capital was 2 to the power of n minus 1. So, in this case, I'm losing this. All these cases, I'm winning one. All right. So, let's just calculate the expectation by, you know, the natural way straightforward. If you have a random variable which takes value x1 with the probability of p1, x2, x2 with probability of p2, etc. xn with probability pn, pn, then this is the definition of expectation, right? This is expected value. Now, our random variable takes value of 1 with probability of this, takes value of 1 with probability of this, takes value of 1 with probability of this, etc., etc. And takes the value minus 2 to the power of n minus 1 with this probability. So, let's just summarize and see how it works. So, it's 1 and 1 and 1 and 1 and 1. So, I'm basically multiplying the whole sum 1 plus 2038 plus 2038 square plus 2038 to the power of n minus 1. I have to multiply by 1. Well, I don't have to write it, right? It's multiplication by 1. And this one, 2038 to the power of n, I have to multiply by this with a negative sign, which is minus 2 to the power of n plus 1. So, that's my expectation. Okay, I don't need this anymore. Now, you remember, we were just calculating a second ago, we were calculating this. And this is 1838 here, 1 minus 2038 to the power of n divided by 1 minus 20 over 38. Now, this is cancelling out, right? And that's what I have. Now, this plus, well, actually it's plus 1. I can say plus 1, right? Plus 1. This is 1. And minus 20 times 2 is 4038 to the power of n. Something like this, right? Oh, no, that's not 1, sorry. This is 1 multiplied by 2038. So, 2038 to the power of n. That's what it is, right? 2038 times 1 and 2038 times minus 2 to the power of n. So, it's 20 times 2, which is 40 to the power of n. So, now this is out. Now, this is cancelling with this one. So, I have 1 minus 4038 to the power of n. So, that's my expectation. And now, let's go to my last problem, which is, is it good or bad? Well, obviously this is bad. Why? Because 4038 is slightly more than 1, obviously, right? To the power of n, it's even more than that. So, this is negative. Since expectation is negative, well, the more you play, the more you lose. What's important, actually, is that if you have a lot of money and you're playing, you're thinking that if you have a lot of money, your chances to win are more. Well, your chances to win are more, but your chances to lose are also very important. And if you lose, you lose more, right? So, the proper way is to compare the expectations. So, the more money you have, so the more the number of games you can, in theory, play losing all the time, the less your expectation is. Because you see this power of n, which means this is greater than 1, it's increasing and increasing and increasing. So, this becomes more and more negative. Which means this game is number one, not good for you. And number two, if you think that with more money, you are bettering your chances, you're wrong. Your chances are on average, I mean, obviously, things happen, but on average, if you play with a million dollars once, twice, twice, etc., etc., etc. On average, you will lose more than if you play with a dollar, right? So, that was my last problem. That number one, that's not a good strategy. And number two, the more money you have, the more you lose on average. And, well, you can conclude that any kind of gambling is not a good thing for you. Because it always wins. Well, unless you cheat or something like that. Alright, anyway, that was my first problem among so-called advanced probability problems. It's not as advanced, but it's kind of involved. And I think it pays to go through all the details, to go through all the elementary events, their probabilities, the expectation, etc., etc. That's it for today. Thank you very much and good luck.