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Contest Design In head-to-head wrestling match. Georg can beat Jakob with probability p-0 and Georg can beat Ulf with probability q?0. Ulf is tougher opponent than ...

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Contest Design In head-to-head wrestling match. Georg can beat Jakob with probability p-0 and Georg can beat Ulf with probability q?0. Ulf is tougher opponent than Jakob so p>q To win contest. Georg has to win at least rOw out of 3 head- head matches against Jakob or Ulf and Georg cannot play the same person twice TOW That is. Georg can play either Jakob-Ulf-Jakob or Ulf-Jakob-Ulf: Show that Georg maximizes his probability of winning the contest by playing the better opponent; Ulf; [Wice_

Contest Design In head-to-head wrestling match. Georg can beat Jakob with probability p-0 and Georg can beat Ulf with probability q?0. Ulf is tougher opponent than Jakob so p>q To win contest. Georg has to win at least rOw out of 3 head- head matches against Jakob or Ulf and Georg cannot play the same person twice TOW That is. Georg can play either Jakob-Ulf-Jakob or Ulf-Jakob-Ulf: Show that Georg maximizes his probability of winning the contest by playing the better opponent; Ulf; [Wice_



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A player of a video game is confronted with a series of opponents and has an $80 \%$ probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents. (a) What is the probability mass function of the number of opponents contested in a game? (b) What is the probability that a player defeats at least two opponents in a game? (c) What is the expected number of opponents contested in a game? (d) What is the probability that a player contests four or more opponents in a game? (e) What is the expected number of game plays until a player contests four or more opponents?

The scenario described for this exercise is somebody playing a video game In which they confront opponents and have an 80% probability of defeating each one if they keep playing until they are defeated by their opponent. And so the probability successes always 80% or the probability of defeating the opponent is always 80% and it's independent of previous encounters. So basically what's being described our Bernoulli trials and so the game goes on until the player is defeated. So let's call the probability P of success 0.2. So that is success. Is the player being being defeated by their opponent. And let's describe the random variable X as the number of opponents required to play against before being defeated. So basically the number of trials until a success. And so X is a geometric random variable. And so for part A. Were asked what is the probability mass function of the number of opponents contested in a game? So that is simply the probability mass function for a geometric random variable. Now, for part B, we are asked for the probability that the player defeats at least two opponents in the game, which means that it takes At least three trials to face defeat. That is we need at least three trials for the success. So we're looking for the probability that X Is at least three, just the one minus the probability that X is less than or equal to two. So the for the probability of X equals one, that is simply P And the probability that x equals two Is 1 -0.2 times 0.2. So this is one 0.36. Which comes out to 0.64 for part C. Were asked for the expected number of components contested in the game. So this is the expected number of trials until the success This is given by one overpay. So the expected number of opponents faced before losing is five for part D. We are asked for the probability that the player contests four or more components opponents in the game. This is 1- the probability that X is at most three. So just borrowing from part B, we have the probability that X equals one and the probability that X equals two. So those some 2.36. And now we need the probability that x equals three added to that. So that will be .8 squared times 0.2. This comes out 2.488. And then for part D. Were asked what the expected number of game plays until a player contests four or more components is. So up until now each trial was facing an opponent within the game. And we were talking about the number of opponents that the player would face before being defeated. Now we are talking about the number of games played Until the player contests at least four opponents. So this is a new random variable. Now we know what the probability successes. It's already given to us from part D. So that's .488. That is the probability that a player will face at least four opponents in any given game. So now let y be the number of games played until The player faces at least four opponents. So what this question is asking for the expectation on why? And since this is a geometric random variable. So what I'm saying is the random variable why is also a geometric random variable? And it's probability successes .488. So how many trials or how many games are expected to be played until our first success? That is one of our P. And that comes out to 2.049. So the number of games we expect to be played before the player faces, at least for opponents within any game Is 2.049.

The scenario described for this exercise is somebody playing a video game In which they confront opponents and have an 80% probability of defeating each one if they keep playing until they are defeated by their opponent. And so the probability successes always 80% or the probability of defeating the opponent is always 80% and it's independent of previous encounters. So basically what's being described our Bernoulli trials and so the game goes on until the player is defeated. So let's call the probability P of success 0.2. So that is success. Is the player being being defeated by their opponent. And let's describe the random variable X as the number of opponents required to play against before being defeated. So basically the number of trials until a success. And so X is a geometric random variable. And so for part A. Were asked what is the probability mass function of the number of opponents contested in a game? So that is simply the probability mass function for a geometric random variable. Now, for part B, we are asked for the probability that the player defeats at least two opponents in the game, which means that it takes At least three trials to face defeat. That is we need at least three trials for the success. So we're looking for the probability that X Is at least three, just the one minus the probability that X is less than or equal to two. So the for the probability of X equals one, that is simply P And the probability that x equals two Is 1 -0.2 times 0.2. So this is one 0.36. Which comes out to 0.64 for part C. Were asked for the expected number of components contested in the game. So this is the expected number of trials until the success This is given by one overpay. So the expected number of opponents faced before losing is five for part D. We are asked for the probability that the player contests four or more components opponents in the game. This is 1- the probability that X is at most three. So just borrowing from part B, we have the probability that X equals one and the probability that X equals two. So those some 2.36. And now we need the probability that x equals three added to that. So that will be .8 squared times 0.2. This comes out 2.488. And then for part D. Were asked what the expected number of game plays until a player contests four or more components is. So up until now each trial was facing an opponent within the game. And we were talking about the number of opponents that the player would face before being defeated. Now we are talking about the number of games played Until the player contests at least four opponents. So this is a new random variable. Now we know what the probability successes. It's already given to us from part D. So that's .488. That is the probability that a player will face at least four opponents in any given game. So now let y be the number of games played until The player faces at least four opponents. So what this question is asking for the expectation on why? And since this is a geometric random variable. So what I'm saying is the random variable why is also a geometric random variable? And it's probability successes .488. So how many trials or how many games are expected to be played until our first success? That is one of our P. And that comes out to 2.049. So the number of games we expect to be played before the player faces, at least for opponents within any game Is 2.049.

Okay, The team starts down by 14 points, but then they score two touchdowns worth six points each. So then they'll only be done by two points. Yeah, that means that among the points they might score in the decisions that we have to make between one and two points. If they score 0 to 1 point, they will lose. If they score two points, they will tie, and if they score three or more points, they will win. So thes are the probabilities that we need to find out the probability that they will get 01 points two or more than three. So Part A says, what if they try both times to score one point? So with a 99% chance of success twice, then the probability that they get zero points, meaning that they fail both of those attempts is 0.1 That's the opposite of this time, 0.1 That's failing both times probability that they get one point would be the probability that they succeed on one of those to attempt come yeah, and fail on the other one. And since there's two orders that you could do that in. They could succeed and then fail or fail and then succeed. We have to multiply this by two. And that probability is just 0.198 which is 1.98 her site and the probability that they get two points. They succeed on both of these attempts, 0.990 point 99 is 0.9801 Yeah, so back to our understanding of how many points you need to win how money you need to lose. Yeah, zero points would be a loss. So that's 0.1%. Lose one point. Also a loss. That total is right. The probability that they tie two points is 98.1%. Yeah, and they can't possibly one because they can't get more than two points if they're trying to get one point twice. Okay, so those three probabilities are the answer to part A. I'm going to erase this because there's not a lot of room here. Look, that's it

Mhm. Yeah, the team starts. That's down by 14 points. Mhm. And we know that they'll score two touchdowns, six points each, so they're going to gain some points. Okay, they gained 12 points, and that means that they're down by two. We've been Yeah. Okay, aside from the choices that we have to make next, so if the coach decides to attempt to get one point twice, Okay? Mhm. Yeah. Then we know that the probability that they win is zero, because, yeah, they can't possibly Even if they get both of the points that they attempt, they'll have tied the game so they can't possibly win. The probability that they tie okay is the probability that they get exactly two points. So they succeed in both of their attempts. But each attempt has a 99% chance of success, so the probability that they both succeed mhm is 99% times 99%. This is 98.1 Perfect. Mhm. And the probability that they lose. Yeah, well, that's the only other option. So we can just subtract the probability of the other things, which is zero plus 0.9801 mhm. Okay, so that's 1.99%. These three probabilities are the answers for part day. No, part B. Cool. Yeah. Suppose that the coach decides to attempt to points twice? Yeah. Okay. So the probability that they win is the probability that they get more than two points because they're down by two. If they get exactly two, they tie, so they need more than two points to win. So they would have to get all four points because it's not possible to get three if you attempt to twice, so they have to get all four points. That's the only way that you can get more than two. So they have to succeed at this On both drives, the probability of success is 45%. Okay, Yeah, yeah, mhm probability that they tie because the probability that they get exactly two points yeah, which could happen in one of two ways, either by getting the first attempt of two points on failing the second or the other way around. You say all the first and succeed on the second, so there's two ways to do this and then the probability of either one of them is okay. A probability of succeeding and of failing. Mhm. Okay. And then the probability that they lose, we could dio the probability that they get zero points so that they fail this twice. And that's 0.55 times 0.55 But uh huh, I would rather use the complementary probability. So one minus these other two answers. So 0.20 to 5 minus zero point 495 which is gonna make more room for that. So these three probabilities, our answers for part B and in part C asks if it's possible to make a strategy where you are more likely to win than to lose. And we can see that in neither of the previous cases. Is that true? In both of the previous cases, it's more likely that you lose than that you win. And there's only one other possible strategy which is to attempt one point and two points. So we're just going to see if that strategy meets the conditions that this part of the problem asks for. Okay, Okay, Okay. Okay. That's okay. Okay. Yes. Mhm. I'm gonna start by looking at the probability that the team gets one point using the strategy, and that probability is, yeah, the probability that you succeed on the one point shot and fail on the two point shot, and that probability is 54.45 percent, and that's a loss. So if you only get one point, the team loses and the probably that they lose is actually more than 54.45 because there's also the possibility that they get zero points. But we don't need to know all of those probabilities to explain why this strategy does not give you a better chance of winning than losing, because the probability of losing is more than half because it's more than 54.45 So there's just no way for the probability of winning to be greater. And this was the only other strategy that existed besides the ones in part A and part B. So, no, there's no such strategy that fits the conditions of this problem. Okay, Mhm, and that's it. That's all three parts. You could calculate more of Part C if you wanted to, but it's not necessary


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