thank you for starting this confusion to distract from the correct answers, but I have to spoil it.
If each of 4 runners has, independently, a 75% chance of making the team, then there's a .25^4 > 0 chance that none do. Unless, as the previous poster suggested, there are no more than 4 runners in the race.
I don't think so. Even if there are other runners, if there is any probability that any other runner beside these four gets top three, then the probability of at least one of the four runners has to be less than 75%. This is because there are only three spots available.
Think of the limit where the probability of each runner is 99%. Clearly this is an impossible scenario, because we know for certain that, in any possible solution, at least one of the runners will be out. There is not a distribution of possible outcomes where the probability of all 4 runners is 99%. I am too lazy to prove it mathematically, but my intuition is that this limit is 75% and in this case, all possible outcomes must involve 3 of the 4 runners on the podium.
I can see why you are thinking along these lines, but it misunderstands the premise, which again is that each runner independently has a 75% chance of making the team. All deductions proceed from that.
I'm going to avoid the rabbit hole here and just say even if they all independently had only a 25% chance of making the team, then the probability of all 4 making the team would be greater than 0 (.25^4), disproving the premise by contradition. In general, it's not possible for all 4 to independently have a > 0 chance when there's only 3 spots. They can't all even have a 1% chance, or they might still all make it. That math cannot describe this race accurately.
The independence assumption for Rojo’s scenario is reasonable as one runner having a good race doesn’t (shouldn’t) impact another runners performance being good or bad (i.e. blowing up or not). However, assuming independence when the probability is defined as making the team doesn’t work. As it doesn’t realistically model the circumstances mathematically.
Assume a race of N total runners. Each runner has a probability density function assigning a probability to each of the potential N finishing places. The probability can’t be zero for any Nth place and the probabilities have to sum to 1 for each runner. Additionally though, the probabilities for each finishing place, 1 through N have to sum to 1 over all runners (assume no DNF’s).
Under these conditions, if 4 runners have a 75% chance of making the team, that’s because they each have a 25% probability of finishing 1st, 2nd, 3rd, or 4th. This is a uniform distribution btw, and in general scales based on the number of entrants, where to maintain uniformity across the N finishing places and participants, the probability assigned to each of the N participants finishing in places 1 through N would result in a probability of (N-1)!/N!.
The distribution does not need to be uniform though. In fact you’d likely have severely right skewed pdf’s for the favorites and severely left skewed pdf’s for the runners who just dipped below the qualifying standard.
Also, to be technically correct, these are probability mass functions since we are dealing with discrete variables. Each runner finishes in a position, 1 through N.
I don't think so. Even if there are other runners, if there is any probability that any other runner beside these four gets top three, then the probability of at least one of the four runners has to be less than 75%. This is because there are only three spots available.
Think of the limit where the probability of each runner is 99%. Clearly this is an impossible scenario, because we know for certain that, in any possible solution, at least one of the runners will be out. There is not a distribution of possible outcomes where the probability of all 4 runners is 99%. I am too lazy to prove it mathematically, but my intuition is that this limit is 75% and in this case, all possible outcomes must involve 3 of the 4 runners on the podium.
I can see why you are thinking along these lines, but it misunderstands the premise, which again is that each runner independently has a 75% chance of making the team. All deductions proceed from that.
I'm going to avoid the rabbit hole here and just say even if they all independently had only a 25% chance of making the team, then the probability of all 4 making the team would be greater than 0 (.25^4), disproving the premise by contradition. In general, it's not possible for all 4 to independently have a > 0 chance when there's only 3 spots. They can't all even have a 1% chance, or they might still all make it. That math cannot describe this race accurately.
The premise is each of the four (independently) has a 75% of winning won of the top 3 spots. This takes up all three spots, leaving 0% for the rest of the field.
Say 3 runners each had 100% chance of making the team, you can see that the rest of the field would have 0% chance, no?
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Under these conditions, if 4 runners have a 75% chance of making the team, that’s because they each have a 25% probability of finishing 1st, 2nd, 3rd, or 4th. This is a uniform distribution btw, and in general scales based on the number of entrants, where to maintain uniformity across the N finishing places and participants, the probability assigned to each of the N participants finishing in places 1 through N would result in a probability of (N-1)!/N!.
Thanks for putting it down in math terms. Only one correction: to make the team you need to get top 3, not top 4. Thus, if 4 runners have a 75% probability each to get one of the top three spots, their combined probability for the first three spots is 300% (75%*4). It does not really matter how you distribute it, there is no probability left in the first three spots for any other runner.
These bots really can't be trusted. I just rephrased it as follows.
4 people are going to the casino. They each have a 25% chance of returning home with zero dollars. What percent of the time will 2 or more of them return home with zero dollars?
Bard said this: There are several ways to solve this problem. Here's one method: Calculate the probability of no one returning with zero dollars: This is the opposite of what we're looking for, but it's easier to calculate first. The probability of one person NOT returning with zero dollars is 1 - 0.25 = 0.75. Since the outcomes are independent (their results don't affect each other), the probability of all 4 not returning with zero dollars is 0.75^4 = 0.31640625. Subtract from 1: Since the two events are complementary (all 4 not losing with 2 or more losing), the probability of 2 or more losing is 1 - 0.31640625 = 0.68359375. Convert to percentage: Multiply by 100% to express the answer as a percentage: 0.68359375 * 100% ≈ 68.36%. Therefore, there is approximately a 68.36% chance that 2 or more of the 4 people will return home with zero dollars.
Chatgpt acutally agreed but phrased it differently.
To solve this problem, we can use the complement rule. The probability of at least 2 people returning home with zero dollars is the complement of the probability that none or only one person returns home with zero dollars. Let's calculate the probability that none or only one person returns home with zero dollars: Probability that all four return home with money: �(All return with money)=(0.75)4P(All return with money)=(0.75)4 Probability that exactly one returns home with zero dollars (and the other three return with money): �(One returns with zero dollars)=(41)×(0.25)1×(0.75)3P(One returns with zero dollars)=(14)×(0.25)1×(0.75)3 Now, let's find the combined probability of these two events: �(None or only one returns with zero dollars)=�(All return with money)+�(One returns with zero dollars)P(None or only one returns with zero dollars)=P(All return with money)+P(One returns with zero dollars) �(None or only one returns with zero dollars)=(0.75)4+4×0.25×(0.75)3P(None or only one returns with zero dollars)=(0.75)4+4×0.25×(0.75)3 Now, the probability that 2 or more return home with zero dollars is the complement: �(2 or more return with zero dollars)=1−�(None or only one returns with zero dollars)P(2 or more return with zero dollars)=1−P(None or only one returns with zero dollars) Let's calculate this: �(2 or more return with zero dollars)=1−[(0.75)4+4×0.25×(0.75)3]P(2 or more return with zero dollars)=1−[(0.75)4+4×0.25×(0.75)3] �(2 or more return with zero dollars)≈1−0.68359375P(2 or more return with zero dollars)≈1−0.68359375 �(2 or more return with zero dollars)≈0.31640625P(2 or more return with zero dollars)≈0.31640625 So, approximately 31.64% of the time, 2 or more out of the 4 people will return home with zero dollars.
Its incredible to see you keep rephrasing it to bots, and complain over people giving you different answers rather than being able to pick out the obvious correct answers here.
Under these conditions, if 4 runners have a 75% chance of making the team, that’s because they each have a 25% probability of finishing 1st, 2nd, 3rd, or 4th. This is a uniform distribution btw, and in general scales based on the number of entrants, where to maintain uniformity across the N finishing places and participants, the probability assigned to each of the N participants finishing in places 1 through N would result in a probability of (N-1)!/N!.
Thanks for putting it down in math terms. Only one correction: to make the team you need to get top 3, not top 4. Thus, if 4 runners have a 75% probability each to get one of the top three spots, their combined probability for the first three spots is 300% (75%*4). It does not really matter how you distribute it, there is no probability left in the first three spots for any other runner.
Correct, to make the team you need to finish top 3 and it does not matter how the probability is distributed among the 4 runners, as long as each of their probabilities sum to 75% for place 1, 2, and 3. All 4 of them will have a 25% probability of finishing 4th though. I was assuming uniformity of probability across place for simplicity. Though, as I pointed out after that, the distribution doesn’t need to be uniform.
The main point I was making is in the scenario you suggested (4 runners with 75% chance to make the team, what is the probability that the 3 spots are taken by these 4), the field can only be made up of 4 runners, so makes the answer self-evident as of course the team will be made up of the runners in the race, which I don’t think was your intention?
Thanks for putting it down in math terms. Only one correction: to make the team you need to get top 3, not top 4. Thus, if 4 runners have a 75% probability each to get one of the top three spots, their combined probability for the first three spots is 300% (75%*4). It does not really matter how you distribute it, there is no probability left in the first three spots for any other runner.
Correct, to make the team you need to finish top 3 and it does not matter how the probability is distributed among the 4 runners, as long as each of their probabilities sum to 75% for place 1, 2, and 3. All 4 of them will have a 25% probability of finishing 4th though. I was assuming uniformity of probability across place for simplicity. Though, as I pointed out after that, the distribution doesn’t need to be uniform.
The main point I was making is in the scenario you suggested (4 runners with 75% chance to make the team, what is the probability that the 3 spots are taken by these 4), the field can only be made up of 4 runners, so makes the answer self-evident as of course the team will be made up of the runners in the race, which I don’t think was your intention?
Yes! Correct, correct, correct. So that can't be what rojo's question meant, can it?
Correct. Rojo's question is not asking the same thing as the scenario posed by mid D Guy. Rojo was asking if there are 4 runners who have a good race 75% of the time, what is the probability that 2 or more of the 4 runners do not have a good race. The binomial theorem is the correct distribution to use in that situation.
I was a math major so maybe I’m a bit delusional about most people’s basic math skills, but this question is so simple that any adult human should be able to calculate for themselves in a few seconds. Embarrassing that the OP is an Ivy League grad.