Not sure if this is helpful (or correct), but perhaps it can lead you to the answer you are looking for.
The favorites are individually A, B, C and D. All longshots are X.
I believe these are the permutations where:
All 4 favorites finish in the top 4, no longshots finish in the top 4. 3 favorites finish in the top 4, 1 longshot finishes in the top 4. 2 favorites finish in the top 4, 2 longshots finish in the top 4. 1 favorite finishes in the top 4, 3 longshots finish in the top 4. No favorites finish in the top 4, 4 longshots finish in the top 4.
Here's a slightly more sophisticated perspective. Suppose instead of a bad race being fixed at 25% likely, it's randomly chosen between 0% and 50% (think of this as being a proxy for the conditions). If each of the 4 runners has a probability of p of "bombing", we can redo Hardloper's calculations:
The number for 2 or more is similar, but it is quite a bit more likely under this framework to have 3 or 4 of them bomb. The reason is that some races just have a lot of people bomb, and if that happens, it wouldn't be surprised to see 3 or 4 of these runners be in that category.
I've narrowed down the field to 4 studs whom I think will make the team. If they run well, they will beat everyone else in the field.
For simplicity sake, I decided to assume that if someone is in shape like these 4 that 75% of the time they will run good a marathon. Well what percent of the time do 2-3 or 4 of them run bad? I want to know because if 2 or more of them run a bad race, then my longshots have a chance.
Despite being a former HS math teacher, I am not going to try to figure it out myself. I've put the scenario into chatgpt and bard and get wildly different answers.
Initially I asked Bard and ChatGPT the following.
Calculate the odds of the following for me please. Four people are running the same race. Their odds of success are independent of each other. They each have a 75% of success. What percentage of the time do 2, 3 or 4 of them run a bad race?
ChatGPT said the "scenario you've described involves a binomial distribution" and said the answer was 30.86% whereas Bard said it was "32.42%".
I also asked the question in a different format and got wildly different results.
When I worded the problem as "Calculate the odds of the following for me please. I'm going to make 4 different investments. They each have a 75% of success. What percentage of the time do 2, 3 or 4 end up in failure?" Chat GPT said "approximately 30.86% of the time, 2, 3, or 4 out of the 4 investments will end up in failure" while Bard said the answer was just 5.08%.
So roughly what is the answer and please tell me why.
Trust LR to carry a simple probability question to three pages.
If only 4 people were running and 3 make the team and all 4 are of equal fitness, then and only then can you say that each has a 75% chance of making the team.
The odds of any of them bombing is unknown on the facts given.
This post was edited 2 minutes after it was posted.
If only 4 people were running and 3 make the team and all 4 are of equal fitness, then and only then can you say that each has a 75% chance of making the team.
The odds of any of them bombing is unknown on the facts given.
OMFH, yes, finally. Newname & automorphic seem to be the only ones who recognize that the probability of bombing HAS NEVER BEEN SPECIFIED in the question.
The 75% never addresses that at all. It refers to 4 people of roughly equal ability competing for 3 spots with roughly equal random error amongst those 4, and assumes they're all a significant tier above the rest.
Bombing refers to performing so poorly that they're not racing up to their normal tier, and a lower tier runner could beat them. The probability of that is separate from the 3/4 problem, and there is no way to calculate it without assuming how often a great runner bombs down way below his/her inter-trial fluctuation, such that they're at least as bad as the second tier. IOW, it is an outlier amongst that person's own distribution of relatively reliable high-tier performance. Yes, it could be p(bombing)=.25 (which would be only a coincidence, and that seems way too high) or it could be .09 or it could be .01 or it could be .15 or any other probability (but presumably low, hence why you'd call that outlier performance "bombing." But we don't know that probability, and it CANNOT be calculated from the information given.
This is similar to the notorious Monty Hall problem, in that so many smart people still often miss it by not thinking it through correctly. All of you are using faulty logic by failing to recognize that the key probability needed for calculation was never given, as poster automorphic noted above.
OMFG people.
This post was edited 7 minutes after it was posted.
A correct answer has been provided, but only if we assume that the probability of 75% refers to having a good race (i.e. not bombing). If the question asked is instead: "assume that there are 4 runners that each have a 75% probability of making the team, what is the probability that all three slots in the team are taken by one of these four athletes? Then it gets a little more tricky, because the slots available are only three and each person that makes the team reduces the probability of the others to make the team as well. in fact, the probability that all 4 make the team is zero. It is too early in the morning and I feel too lazy to work out the math, I'll leave that to the experts.
A correct answer has been provided, but only if we assume that the probability of 75% refers to having a good race (i.e. not bombing). If the question asked is instead: "assume that there are 4 runners that each have a 75% probability of making the team, what is the probability that all three slots in the team are taken by one of these four athletes? Then it gets a little more tricky, because the slots available are only three and each person that makes the team reduces the probability of the others to make the team as well. in fact, the probability that all 4 make the team is zero. It is too early in the morning and I feel too lazy to work out the math, I'll leave that to the experts.
Under your scenario, if each of 4 runners have a 75% chance of making the team, then all 3 slots will go to 3 of these four. Why? Because the rest of the field has 0% chance of making the team.
"Despite being a former HS math teacher, I am not going to try to figure it out myself."
Well, that explains why American students lag behind so much of the world in math...that and expecting math to answer questions that are not governed by the rules of math.
A correct answer has been provided, but only if we assume that the probability of 75% refers to having a good race (i.e. not bombing). If the question asked is instead: "assume that there are 4 runners that each have a 75% probability of making the team, what is the probability that all three slots in the team are taken by one of these four athletes? Then it gets a little more tricky, because the slots available are only three and each person that makes the team reduces the probability of the others to make the team as well. in fact, the probability that all 4 make the team is zero. It is too early in the morning and I feel too lazy to work out the math, I'll leave that to the experts.
Under your scenario, if each of 4 runners have a 75% chance of making the team, then all 3 slots will go to 3 of these four. Why? Because the rest of the field has 0% chance of making the team.
Yes, this is correct. The 75% case is trivial as the only distribution that allows for 75% probability to each of the 4 involves three of them being on the team, if the probability is higher, there is no solution, if it is lower, then it gets more interesting.
The probability that at least 2 of them bomb is exactly 0.26172.
I have a PhD in statistics and (assuming whether any runner bombs is independent of the others bombing, which I think is reasonable in this case), this is a binomial distribution with n = 4, p = 0.25.
Probability no one bombs: 0.75^4 = 0.3164
Probability exactly 1 person bombs: (4 choose 1)*(0.25^1)*(0.75^3) = 0.4219
Probability exactly 2 people bomb: (4 choose 2)*(0.25^2)*(0.75^2) = 0.2109
Probability exactly 3 people bomb: (4 choose 3)*(0.25^3)(0.75) = 0.0469
Probability everyone bombs: 0.25^4 = 0.0039
Probability at least two bomb: 0.2109 + 0.0469 + 0.0039 = 0.2617
As a hs math teacher and running nerd, this may be one of my favorite threads ever on letsrun. I also find it ironic that the probability in this scenario rounds to 26.2% for a marathon related question.
Under your scenario, if each of 4 runners have a 75% chance of making the team, then all 3 slots will go to 3 of these four. Why? Because the rest of the field has 0% chance of making the team.
Yes, this is correct. The 75% case is trivial as the only distribution that allows for 75% probability to each of the 4 involves three of them being on the team, if the probability is higher, there is no solution, if it is lower, then it gets more interesting.
Your scenario only allows there to be 4 runners in the entire field
Under your scenario, if each of 4 runners have a 75% chance of making the team, then all 3 slots will go to 3 of these four. Why? Because the rest of the field has 0% chance of making the team.
Yes, this is correct. The 75% case is trivial as the only distribution that allows for 75% probability to each of the 4 involves three of them being on the team, if the probability is higher, there is no solution, if it is lower, then it gets more interesting.
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.
If this 75% chance is not independent, then it's not a meaningful premise, because we haven't been given information about that dependence.
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.