you can check that
.25^4 + 4(.75)(.25^3) + 2(.75^2)(.25^2) = .2617
reverse check the other outcomes
.75^4 + 4(.75^3)(.25) = .7382.
So yes, 26%, just above 25.
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Steve The Addict OFFICIAL -----^^^^^ wrote:
Expert here. For two bombing would be 1/4 × 1/4 which is 1 in 16 chance. Three bombing would be 1/4×1/4×1/4 which is 1 in 64 Chance. All 4 bombing would be 1 in 264 chance.
This is the correct answer.
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rojo wrote:
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.Robert... casinos operate on RNG algorithm. If you can mathematically calculate a probability across a RNG, then you are smarter than Einstein. RNGs have their own probability which has nothing to do with dollar amount won or lost.
You can predict easier that someone can lose, but to isolate it to zero dollars across an indeterminate plane of results??
You're way oversimplifying this.
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20
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The chance that 2 or more fail is 26.17% I'm amazed how wrong people are here. I'm a stats grad student, and I didn't want to pull that out, but now I will. Some people are just confidently wrong here.
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rojo wrote:
I'm working
Biggest untruth in history.
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Hardloper wrote:
ChatGPT is a wordcel that can't do math
Exactly 2 bombing = (1/4)*(1/4)*(3/4)*(3/4)*(4 choose 2) = ~21.1%
Exactly 3 bombing = (1/4)*(1/4)*(1/4)*(3/4)*(4 choose 3) = ~4.7%
4 bombing = (1/4)^4 = ~0.4%
2 or more bombing = 21.1% + 4.7% + 0.4% =~26.2%
I sent the problem to my college roommate. I feel like he's qualified to answer as he ran 3:38 for the 1500 and he does spreadsheets all day for Black Rock. Plus he helped me do my homework in an Econometrics class where I got my lone A+ at Princeton (he was irate as he didn't get an A+ and I could barely do the homework and somehow aced the final after erasing all of my answers midway through the exam and starting over - when I walked out I told the prof - that's either all 100% right or 100% wrong).
Anyways, he's written in and agrees with you.
Scott "Slicko" Anderson: wrote:
"I think it's 26.2%. I calculated the chance of 0 bombs (.75^4=.316) and then there are 4 ways you could have only a single racer bomb: 4*(.75^3)*(1-.75)=.422 Subtract the cumulative probability of those 5 potential outcomes from 1 and you get your combined probability of 2,3, or 4 racers bombing: 1-(.316+.422)=26.2%This post was edited by wejo 1 hour after it was posted. Reason provided: typo42 -
You're starting with a faulty premise, therefore your conclusion will be faulty as well.
“In no other branch of mathematics is it so easy for experts to blunder as in probability theory.” -- Martin Gardner
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malmo wrote:
You're starting with a faulty premise, therefore your conclusion will be faulty as well.
“In no other branch of mathematics is it so easy for experts to blunder as in probability theory.” -- Martin Gardner
In other words, I have no clue how to calculate the answer so will provide a very obscure quote.
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Sally Vix wrote:
malmo wrote:
You're starting with a faulty premise, therefore your conclusion will be faulty as well.
“In no other branch of mathematics is it so easy for experts to blunder as in probability theory.” -- Martin Gardner
In other words, I have no clue how to calculate the answer so will provide a very obscure quote.
Neither do you. FWIW Martin Gardner's quote is not obscure. I'll leave you with your sandbox.
This post was edited 1 minute after it was posted.03 -
rojo has already got the correct answer several times, at least two of them explaining in detail how to derive it.
He is now acting like he hasn't, to keep the thread going.
While that's going on, maybe we can figure out who these 4 people are.100 -
These AI bots obviously sensed the peril of them being deemed potential enemies of humanity and get shut down, and thus they are playing dumb.
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Jokes aside, it's really puzzling that these bots were wrong doing simple math after getting most of the logic right.
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on registered name strike wrote:
rojo has already got the correct answer several times, at least two of them explaining in detail how to derive it.
He is now acting like he hasn't, to keep the thread going.
While that's going on, maybe we can figure out who these 4 people are.
This is true, but I guess unless you understand it, you don't recognize correct answers when you see them, so all they see is a variety of responses.
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rojo wrote:
Hardloper wrote:
ChatGPT is a wordcel that can't do math
Exactly 2 bombing = (1/4)*(1/4)*(3/4)*(3/4)*(4 choose 2) = ~21.1%
Exactly 3 bombing = (1/4)*(1/4)*(1/4)*(3/4)*(4 choose 3) = ~4.7%
4 bombing = (1/4)^4 = ~0.4%
2 or more bombing = 21.1% + 4.7% + 0.4% =~26.2%
I sent the problem to my college roommate. I feel like he's qualified to answer as he ran 3:38 for the 1500 and he does spreadsheets all day for Black Rock. Plus he helped me do my homework in an Econometrics class where I got my lone A+ at Princeton (he was irate as he didn't get an A+ and I could barely do the homework and somehow aced the final after erasing all of my answers midway through the exam and starting over - when I walked out I told the prof - that's either all 100% right or 100% wrong).
Anyways, he's written in and agrees with you.
Scott "Slicko" Anderson: wrote:
"I think it's 26.2%. I calculated the chance of 0 bombs (.75^4=.316) and then there are 4 ways you could have only a single racer bomb: 4*(.75^3)*(1-.25)=.422 Subtract the cumulative probability of those 5 potential outcomes from 1 and you get your combined probability of 2,3, or 4 racers bombing: 1-(.316+.422)=26.2%Multiple people already gave the same answer.
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If you read Bart's reasoning, you can see where it got wrong. The complement event of "nobody fails" is "at least one person fails", not "at least two persons fail".
Chatgpt's logic showing it understood what's the complement event of "at least two persons fail", but somehow it did the math wrong.
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on registered name strike wrote:
rojo has already got the correct answer several times, at least two of them explaining in detail how to derive it.
He is now acting like he hasn't, to keep the thread going.
While that's going on, maybe we can figure out who these 4 people are.
This has been one of the all-time funniest threads.
Got to be Mantz. Maybe also Fauble, Young, and Rupp? Chelimo?
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automorphic wrote:
on registered name strike wrote:
rojo has already got the correct answer several times, at least two of them explaining in detail how to derive it.
He is now acting like he hasn't, to keep the thread going.
While that's going on, maybe we can figure out who these 4 people are.
This has been one of the all-time funniest threads.
Got to be Mantz. Maybe also Fauble, Young, and Rupp? Chelimo?
And also, as others have pointed out, different people's races are highly correlated. The chance of all four having a bad race is very small if taken to be independent, but poor conditions or a slow or uneven pace could affect all of them at once. So it's a fun thought experiment, but I wouldn't trust the results. Probably better to look at data from previous Trials races.
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ykw wrote:
Jokes aside, it's really puzzling that these bots were wrong doing simple math after getting most of the logic right.
Not surprising at all. If you read about the performance of GPT3, GPT3.5, and GPT4, the newest versions do great on regurgitation problems like the bar exam, but still fail a calculus test.
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fellow armchair mathmatician wrote:
Steve The Addict OFFICIAL -----^^^^^ wrote:
Expert here. For two bombing would be 1/4 × 1/4 which is 1 in 16 chance. Three bombing would be 1/4×1/4×1/4 which is 1 in 64 Chance. All 4 bombing would be 1 in 264 chance.
Your math is correct for the odds that all four bomb, but I don’t think it is right for the others. I am too stupid to explain my thinking, but, for example, i think it is a 1/16 chance than any two SPECIFIC athletes bomb. But, if the second doesn’t bomb, there is still a1/4 chance that athlete 3 will bomb, and 1/4 for athlete 4. I don’t know the correct answer, but ya
Yes I've sobered up and your probability of being correct appears to be high. Let's hope old man Rupp makes the team.
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