Live data from Hacker News

A confusing probability question: Red and green balls in an urn

colab.research.google.com

91–100 of 169 posts

Re: A confusing probability question: Red and green balls in an urn

#91
post #83

Early in my career, when I first started interviewing, I used to ask a version of this to recent grads. It was never a make-or-break question, but I found it to be a great way to a.) see how people approach problem solving and probability and b.) see how they respond when you start asking whys (even if they answered/guessed 1/3). It's something that takes zero code to answer, and the intuition is easy to grok once ex…

> Two points are randomly and uniformly selected on a line 0.0 to 1.0. What is the most probable distance between the two points?

Unless I am reading this wrong, I think all values between 0 and 1 have an equal probability (of 0).

The probability that a random uniform variable will equal any number between 0 and 1 is zero. It seems to follow that the probability of the difference between two uniform variables equaling any exact value would also be 0.

Have I missed something obvious? If zero really is the correct answer, that is pretty tricky.

Re: A confusing probability question: Red and green balls in an urn

#92
post #75

Never liked the way these problems are worded. `You take a random ball out of the urn—it’s red—and discard it.` How normal people read it: Given this specific instance where you just discarded a red ball from this urn, what's the probability of the next ball? How it expects you to read it: Given infinitely many random samples from the urn. For cases where you get red, remove it, then take a second sample. What's the…

I would go further and say that the second reading is in fact incorrect interpretation of the problem in the English language. Being a mathematician doesn’t give some special right to gaslight people on their knowledge of English. This problem is in a similar category as badly explained monty hall problems where the statement of the problem is so bad that it ends up changing the answer. For example I have seen the Mo…

Love this, can we mangle the problem some more? Like, you open one door and it’s a goat, the host then says he will open another door containing the trip to Bali. Do you want to open the other door or stay with the goat you already have?

Re: A confusing probability question: Red and green balls in an urn

#93

Imagine you have to guess a random number between 0 and 255, i.e. you have to guess 8 bits. You already know that it's less than 128 so the most significant bit is 0. Can you derive any knowledge about whether the second bit is 0 or 1?

Watch out. Because your bits are indexed you have a different problem than before. The first red ball does not give you a significant bit.

Instead: Make a sample and now you know that it is greater than a random number unknown to you. Can you derive any knowledge about your next sample?

Re: A confusing probability question: Red and green balls in an urn

#94
post #76
post #22

Earlier quoted context omitted.

Why would n be "uniformly randomly chosen" but then suddenly be known? I would mark this mis interpretation simply as wrong if I was a teacher.

Another way of making your point: it would also make the information about the distribution of n irrelevant, so it's a bad interpretation of the question.

So? There's plenty of problem statements with irrelevant data. Children are also getting taught to determine which is relevant in class.

But also, it's not irrelevant because for two cases the probability stays equal. Calculating the exact chance is still an interesting problem and it could have been dumbed down for twitter only.

Re: A confusing probability question: Red and green balls in an urn

#95
post #22
post #12

I think this is less about Bayesian thinking and more about misinterpreting the question, as another poster mentions: https://twitter.com/farrwill/status/1751788706355392639 ie i think many of the More Likely to be Green people are doing the math of if you pull from an urn with n/100 odds of getting red, your second pull will have odds (n-1)/99, which is less than n/100 for all n except n=100. Which is obviously a di…

Why would n be "uniformly randomly chosen" but then suddenly be known? I would mark this mis interpretation simply as wrong if I was a teacher.

You have an urn of 50 red and 50 green balls. Roll a die and add a red ball for every pip. What is the probability of the next ball you take being red?

I say both can be a question in Statistics I where you calculate probabilities based on combinations.

So I'll say it depends on the level of education for how the question gets interpreted and saying those who never took Bayesian reasoning and consider n unknown are wrong gets elitist.

Re: A confusing probability question: Red and green balls in an urn

#96
post #76
post #22

Earlier quoted context omitted.

Why would n be "uniformly randomly chosen" but then suddenly be known? I would mark this mis interpretation simply as wrong if I was a teacher.

Another way of making your point: it would also make the information about the distribution of n irrelevant, so it's a bad interpretation of the question.

Because including some pieces of irrelevant information as red herrings in probability problems is totally unheard of?

Re: A confusing probability question: Red and green balls in an urn

#97
Seems to be fully depending on your interpretation of "more likely". If you take that as "more likely than it was before", then the answer is obviously "more likely to be green". If you take it as "more likely to be green or red this pull", then that answer would depend fully on the unknown distribution and the closest least wrong answer would be "don't know".

Backtranslating from the interpretation/answer pairs, if this was on a test I would go with the former interpretation and answer green.

Re: A confusing probability question: Red and green balls in an urn

#98

Seems to be fully depending on your interpretation of "more likely". If you take that as "more likely than it was before", then the answer is obviously "more likely to be green". If you take it as "more likely to be green or red this pull", then that answer would depend fully on the unknown distribution and the closest least wrong answer would be "don't know". Backtranslating from the interpretation/answer pairs, if…

But there is no unknown distribution. There is a latent parameter (the n that's uniformly sampled) whose distribution is known; so the distribution of colors for the next draw is well defined. "Depend fully on the unknown distribution" is a statement about the distribution conditioned on that latent parameter; even in that case, the answer isn't "don't know" but "whatever the ratio of remaining balls given that parameter, and the one drawn ball, is."

"Don't know" isn't a valid answer in this case; "Don't know" would be valid if we were not told that n has that uniform distribution.

Since (unconditionally) red and green are symmetric, and the first draw is 50-50, the 'more likely/less likely' has the same meaning whether you interpret it as 'more/less than 50-50' or 'more/less than before.' This is convenient; if not for this symmetry, the answer would depend on interpretation of the 'more/less likely' language and could be ambiguous. The way the problem is stated, it is not ambiguous because either reading leads to the same result.

Re: A confusing probability question: Red and green balls in an urn

#99
Seems that the notion of `uniformly` is the key aspect here, i.e. "the sets are ordered" in my intuitive notion.

So if the first one you've picked is red, then the only way for the next one to be green is if you started in the set of one red ball + 99 green balls; so you'd have to "pick" one specific set out of 100 possible (set of "all are green" is already impossible at this point), which is why it is very unlikely, and so you should expect the next one to be red.

And the more reds in a row you pick, the less likely you are to see a green one.

Re: A confusing probability question: Red and green balls in an urn

#100

In these cases it's always good to think of extremes. For example for the Monte Hall problem imagine a million doors and all but one of them are opened, all empty, in any case. Then it's clear you originally had one in a million chances of picking the right door. Two ways to think about this (same conclusion): - Without loss of generality, take the case of just one red ball, rest are green. If you pick the one red ba…

>Without loss of generality, take the case of just one red ball, rest are green. If you pick the one red ball on your first draw the chances are now 100% that the next ball is green (since there aren't any red ones left) whereas before you had picked the red one you had just a 99% chance of picking a green one.

Your reasoning can conclude that picking a green ball next time is more likely for any specific bag contents. But you can't then conclude that picking a green ball the next time is more likely, period. Picking red the first time gives you information that changes the probability distribution of likely bag contents.

Post reply on HN