71–80 of 169 posts
To try to put it into words: We don't actually know the mix-rate of what's in the urn. It's possible we just pulled the only red ball and all the others are green... However the fact that our first n=1 sample happened to be red (and not something else) gives a small (and varying) amount of confidence towards red-heavier mixes rather than the red-scarce ones. We can use that to figure out which mixes are slightly more…
Follow up question: > You are given an urn containing 100 balls; n of them are red, and 100-n are green, where n is chosen uniformly at random in [0, 100]. You select 99 balls at random and find that they are all red. What is the probability the last ball is red?
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.
The mathematical term for this is a "Laplacian urn" and the probability is governed by https://en.wikipedia.org/wiki/Rule_of_succession In this specific case, P(redOnKthSample) = (numberOfRedSamples + 1) / (totalNumberOfSamples + 2) = (1+1)/(1+2) = 66% red. On the second draw, if the ball is green, then you get P(redOnThirdSample) = (1 + 1) / ( 2 + 2) = 50%
This answer should go to the top.
`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 probability of the next ball, given all the samplings?
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.
Earlier quoted context omitted.
That question is quite a bit of handholding. A closer analogue would be “would you bet the next ball is red / green / they’re equally likely?”. Personally I’d word it as “The next ball is… more likely to be red than green / more likely to be green than red / equally likely to be red or green / I don’t know.”
But isn't that exactly how it is worded? Just showing it a bit?
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 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.
Hence after picking a red ball you are slightly more likely to pick a green ball next.
- Another way to think of it, again without loss of generality consider if there was just one green ball and 99 red, then if you pick every one of them in some order there are a hundred different positions the green one could end up as. If you consider the one where it's last, after you have picked all but one (and they were all red), now you have a 100% chance to pick the green ball, even though you started out with 1% chance on the first draw: without loss of generality in the case of drawing until you reach the green ball when it is 1 in 100, each red ball you have drawn so far increases the chance of drawing the green ball next, up to a 100% chance when it is the only one left. Visualizing the chance going from 1% to 100% in this case as you draw more red balls should drive home the point that it becomes more and more likely you'll draw a green one next, in the cases that you've made it that far drawing red ones.)
We have N+1 people, also numbered 0 through N. Person k is holding jar k.
We ask each person to draw a ball at random from their jar. If they draw a green ball they take their jar and leave. If they draw a red ball they stay.
We expect on average (N+1)/2 people will remain.
We now ask them to draw another ball.
The expected number of red balls on this second drawing is (N+1)/3.
The expected number of green balls on this second drawing is (N+1)/6, half the expected number of red balls.
That's because on the first drawing, the more green balls in your jar the more likely you will draw green on the first round and thus take your jar and go home. The second round is biased toward people who have jars with more red balls.
Here's the math behind those results. The probability that person k makes it to round 2 is k/N. The expect number that will make it is sum {k=0, N) = (N+1)/2.
Once in round 2 the probability they draw red is (k-1)/(N-1). The probability they draw green is (N-k)/(N-1).
The probability then that they make it to the second round and then draw red is k/N (k-1)/(N-1).
The probability that they make it to the second round and then draw green is k/N (N-k)/(N-1).
The expected number of reds we see in the second round is sum {k=1, N} k (k-1) / N / (N-1).
The expected number of greens is sum {k=1, N} k (N-k) / N / (N-1).
Use the well known formulas 1 + 2 + 3 + ... + M = M (M+1) / 2 and 1^2 + 2^2 + ... + M^2 = M (M+1) (2M+1) / 6 and simplify and you'll get the (N+1)/3 expected red and (N+1)/6 expected green I stated above.
Earlier quoted context omitted.
Why is that the question? Why not "more likely than the last draw" ?
How could the next draw be more likely to be red than the draw already known to be red?
Like: you have a box with a whole bunch of red and green balls. Would removing a red ball increase or decrease the likelihood of drawing a red ball from the box in the future?
I wouldn't buy that as a reasonable misinterpretation of the question in most cases, but I don't know what standard of analysis holds for Twitter polls.
I think it's a matter of knowing that the phrase "where n is chosen uniformly at random" means something very different in a probability question than "where n is an arbitrary value".