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

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

#61

Post by bo1024 »

chrismcb wrote: post #54
danielmarkbruce wrote: post #19

Earlier quoted context omitted.

It's definitely a poorly worded question. "More likely" is ill-defined. More likely than what? Than the last draw? Than drawing the other color just on this pick?

How is "more likely"I'll defined? They are two options red or green. Are you more likely to draw red or are you more likely to draw green.. how is it poorly worded?

My first interpretation of "more likely" was also "more likely than the previous ball". It makes more sense in the context of the previous sentence.

For example:

> The first time you shoot a basketball, you will almost certainly miss. The second time, you are more likely to make it.

Does the second sentence mean "more likely to make it than to miss"? No.

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

#63

Post by mewpmewp2 »

chrismcb wrote: post #55
anigbrowl wrote: post #25

I think the wording is also biasing the outcome. The next ball is more likely to be green than the previous ball , and a lot of people will regard that posterior probability statement as implicit. If the question had been phrased differently eg, 'what is the most likely color for the next ball selected' I suspect more people would have voted for equal likelihood.

What does the previous ball have to do with it? We KNOW the previous ball is red. Then it asks "more likely to be red" It seems pretty obvious that it is saying draw the next ball, which is more likely? Red, green, equal. Why would you think it is comparing the second pick probability to the first pick probability?

Agreed. The first likelihood of green is 50 percent, and of the first is red then the next ball likelihood of being green is smaller than 50 percent, so the above statement makes no sense.

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

#64

Post by danielmarkbruce »

bo1024 wrote: post #61
chrismcb wrote: post #54

Earlier quoted context omitted.

How is "more likely"I'll defined? They are two options red or green. Are you more likely to draw red or are you more likely to draw green.. how is it poorly worded?

My first interpretation of "more likely" was also "more likely than the previous ball". It makes more sense in the context of the previous sentence. For example: > The first time you shoot a basketball, you will almost certainly miss. The second time, you are more likely to make it. Does the second sentence mean "more likely to make it than to miss"? No.

Great example.

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

#65

Post by bo1024 »

chrismcb wrote: post #55
anigbrowl wrote: post #25

I think the wording is also biasing the outcome. The next ball is more likely to be green than the previous ball , and a lot of people will regard that posterior probability statement as implicit. If the question had been phrased differently eg, 'what is the most likely color for the next ball selected' I suspect more people would have voted for equal likelihood.

What does the previous ball have to do with it? We KNOW the previous ball is red. Then it asks "more likely to be red" It seems pretty obvious that it is saying draw the next ball, which is more likely? Red, green, equal. Why would you think it is comparing the second pick probability to the first pick probability?

"You shoot a basketball for the first time -- you miss. The second time, you are more likely to score."

A common phrasing. It doesn't mean that scoring is more likely than missing, it means the second time is more likely than the first was.

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

#66

Post by sampo »

thfuran wrote: post #26
danielmarkbruce wrote: post #24

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?

> How could the next draw be more likely to be red than the draw already known to be red?

Under this interpretation, the second ball is more likely to be green.

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

#67

Post by zharknado »

This is much easier to intuit if I imagine 101 urns.

One is 0% red, one 1% red, one 2% red, etc.

If I choose one urn at random, and then the first ball I pick is red, odds are good I picked a red-heavy urn.

I’d be interested to see if the general public answers differently when it is framed more concretely.

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

#68

Post by andreimatei1 »

sxp wrote: post #51

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.

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

#69

Post by Syzygies »

One gets Bose-Einstein statistics from the same urn setup: if you draw two marbles from such an urn, the three combinations RR, RG, GG are equally likely.

An obvious visual proof: Line up 101 marbles. Choose one to be the divider; to the left is red, to the right is green. Now draw two more marbles. Or just pick three, then decide which one is the divider.

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