# A question for the mathematically minded jellies, involving statistics and limits.

Asked by Mariah (25301) March 11th, 2011

Say you have a bucket filled with a very large number of red jellybeans and a smaller, but still very large number of green jellybeans. You reach in and grab a bean at random and eat it – as you keep doing this, does the ratio of red to green jellybeans left in the bucket approach 1:1? Common sense is telling me it would, because when you start it is more likely you’ll grab red beans often, thus depleting the supply of red beans faster and evening out the ratio. But, I’m finding it difficult to approach in a mathematical manner. I don’t think I can make the iterations approach infinity because that would assume infinite beans, and I’m having trouble wrapping my mind around there being an infinite number of beans, exactly half of which are red…

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The ratio would be expected to stay at the original ratio, whatever that was. For example, if there are 12000 red and 10000 green and you took out a 220, you’d expect to get 120 red and 100 green and the ratio would remain 1.2:1.

6rant6 (13672)

I think your intuition is correct, but I do not know how to show it either. I can show that for a small number of selections, the ratio moves toward 1:1, by arguing as follows. To make things easier, suppose there are 8000 red jellybeans and 2000 green ones. What can we say about the first 20 selections? If we remove 20 red ones that gives 7980. We can therefore say that for each of the first 20 selections, the probability of choosing red is greater than 7980/8000. The expected number of red jellybeans that will be taken is therefore greater than 20*(7980/8000). Unfortunately, this reasoning does not work for a large number of selections, because the lower bound that I used on the probability of taking a red jellybean will move toward 0.

The ratio would stay the same.

Lightlyseared (30761)

I’m approaching this from a strictly logical POV, with the added benefit of having recently studied my ass off for a statistics exam (which I failed, so make of that what you will).

Assuming that, in the beginning, we have a different amount of beans for each colour we would also have different Probability values for drawing either colours.

To me, it seems fairly clear that, since drawing one colour is more likely than the other one would decrease faster than the other, which would still decrease, though, and therein lise the problem.

Drawing an x number of times, possibly a high number, from a very large pool means that you’ll in the end, have drawn a number of green and red jellybeans whose proportion will closely resemble the one of the pool, that’s why random statistical survey is considered accurate (as accurate as the amount of single experiments performed, to be precise).

So, no, while it might seem believable, your intuition would verify only in an anomalous case, not as a rule. (as @6rant6 and @Lightlyseared already pointed out)

@LostInParadise Your demonstration doesn’t work with high numbers precisely because the overall idea is based on a statistical anomaly, which is drawing mainly red jellybeans. Yes, it is more likely to draw red jelly beans on a small scale, but on a large number of attempts (even with the same pool, resetting it after every attempt) you will draw more and more closely to an average of 16 red to 4 green because of the law of large numbers

Thammuz (9262)

I was wrong. The ratio will tend to stay the same. It is true that choosing a red jellybean is more probable, but the probability is proportional to the ratio. If there is a 2:1 ratio of red to green then you would expect to choose two red for every green, preserving the ratio. You can think of it this way. Imagine that the jellybeans are placed in a large sack and are thoroughly mixed. If you put a scoop in the sack to remove jellybeans, you would expect the red:green:ratio in the scoop and in the remaining jellybeans to be the same as in the whole sack.

If you take the long view (the very loooong view) there will eventually be a point where you only have two beans so statisitaclly at that point (and that point alone) you may have a ratio of 1:1. Of course you’ll probably end up with 2 of those weird jelly beans that taste foul and you can’t quite identify but hey ho.

Lightlyseared (30761)

Okay, thanks everyone! Yeah, what my mind didn’t come up with at two in the morning is that, while more red beans will be picked, for every green bean picked the percentage of green beans drops more dramatically. If there are 400 red beans and 100 green beans, I’ll grab red beans four times as often, but when I pick a red bean I am only taking away ¼00 of the red beans, and when I pick a green bean I am taking away 4/400 of the green beans. In a selection of 5 beans I pick an average of 4 red ones and 1 green one, and the percentage of each drops by 1/100.

Mariah (25301)

@Mariah while more red beans will be picked, for every green bean picked the percentage of green beans drops more dramatically

This was the first thing that popped into my head, and I think that it sums up the issue succinctly.

In terms of the situation where there’s a limit – it’s also helpful to think atbout the last choice. It’s possible that, after your second to last choice, that you’ll be left with a ratio of 1:1. As @Lightlyseared mentions, as you reduce the pool, the likelihood that you’ll reach 1-to-1 could increase – and it’s less likely that each choice you make will result in a ratio that will be equivalent to the previous ratio. At the last two choices, if your penultimate choice has a ratio of 2 red to 1 green, then it’s likely that you’ll end up with a ration of 1 to 1. However, there’s a 33% chance that you will never reach that ratio.

iamthemob (17147)

You can think of this in terms of taking samples. The beans selected form a sample, but so do the remaining beans. Since the question is about the remaining beans, then that is the sample being focused on. The case where there are just a few beans left is equivalent to taking a small sample. The laws of large samples does not apply in this case. Looking at what happens when there are only one or two beans left is examining a sample of size 1 or 2 and for small samples there is a high probability that the ratio of red to green will differ significantly from the overall ratio.

This would completely depend on the availability of the ‘buckets’ to show a 50/50 choice differential. A person, knowing where the buckets are place and being aware of the experiment, can have an impact of where they choose. The same person, knowing if they choose a bit to the left or right knows where the buckets are placed, will then THEMSELVES determine how the experiment plays out.

IF, the chooser has to roll a die…. and 1 to 3 determines the choice from one bucket and 4 to 6 determines the choice from the other bucket… only THEN will you have a proper random selection to determine the end result.

cazzie (24503)

(I am enjoying this question and the answers very much. Thanks, all.)

gailcalled (54512)

@Mariah There is a really good book I’m reading which you might enjoy.
The Drunkard’s Walk: How Randomness Rules our Lives

Rarebear (25144)

@Rarebear Ooh thank you for that link, it looks fascinating! I’ll make a related book recommendation right back at you: Chances Are… Adventures in Probability

Mariah (25301)

@Mariah The books look similar. I just found it for 4 bucks online used, thanks!

Rarebear (25144)

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