The simplest proof that I have that there's only one solution uses the setting instead of the math: It seems safe to assume that Phil can't hold a bag that has more than 10,000,000 balls in it, and checking all solutions up to that size is computationally trivial.
I started studying the variant of the problem where you pull out 2 blue balls instead of 4; that variant has multiple solutions, which intuitively implies that you're not going to show uniqueness in the 4-ball case from very simple properties of the obvious equation. [Computationally, the 3 ball case is also unique up to 10,000 balls.] If there's a simple answer, it's very clever, but I'm starting to suspect this requires deeper knowledge of Diophantine Equations than I've got?
Actually, I think you can make a lot of progress by playing with evens and odds. For instance, defining t = r + b (t is the total number of balls in the sack), we are looking for solutions of:
Now, consider both the numerator and the denominator separately. They each contain one term which is a multiple of 4, one term which is a multiple of 2 but not a multiple of 4, and two odd terms. Defining n and d to be the multiple of 4 in the numerator and denominator respectively, match up the multiples of 4, pull out the multiples of 2 from the other even term and divide, and then you've got (n*(product of 3 odds)) / (d*(product of 3 odds)) = 2. This implies that for any solution, n/d = 2 and the product of the 3 odds in the numerator and denominator must be equal. You also know that the 3 odds in the numerator and the denominator are each a pair of numbers separate by 2, and a third number which is roughly half as large. I haven't quite turned this into a full proof though.
OK, this is enough to finish the proof. Recalling that n and d are the multiple of 4 in the numerator and denominator respectively, we've shown that n = 2d
( ... )
It's a property of the solution. For most values of r and b [in fact for all values except r=1 and b=7], the odd terms in n/d won't be the same, so we don't get a valid solution. I'm using the original constraint on any solution (b/(b+r))*((b-1)/(b+r-1))*((b-2)/(b+r-2))*((b-3)/(b+r-3)) = 1/2 [and then taking the reciprocal of both sides and defining t = b+r.]
By the way, the way I defined, the 2^(x+1) is in n, not in d.
FWIW, I posted this on the Google internal math list which included a large number of hardcore mathematicians, some of whom have PhD's in number theory and related, and nobody had an answer. So I think there's unlikely to be any sort of cute simple easily understandable trick to the proof.
Well, both. Pleased with your own relative abilities, but disappointed because if there's no cute answer it's not *actually* a good puzzle. It's like telling someone the Collatz Conjecture but framing it as "Oh, I heard this cool brainteaser at work the other day." Good puzzles have good answers. Who gave this monster to you?
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The simplest proof that I have that there's only one solution uses the setting instead of the math: It seems safe to assume that Phil can't hold a bag that has more than 10,000,000 balls in it, and checking all solutions up to that size is computationally trivial.
I started studying the variant of the problem where you pull out 2 blue balls instead of 4; that variant has multiple solutions, which intuitively implies that you're not going to show uniqueness in the 4-ball case from very simple properties of the obvious equation. [Computationally, the 3 ball case is also unique up to 10,000 balls.] If there's a simple answer, it's very clever, but I'm starting to suspect this requires deeper knowledge of Diophantine Equations than I've got?
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(t * (t-1) * (t-2) * (t - 3)) / (b * (b-1) * (b-2) * (b-3)) = 2
Now, consider both the numerator and the denominator separately. They each contain one term which is a multiple of 4, one term which is a multiple of 2 but not a multiple of 4, and two odd terms. Defining n and d to be the multiple of 4 in the numerator and denominator respectively, match up the multiples of 4, pull out the multiples of 2 from the other even term and divide, and then you've got (n*(product of 3 odds)) / (d*(product of 3 odds)) = 2. This implies that for any solution, n/d = 2 and the product of the 3 odds in the numerator and denominator must be equal. You also know that the 3 odds in the numerator and the denominator are each a pair of numbers separate by 2, and a third number which is roughly half as large. I haven't quite turned this into a full proof though.
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I can see that
n = 2^x * (odd number)
d= 2^(x+1) * (odd number)
But do we know the odd factors are the same?
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By the way, the way I defined, the 2^(x+1) is in n, not in d.
Does this make sense?
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It sounds straightforward enough, though, doesn't it?
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