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# The "aha!" problems 7. Urning the balls

Best Answer m00li, 06 May 2014 - 01:38 AM

Spoiler for no aha here

continuing...

Spoiler for continuing
Go to the full post

19 replies to this topic

### #11 bubbled

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Posted 12 May 2014 - 04:13 AM

Spoiler for Aha hint

"aha" still eludes me. I ran a simple Python simulation as well. But I thought I'd share my friend's elegant recursive approach to this problem. The nice thing was he got exact answers, which helps when looking for patterns.

Spoiler for Python recursion

Edited by bubbled, 12 May 2014 - 04:16 AM.

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### #12 bonanova

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Posted 12 May 2014 - 09:02 AM

Spoiler for Aha

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Vidi vici veni.

### #13 bubbled

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Posted 14 May 2014 - 03:02 AM

Spoiler for Aha

I'm not totally convinced by this argument. I obviously believe it. But don't see the logical argument why the number of black balls left scales linearly with the starting number of black balls, given a fixed number of white balls.

For instance. It's obvious that if w = 10 and b = 1, then on average, there will be 1/11 black balls left. But let's look at w = 10 and b = 2. I would use this formula:

1/12 * 1/11 * 2 * 2 + 1/12 * 10/11 * 2 * 1 which does equal the expected result of 2/11. The first term is chances a particular black ball is in last position * chances the other black ball is in second-to-last position * 2 ways for that happen * 2 black balls left. The second term is chances a particular black ball is in last position * chances the second black ball is not in second-to-last position * 2 ways for that to happen * 1 black ball remaining.

But the statement "If there are originally b black balls, this expectation becomes b/(w+1)" does not convince me of this fact.

Maybe I'm missing a simple identity, but I haven't been "aha'ed" just yet.

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### #14 bonanova

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Posted 14 May 2014 - 03:30 AM

What is true of any black ball is true of every black ball.

They are indistinguishable.

They can't have distinct probabilities.

Every black ball has a probability of 1/(w+1) of being in the urn after the last white ball has been drawn.

If you like, it's precisely this awareness that constitutes the aha moment.

If you want to say this black ball has this probability and that black ball has another probability,

you'd have to say why this is so, and you'd have to say how you can tell them apart.

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Vidi vici veni.

### #15 bubbled

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Posted 14 May 2014 - 03:53 AM

What is true of any black ball is true of every black ball.

They are indistinguishable.

They can't have distinct probabilities.

Every black ball has a probability of 1/(w+1) of being in the urn after the last white ball has been drawn.

If you like, it's precisely this awareness that constitutes the aha moment.

If you want to say this black ball has this probability and that black ball has another probability,

you'd have to say why this is so, and you'd have to say how you can tell them apart.

Aha!!

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### #16 bonanova

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Posted 14 May 2014 - 06:24 AM

Lovely, isn't it?

I was going to add ... think of the string of black balls like a stick.

And the white balls like cutting the stick in w random locations.

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Vidi vici veni.

### #17 bubbled

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Posted 14 May 2014 - 08:48 AM

Lovely, isn't it?

I was going to add ... think of the string of black balls like a stick.

And the white balls like cutting the stick in w random locations.

Indeed.

One really interesting aspect of this problem is that by removing the balls, you even out the odds of pulling one color or another as the numbers get big.

For example, I ran a simulation 1,000 times where I flipped a fair coin 1,000,000 times and noted abs( H - T ). The average value was 783. But if we put 500,000 black balls and 500,000 white balls in the urn, the average number of balls left after exhausting one color is very close to 2!

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### #18 bonanova

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Posted 14 May 2014 - 06:47 PM

Lovely, isn't it?

I was going to add ... think of the string of black balls like a stick.

And the white balls like cutting the stick in w random locations.

Indeed.

One really interesting aspect of this problem is that by removing the balls, you even out the odds of pulling one color or another as the numbers get big.

For example, I ran a simulation 1,000 times where I flipped a fair coin 1,000,000 times and noted abs( H - T ). The average value was 783. But if we put 500,000 black balls and 500,000 white balls in the urn, the average number of balls left after exhausting one color is very close to 2!

Coin outcomes remain 50-50. Ball outcomes readjust to b/(b+w) and w/(b+w).

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Vidi vici veni.

### #19 bubbled

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Posted 16 May 2014 - 02:38 AM

Lovely, isn't it?

I was going to add ... think of the string of black balls like a stick.

And the white balls like cutting the stick in w random locations.

Indeed.

One really interesting aspect of this problem is that by removing the balls, you even out the odds of pulling one color or another as the numbers get big.

For example, I ran a simulation 1,000 times where I flipped a fair coin 1,000,000 times and noted abs( H - T ). The average value was 783. But if we put 500,000 black balls and 500,000 white balls in the urn, the average number of balls left after exhausting one color is very close to 2!

Coin outcomes remain 50-50. Ball outcomes readjust to b/(b+w) and w/(b+w).

This puzzle has revealed so many nice symmetries, I thought I'd share another one. If you have b black balls and w white balls in an urn, the expected number of monochromatic balls you will pick out of the urn before you pick a ball of the other color is the same as the expected number of balls remaining after you have exhausted one color.

Pretty interesting.

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### #20 bonanova

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Posted 16 May 2014 - 08:44 AM

The other end of the stick in post 18.
Nice.
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Vidi vici veni.

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