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Suppose you have a bag with R red marbles and B blue marbles.

R + B is even.

You reach in and pull out two marbles and place them side by side.

You continue pulling pairs until all the marbles have been taken from the bag.

What is the expected number of Red-Blue "couples"?

Edit - removing spoilers for the moment ...

Edit - to make it clear the couples have one Red and one Blue marble.

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Suppose you have a bag with R red marbles and B blue marbles.

R + B is even.

You reach in and pull out two marbles and place them side by side.

You continue pulling pairs until all the marbles have been taken from the bag.

What is the expected number of Red-Blue "couples"?

Edit - removing spoilers for the moment ...

Wouldn't the expected number of couples vary depending on the total amount? I.e. if you start with 3R and 5B compared to 300R and 500B. The bag with 800 marbles would have way more expected couples, right?

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Suppose you have a bag with R red marbles and B blue marbles.

R + B is even.

You reach in and pull out two marbles and place them side by side.

You continue pulling pairs until all the marbles have been taken from the bag.

What is the expected number of Red-Blue "couples"?

Edit - removing spoilers for the moment ...

We cannot say what the expected amount of pairs is because we don't know the amount of marbles. If there was 1 Red and 99 Blue, it would be different than a 50/50 split.

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Sure you can.

The expected number of couples is an expression involving R and B.

Here's a thought question as you think about the solution:

Does the expected number of couples double if you double both B and R?

If you want numbers, suppose R=3 and B=4.

Or another: R=6 and B=10.

If you get these answers, you probably know the general formula.

Edit: R+B is even.

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Suppose you have a bag with R red marbles and B blue marbles.

R + B is even.

You reach in and pull out two marbles and place them side by side.

You continue pulling pairs until all the marbles have been taken from the bag.

What is the expected number of Red-Blue "couples"?

Edit - removing spoilers for the moment ...

R+B is even.I can assume that R=B,hence the R-B couples will be 1/2*(R+B)

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Wouldn't the expected number of couples vary depending on the total amount?

I.e. if you start with 3R and 5B compared to 300R and 500B.

The bag with 800 marbles would have way more expected couples, right?

Correct. The answer depends on R and B.

For starters, try R=6 and B=10.

Then try for a formula for arbitrary R and B such that R+B is even.

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It never said that R=B. How can you make that assumption?

It never said that R=B,but said that R+B is even,that's way if I want to make my life easier I can assume that R=B.

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It never said that R=B,but said that R+B is even,that's way if I want to make my life easier I can assume that R=B.

Hi ash,

You may have missed the idea of the question in the OP.

A pair of marbles is a "couple" only if there is one Red and one Blue marble.

Your answer [and method] makes sense if they can be any color,

but that's a trivial puzzle. This one requires a little more thought.

See what you can come up with....

- bn

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(2RB/(R+B)2) * 100

Yes, basically. B))

When the 2nd marble is drawn there is one fewer marbles to choose from,

so it changes slightly:

(2RB/(R+B)(R+B-1))

Now we're going to draw all the marbles - what's the expected number of red-blue pairs?

That's a simple modification of the formula for drawing only one pair. ;)

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