300 balls, 150 out, order!

Thanks for your help!

We have a box with balls, numbered from 1 to 300.

We draw the first ball at random. Let's imagine is 24, then we write 1-24. Then we go with the second, number 278, then, 2-278... And so on until we get 150 balls - so, there 150 remaining in the box!

How many 'perfect pairs' -this is, the ordered number is the same that the number itself- could be expected?

Thanks!

lau_laura said:

Thanks for your help!

We have a box with balls, numbered from 1 to 300.

We draw the first ball at random. Let's imagine is 24, then we write 1-24. Then we go with the second, number 278, then, 2-278... And so on until we get 150 balls - so, there 150 remaining in the box!

How many 'perfect pairs' -this is, the ordered number is the same that the number itself- could be expected?

Thanks!

Click to expand...

What are your thoughts?

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Denis said:

I'm confused...probably need another coffee...
BUT why have the balls numbered 151 to 300?
Every time you draw one of those, you're automatically out...

Click to expand...

Example of the way I understand the problem but with smaller numbers: Six balls marked 1b, 2b, 3b, 4b, 5b, 6b. You draw three balls in the order of 3b, 2b, and 1b with which you create a set of three pairs: {(1,3b), (2,2b), (3,1b)} This set contains one 'perfect pair', the pair (2, 2b) because 2b was the second ball drawn and none of the other balls drawn matched their order drawn. What is the expected value of the number of perfect pairs?

Is that correct?

Denis said:

...Sooo...I guess that throwing in the extras is to
"reduce" the above expectation...

Click to expand...

I would think so plus it makes the problem somewhat harder.