OT - Probability Problem

[aostling]

Barrington's logic is flawed because he used an odd number of cards. Of course it will produce a bias.

Write out Barrington's combinations, for two red queens in six cards. It shows that a red Queen will be the first card 5 times, the second card 4 times, the third card 3 times, the four card 2 times, the fifth card once. So the logic holds, with an even number of cards in the deck.

[Barrington]

Barrington's logic is flawed because he used an odd number of cards. Of course it will produce a bias.

Here's a 'deck' of 8 cards - the argument is no different:-

Position ->>
1 2 3 4 5 6 7 8

Q Q x x x x x x
Q x Q x x x x x
Q x x Q x x x x
Q x x x Q x x x
Q x x x x Q x x
Q x x x x x Q x
Q x x x x x x Q
x Q Q x x x x x
x Q x Q x x x x
x Q x x Q x x x
x Q x x x Q x x
x Q x x x x Q x
x Q x x x x x Q
x x Q Q x x x x
x x Q x Q x x x
x x Q x x Q x x
x x Q x x x Q x
x x Q x x x x Q
x x x Q Q x x x
x x x Q x Q x x
x x x Q x x Q x
x x x Q x x x Q
x x x x Q Q x x
x x x x Q x Q x
x x x x Q x x Q
x x x x x Q Q x
x x x x x Q x Q
x x x x x x Q Q

7 in position 1, 6 in position 2, 5 in position 3 etc.

Note also that if your turn the deck over the last position is equally likely.

- But if you turn the deck over the 'last position' is the one you encounter first !

<<- Position
8 7 6 5 4 3 2 1

Q Q x x x x x x
Q x Q x x x x x
Q x x Q x x x x
Q x x x Q x x x
Q x x x x Q x x
Q x x x x x Q x
Q x x x x x x Q
x Q Q x x x x x
x Q x Q x x x x
x Q x x Q x x x
x Q x x x Q x x
x Q x x x x Q x
x Q x x x x x Q
x x Q Q x x x x
x x Q x Q x x x
x x Q x x Q x x
x x Q x x x Q x
x x Q x x x x Q
x x x Q Q x x x
x x x Q x Q x x
x x x Q x x Q x
x x x Q x x x Q
x x x x Q Q x x
x x x x Q x Q x
x x x x Q x x Q
x x x x x Q Q x
x x x x x Q x Q
x x x x x x Q Q

The bold Q is the first Q encountered going 'backward through the deck.
Same frequencies as above - 7 in the first position encountered, 6 in the second etc.

Cheers

.edit: Sorry, cross posted...

[IdahoJim]

No, that isn't addressing the question asked. That is addressing the question "what position has the highest probability the 1st red queen will be found 'AT OR BEFORE'.

In other words, I'm gonna offer you 10 bucks if you can correctly predict which spot will hold the FIRST red queen if we start pulling them off the top.
You, naturally want to maximize your chances for scarfing up my money. If you're not right on the correct spot, then you don't get the $10. Of course you realize going in, that the odds are greatly against you, but after all, it isn't costing you anything to try.

I think what threw me off was the "one card at a time". If you are asking which card, given that you can only draw once, it has to be the top card. That card is the only one with a 100% chance of being above the 2nd red queen.
Jim

[lynnl]

Look at it this way:

If you consider any of the 51 positions (the 1st red Q can't be in #52), then for all positions OTHER THAN #1, two (2) conditions must be met: 1) a red Q must be there, and 2) the other red Q must be later or deeper in the stack.
For postion #1, only the condition 1) must be met, the other condition is automatically satisfied.

[lynnl]

I think what threw me off was the "one card at a time". If you are asking which card, given that you can only draw once, it has to be the top card. That card is the only one with a 100% chance of being above the 2nd red queen.
Jim

Yeahbut..., that's true, but you aren't asked to pick a position with 100%. Only to pick the most likely. Even tho it is the only spot with, as you say, 100% chance of being above the 2nd red Q, it still only has a 1/52 chance of being a red Q. Any other spot also has a 1/52 chance of having a red Q, but to meet the stated requirement, that 1/52 is further reduced by the additional stipulation that it be FOLLOWED by the other red Q.

[lynnl]

The following was going around the internet a few weeks ago, after the death of Martin Gardner. It is the Two Sons problem, with a twist: http://www.sciencenews.org/view/gene...bly_look_wrong

I looked at that discussion. I think some people are full of horse hockey.
If someone tells me he has two children, and that one is a son, that leaves two, and only two, possibilities for the other child. And, ignoring any slight statistical M/F birth rate difference, the probability the other child in question is either M or F is 50%. Doesn't matter what day either was born on.
The probability of multiple independent events is the product of their individual probabilities, so in the case: 100% X 50%. And in this neck of the woods that computes to be 50%.

[ieezitin]

who cares?

[lynnl]

who cares?

Obviously not everyone. But some do. Some enjoy puzzles and problems, even though there's no practical purpose served.

That's true of lots of topics. Not everyone cares.

[Carld]

The probability of it being probable is probably not as probable as you probably would think it probable so it's not probable at all as you observe the probabilities of it.

[aostling]

I think some people are full of horse hockey.

"What you have told us is rubbish. The world is really a flat plate supported on the back of a giant tortoise." The scientist gave a superior smile before replying, "What is the tortoise standing on?" "You're very clever, young man, very clever", said the old lady. "But it's turtles all the way down!"