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Lotto Probability

ynot said:
The woman has a pair of children which could be any one of the following . . .

(1) A pair consisting of two males.
(2) A pair consisting of two females.
(3) A pair consisting of one of each gender.

Not . . .
(1) A pair consisting of two males.
(2) A pair consisting of two females.
(3) A pair consisting of one of each gender.
(4) A pair consisting of one of each gender.
Click to expand...

Yes, there are just three possible gender-combinations. But they are not all equally likely, because there are two ways in which you could have one boy and one girl:

You could have an older girl and a younger boy, or a younger girl an an older boy. These two combinations are equally likley. If you looked at all couple that had one boxy and one girl, you would find that for half these couple, the boy would be older.

Please take a few coins and just try this. There really is no need to argue about it, since the result should be obvious.
 
ynot,

Perhaps you should step back for a moment and ask yourself the following ...

Given n outcomes, does the probability of any single outcome always equal 1/n ?

Example -- given 6 possible outcomes, does any one of the 6 always have a 1 in 6 chance of happening?

What do you think?
 
Just thinking said:
ynot,

Perhaps you should step back for a moment and ask yourself the following ...

Given n outcomes, does the probability of any single outcome always equal 1/n ?

Example -- given 6 possible outcomes, does any one of the 6 always have a 1 in 6 chance of happening?

What do you think?
Click to expand...

Just great, now I will sit here and torture my brain until I can think of a common, everyday yet non-intuitive example .... THANK YOU! :rolleyes:
 
BZZZZTT - The game is over! And the winner is . . . . . . . . . Highlight > Not ynot!



There are three gender combinations possible but the odds that the result is a boy/girl combination are twice as likely as a boy/boy or a girl/girl combination. The probability that both are boys is therefore 1/4. I don’t think this has anything to do with the relative ages of the children. It just represents the actual spread of the odds.

(A) Did I know this all along and I was just testing your abilities to defend the obvious? Or my abilities to defend the ridiculous?
or
(B) Was my stance a knee-jerk reaction based on misguided intuition?

In either case, it was good to see that not one other member sided with the stance I was taking.

PS - What about those hermaphrodite children?
 
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ynot said:
PS - What about those hermaphrodite children?
Click to expand...

They, like any other real child, just mess up the puzzle. Saying that any child has a 50% of being a boy, and a 50% chance of being a girl is a drastic over-simplification.

I think that on average, there is a slight biological preference for girls; i.e. a slight majority of pregnant mothers will carry girls.

Also, we have to look at other factors; I think a Chinese girl has a much higher chance of being aborted than a Chinese boy; then there's (infant) mortality rates that might be different for boys and girls. Hermaphrodites would be one such factor, too.

Oh, I can think of more: Depending on the age of the children wee are looking at, recent locals history would have to be taken into account, too. Wars are often a diminishing factor for the male more than the female population.

Now, excuse me, I am off to complain to all twins I know that there's never one who always speaks the truth and one that always lies ... if i'll ever find the way to their village, that is.
 
Rasmus said:
Just great, now I will sit here and torture my brain until I can think of a common, everyday yet non-intuitive example .... THANK YOU! :rolleyes:
Click to expand...
If you roll two dice, they can total 2-12 for a total of 11 outcomes, but the odds of rolling a 7 is not 1/11.
 
illuminatedwax said:
If you roll two dice, they can total 2-12 for a total of 11 outcomes, but the odds of rolling a 7 is not 1/11.
Click to expand...

I have to pedantic and point out that that's not 6 possible results, though, lest somebody notice I was too dense to think of this trivial example myself .... *grumbles*
 
Rasmus said:
I have to pedantic and point out that that's not 6 possible results, though, lest somebody notice I was too dense to think of this trivial example myself .... *grumbles*
Click to expand...

Well, my loaded dice cube should meet your criteria for 6 outcomes, of which each result does not have to be P = 1/6.
 
Just thinking said:
Well, my loaded dice cube should meet your criteria for 6 outcomes, of which each result does not have to be P = 1/6.
Click to expand...

True - but they aren't really that common, or are they? *S*

Anyways, they are both good examples and I shall be content.
 
Molinaro said:
Before any card is chosen, you would expect the probability to be 1/2 -- that the other side of the card you choose, will be blue.
Click to expand...
I get 3/4....

There are a total of 4 sides: blue, blue, blue, red
The reverse of these 4 sides: blue, blue, red, blue

That's 3/4 that are blue on the other side.

Molinaro said:
However, here we are asking the question AFTER you pick a card and ONLY IF the face up side is blue.
Click to expand...
Of course you have changed the wording of your puzzle so, yes, that now makes me wrong when I said there was no extra information. :mad:
 
ynot said:
I don’t think the question asks what the odds were during the process of determining the result. Rather that now that the result has been determined, what are the odds that the result is what it is.

The woman has a pair of children which could be any one of the following . . .

(1) A pair consisting of two males.
(2) A pair consisting of two females.
(3) A pair consisting of one of each gender.
Click to expand...

This analysis isn't relevant unless all three cases are known to be equiprobable. In this case, they're not only not known to be equiprobable, they're known not to be.

Similarly, a roulette ball lands on a coloured space on the wheel which could be any of the following....

(1) A red space
(2) A black space
(3) A green space

If, from this analysis, you conclude that the probability of landing on green is 1/3, then you're in for some serious financial pain....
 
Rasmus said:
Just great, now I will sit here and torture my brain until I can think of a common, everyday yet non-intuitive example .... THANK YOU! :rolleyes:
Click to expand...

The local baseball team plays. Do they

(1) win?
(2) lose while playing?
(3) tie?
(4) forfeit?
(5) rain-out?
or
(6) have the game cancelled due to volcanic eruption?

All six possibilities have been recorded in the anals of MLB (the volcanic eruption was of course the Seattle Mariners in the aftermath of the 1984 Mt. St. Helens eruption). But they're hardly equiprobable.
 
drkitten said:
This analysis isn't relevant unless all three cases are known to be equiprobable. In this case, they're not only not known to be equiprobable, they're known not to be.

Similarly, a roulette ball lands on a coloured space on the wheel which could be any of the following....

(1) A red space
(2) A black space
(3) A green space

If, from this analysis, you conclude that the probability of landing on green is 1/3, then you're in for some serious financial pain....
Click to expand...
Sorry, but the game has been concluded.

A winner has been found and the sponsors word is final! (see post #545)

Was my position in this game (A) or (B)?
 
BillyJoe said:
I get 3/4....

There are a total of 4 sides: blue, blue, blue, red
The reverse of these 4 sides: blue, blue, red, blue

That's 3/4 that are blue on the other side.

Of course you have changed the wording of your puzzle so, yes, that now makes me wrong when I said there was no extra information. :mad:
Click to expand...

We must be speaking different versions of English, because I don't see any changes in what I said.. oh well.
 
Molinaro said:
We must be speaking different versions of English, because I don't see any changes in what I said.. oh well.
Click to expand...
Your first post on this suggested to me that a blue faced card was going to be drawn for me and then I was to say what the probability was of the reverse being blue. In this case, there is no new information when I actually receive the blue card.

What about the 3/4 bit, though? You say it's 1/2
 
Sorry to be flogging a dead horse, but...

BillyJoe said:
I get 3/4....

There are a total of 4 sides: blue, blue, blue, red
The reverse of these 4 sides: blue, blue, red, blue

That's 3/4 that are blue on the other side.
Click to expand...
Molinaro says 1/2
Anyone else?
 
There are two cards on a table, each covered by an opaque sheet of paper. You're told that one of the cards is blue on both sides, and that the other card is blue on one side and red on the other side. But you aren't given any information about which card is which, nor about which side of the two-color card is facing up.

You pick one of the two cards, while they're still covered. Q: What is the probability that your card is the one that's blue on both sides? A: 1/2.

Next, you uncover your card, and see that its top is blue. (You didn't know this before---it might have been red---so you've gotten new information.) Q: Now, what is the probability that your card is the one that's blue on both sides? A: 2/3.

It makes sense that seeing a blue top increases the probability, as seeing a red top would certainly decrease it. Down to 0, in fact.

I realize that I've changed the wording a bit, from "the bottom of your card is blue" to "your card is the one that's blue on both sides". This is intentional. When I tried doing the math in detail, this was the only way I could get things to work. (Of course, if you know that the top of your card is blue, then the two are equivalent; but if you don't, they aren't.)
 
I'm in full agreement with 69dodge in post 559.

But there is one tiny detail not mentioned that I think is important -- and that is that the two cards were originally placed down randomly with respect as to which sides are facing up.
 
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