19 Jul '09 23:10>
Originally posted by PalynkaThis is so simple.
Originally posted by wolfgang59These things always become easy if you go back to basic.
Palynka I agree with you entirely ... so why do we get the old chestnut of:-
A man has two children.
What is the probability he has one son and one daughter?
We all agree 0.5
You hear him on the phone chatting to one of his children called MARY.
now what is the probabilty he has a son and a daughter?
0.5 (I hope Palynka agrees) but so many puzzles, quizzes, books and (dare I say it teachers) say 0.67 !!!!!!!!!!
Originally posted by wolfgang59Isn't this just straightforward application of Bayes?
I toss two coins, lets say a £1 coin and a £2 coin they land unseen by me.
A friend tells me that I have tossed "at least one Head"
What are the chances that the coins are different? (ie H & T)
My friend's information tells me that the result is one of these three.
£1 head and £2 head
£1 head and £2 tail
£1 tail and £2 head
The c ...[text shortened]... l the time) my chances of tossing a head and a tail are 2/3.
But we all know its 1/2.
Originally posted by MetBierOpThe Monty Hall problem is NOT the same. The game show host would never open a door with a prize, something which is key. That doesn't apply here. In the Monty hall you're conditioning on the game show host opening a door with a goat (something which happens with probability 1). In this case, you're conditioning on something which doesn't have necessarily probability one.
This is so simple.
At first you have a probability of 1/3 of picking the correct door.
Then a WRONG door is eliminated.
This leaves two doors.
Although you do not know if at first you selected the wrong door.
The chances are 2/3 you have it wrong. Thus your change of winning the car is 1/3. However you do know that the car is still there. So changing your choice reverses the odds.
Originally posted by LemonJelloSo you are saying its 2/3?
Isn't this just straightforward application of Bayes?
P(H&T| at least 1 H) = P(at least 1 H|H&T)*P(H&T)/P(at least 1 H)
= 1*0.5/0.75 = 2/3.
In what follows, you are simply not answering the same question. You would be answering P(H&T|at least 1 H or T) = P(at least 1 H or T|H&T)*P(H&T)/P(at least 1 H or T) = 1*0.5/1 = 1/2. So, your mis ...[text shortened]... doesn't work the other way around, so they don't provide the same amount of information.
Originally posted by wolfgang59Hello Wolfgang.
METBIEROP you are wrong!
A palynka has demonstarted. (Just substitute heads and tails for boys and girls in original problem)
Another way of looking at it is this.
A man has two children.
One he is speaking to on the phone, the other he isnt.
What is the probability of the child not on the phone being a boy?
How can knowing the sex of the child on the phone change that?
Originally posted by wolfgang59Yes, I think in the problem as you laid it out, the probability of H&T given you have at least 1 H is 2/3. Same would go for the probability of H&T given you have at least 1 T. Like I mentioned, isn't this what an application of Bayes would tell us?
So you are saying its 2/3?
Originally posted by LemonJelloIt's a subtle point, but what you say is not the same as the stated problem. It assumes that the friend would always say that there's at least one H when the coins are different.
Yes, I think in the problem as you laid it out, the probability of H&T given you have at least 1 H is 2/3. Same would go for the probability of H&T given you have at least 1 T. Like I mentioned, isn't this what an application of Bayes would tell us?