When the Professor of Mathematics feels bored he flips two coins.
If the result is not two heads he gets his assistant to
run over to Student A and inform him: "Not 2 heads"
Student A calculates that there is a 2/3 chance of a head
and a tail and phones you up. You get to place a bet with the Prof
(even money) that the result is a head and a tail. Fantastic!
It's like "insider dealing"

However, whenever the result is not 2 tails, the Prof gets his
assistant to inform Student B.
Student B calculates that there is a 2/3 chance of a head
and a tail and phones you up. You get to place a bet as before.
So whatever the coins are you will get a call informing
you that the chance of a head and a tail is 2/3.
Every time.

Given that you always place your bet after a single phone call.
What are the odds of a head and a tail?

Originally posted by wolfgang59 When the Professor of Mathematics feels bored he flips two coins.
If the result is [b]not two heads he gets his assistant to
run over to Student A and inform him: "Not 2 heads"
Student A calculates that there is a 2/3 chance of a head
and a tail and phones you up. You get to place a bet with the Prof
(even money) that the result is a head and ...[text shortened]... at you always place your bet after a single phone call.
What are the odds of a head and a tail?[/b]

If you want the truth, here it is:
I have absolutely no idea.
BTW- i'm just curious… where do you get all of these posers?

Originally posted by wolfgang59 When the Professor of Mathematics feels bored he flips two coins.
If the result is [b]not two heads he gets his assistant to
run over to Student A and inform him: "Not 2 heads"
Student A calculates that there is a 2/3 chance of a head
and a tail and phones you up. You get to place a bet with the Prof
(even money) that the result is a head and ...[text shortened]... at you always place your bet after a single phone call.
What are the odds of a head and a tail?[/b]

When you say "every time" do you mean that whenever there are not two heads
(in one case) nor two tails (in another case) you will always get a phone call informing
you that the probability of a head and a tail is 2/3?

Originally posted by HandyAndy When you say "every time" do you mean that whenever there are not two heads
(in one case) nor two tails (in another case) you will always get a phone call informing
you that the probability of a head and a tail is 2/3?

YES.
One of the students necessarily MUST call you after every roll of the dice.
(Sometimes both will try to call you but you only get the info from the first)
Whichever one calls you he correctly informs you that you have a
2/3 chance of one of each.

Originally posted by wolfgang59 When the Professor of Mathematics feels bored he flips two coins.
If the result is [b]not two heads he gets his assistant to
run over to Student A and inform him: "Not 2 heads"
Student A calculates that there is a 2/3 chance of a head
and a tail and phones you up. You get to place a bet with the Prof
(even money) that the result is a head and ...[text shortened]... at you always place your bet after a single phone call.
What are the odds of a head and a tail?[/b]

Originally posted by wolfgang59 No.
The paradox is that we all know it is 1/2.
Yet the students are correctly telling us it is 2/3.

?!?!??!?!

There are four possible outcomes: HH, TT, HT or TH. If the professor does not see two heads,
he does see TT, HT or TH. The probability of a head and a tail is 2/3. If he does not see two tails,
he does see HH, HT or TH -- again 2/3.

Originally posted by HandyAndy There are four possible outcomes: HH, TT, HT or TH. If the professor does not see two heads,
he does see TT, HT or TH. The probability of a head and a tail is 2/3. If he does not see two tails,
he does see HH, HT or TH -- again 2/3.

How do you arrive at 1/2?

It would have to be 1/2. It's not the same as the Monty Hall problem, but the part about the students correctly assessing the probability at 2/3 threw me for a loop... I still don't see how that can be correct. Eliminating one of the outcomes will leave only 2 possible outcomes.

Originally posted by HandyAndy There are four possible outcomes: HH, TT, HT or TH. If the professor does not see two heads,
he does see TT, HT or TH. The probability of a head and a tail is 2/3. If he does not see two tails,
he does see HH, HT or TH -- again 2/3.

How do you arrive at 1/2?

The problem doesn't specify left coin/right coin, so I believe HT and TH only counts as one possible combination rather than two.

Originally posted by lemon lime It would have to be 1/2. It's not the same as the Monty Hall problem, but the part about the students correctly assessing the probability at 2/3 threw me for a loop... I still don't see how that can be correct. Eliminating one of the outcomes will leave only 2 possible outcomes.

You are told
EITHER 2 heads has NOT occurred
OR 2 tails has NOT occurred.

In either case there are 3 possibilities left, 2 of which are heads and tails (HT or TH)