- 24 May '16 20:35When 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? - 25 May '16 03:47

If you want the truth, here it is:*Originally posted by wolfgang59***When the Professor of Mathematics feels bored he flips two coins.**two heads he gets his assistant to

If the result is [b]not

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]

I have absolutely no idea.

BTW- i'm just curious… where do you get all of these posers? - 25 May '16 15:14 / 1 edit

When you say "every time" do you mean that whenever there are not two heads*Originally posted by wolfgang59***When the Professor of Mathematics feels bored he flips two coins.**two heads he gets his assistant to

If the result is [b]not

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]

(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? - 25 May '16 19:58

YES.*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?

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. - 27 May '16 04:50

2/3 ?*Originally posted by wolfgang59***When the Professor of Mathematics feels bored he flips two coins.**two heads he gets his assistant to

If the result is [b]not

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] - 27 May '16 21:48

There are four possible outcomes: HH, TT, HT or TH. If the professor does not see two heads,*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.

?!?!??!?!

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? - 27 May '16 23:42

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? - 27 May '16 23:50 / 1 edit

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 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? - 28 May '16 00:07

You are told*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.**

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)