Settle a bet - Probability question

[CanTak3]

Thanks to everybody for the answers. It was only a $1.00 bet, but we’re fighting ignorance even if it’s my own!

[CanTak3]

I should add that I wrote a Delphi app to try this and it confirmed it is 2/3. Probably should have done that in the first place! Any non-believers can email me for code.

[Gary_T]

@Jman

The way the question is phrased, you CAN NOT pull out a tail showing…

Sure you can. You just ignore the 25% of the time when it happens, because it’s not relevant to the question, which is only interested in the 75% of cases where heads was showing. And in those cases, the other side of the coin will be heads 2/3 of the time.

[Malacandra]

It was 2/3 once, in 1960. For about 20 minutes.

[Ximenean]

@Jman

You must not confuse this with the “Monty Hall” problem, where you learn one of the ‘doors.’ The revealing of a door in that situation only eliminates that door, whereas the revealing of a heads in this situation is tied to the probability of another heads being on the reverse. This is a situation where knowing one of the sides alters the probability of the remaining sides showing up, while in a ‘door choice’ problem, it doesn’t.

I think you can only arrive at an answer of 1/2 if you do assume that there is a Monty Hall-type figure who takes a coin, inspects it and shows you a Head. Since he can do this whichever coin he picks, the probability is 1/2.

But if you just get to see a random face of a random coin, the probability is 2/3.

[glee]

It may help Jman if you imagine scratching both sides of the two-headed coin, putting an A on one side and a B on the other.

Now when you look at a head, you may see:

• no scratching (the other side is a tail)
• scratch A (the other side is a head)
• scratch B (the other side is a head)

Therefore it is 2/3 the other side is a head.

[Mathochist]

@Jman

Incorrect…since you’ve already established that the face observed is heads, you have TWO possible outcomes, H or T. You must eliminate the ‘showing heads’ for both coins, not just the one…

You must not confuse this with the “Monty Hall” problem, where you learn one of the ‘doors.’ The revealing of a door in that situation only eliminates that door, whereas the revealing of a heads in this situation is tied to the probability of another heads being on the reverse. This is a situation where knowing one of the sides alters the probability of the remaining sides showing up, while in a ‘door choice’ problem, it doesn’t.

This is the MH problem, since you don’t know which head of the two-headed coin it might be.

[Jman]

As I read the problem originally, I was discounting the inital draw, because I initially interpreted it to mean that drawing the coin showing tails was impossible (given the conditions of the problem). With that thinking, it IS the same as “I give you one of two coins, showing heads, waht’s the probability that the other side is heads.” 1/2.

As I read the problem again today, the correct answer seems obvious, as the problem does clearly ask that on the occasions that the pulled coin is heads, what’s the probability that the other side is heads. I have no idea why I originally interpreted the problem as I did, because given the true question, it is quite obviously 2/3. :smack: