[Guest]

There are two coins in an opaque bag. They are identical except one is a heads and a tail; the other has two heads. What are your odds of getting two heads if you withdraw one coin at random, place it face down on the desk, and continue by doing the same with the second coin?

[Guest]

0 - if you place them both face down, you'll have 1 tail up and 1 head up.

[Guest]

The odds would be 50/50. If you pull out the two-headed coin, you have 100% of placing it heads up, the two-sided coin is a 50/50 shot. So the odds of both being placed heads up are 50/50.

[bonanova]

A coin toss.

Label the four sides A [heads] B [tails] - that's the first coin - and C [heads] D [heads] - that's the second coin.

There are four equally likely events:

A and C -> HH

A and D -> HH

B and C -> TH

B and D -> TH

The question is identical to this:

You have one coin, showing both H and T.

You pull it from a bag and place it on the table.

What is the probability it shows H?

[Guest]

Sorry, the odds are 2 to 1. If the first coin face revealed is a head you stop because there is no way to achieve two tails.

[Guest]

0 - if you place them both face down, you'll have 1 tail up and 1 head up.

Suggest you re-read the problem

[Guest]

Sorry, the odds are 2 to 1. If the first coin face revealed is a head you stop because there is no way to achieve two tails.

That is if it is a tails you stop.

[Guest]

A coin toss.

Label the four sides A [heads] B [tails] - that's the first coin - and C [heads] D [heads] - that's the second coin.

There are four equally likely events:

A and C -> HH

A and D -> HH

B and C -> TH

B and D -> TH

The question is identical to this:

You have one coin, showing both H and T.

You pull it from a bag and place it on the table.

What is the probability it shows H?

The difference comes from drawing them one by one.If the first shown face is a heads you have no reason to continue,

[Guest]

I'd heard "heads" called "faces" before, and thought it was a trick question. Oh, well!

[Guest]

The odds would be 50/50. If you pull out the two-headed coin, you have 100% of placing it heads up, the two-sided coin is a 50/50 shot. So the odds of both being placed heads up are 50/50.

But the point is you are not pulling them out at the same time and that would dictate your next move.

[bonanova]

The difference comes from drawing them one by one.If the first shown face is a heads you have no reason to continue,

The OP asks: What are the odds if both coins are drawn, one at a time. It says to draw the second coin, regardless.

No matter, it's a red herring.

The two coins have four faces. Three faces are H; 1 face is T.

The first face you see is a T 1/4 of the time, whether you stop or not.

The second face you see is also a T 1/4 of the time.

So half of the time you'll see a T.

The other half of the time you'll see 2 heads.

It might help to spell it out in more detail.

Considering order, there are 8 equally likely events.

Referring to my previous post, they are:

HT coin is drawn first:

A then C - HH

A then D - HH

B then C - TH - Here you choose to stop. So it's B [tails] then stop.

B then D - TH - Here you choose to stop. So it's B [tails] then stop.

HH coin is drawn first:

C then A - HH

C then B - HT - Here a T shows on the second coin. You'll always see it, because here the first coin is HH.

D then A - HH

D then B - HT - Here a T shows on the second coin. You'll always see it, because here the first coin is HH.

In 2 cases [1/4 of the time] you see T on the first coin and choose not to draw the second coin.

In 2 other cases [1/4 of the time] the second coin is T.

So, in 4 cases [1/2 of the time] you see T.

The other 1/2 of the time you see 2 heads.