On Friday I set this puzzle….

Imagine that I toss two coins. If at least one of them comes up heads, I show it to you. What is the probability that the other one is also a head?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

There are 3 possible outcomes:

H T

H H

T H

Therefore there is a one in three chance that the other coin is also a head.

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

### Like this:

Like Loading...

*Related*

I’m sure you already asked this question a while ago, albeit in a different form. Slightly less ambiguous this time, if I remember rightly.

I’d say Richard is correct on this one.

3 possibilities

coin 1: H T H

coin 2: H H T

He shows you a H (without identifying it as coin 1 or coin 2).

Maybe he showed you coin 1 and coin 2 is a T

or maybe he showed you coin 2 and coin 1 is a T

or maybe he showed either and both are Hs.

There’s only 1 way out of the 3 that both end up Hs. I think if he showed you a H and said “this is from coin 1”, then the chances are 1 in 2.

I had 100% chance. The coin I chose was an Australian $2 coin, which has two heads. :P

i think it should be 50% cos you will have the following possibilities: 1H-2T, 1H-2H, 2H-1T, 2H-1H.

Not seeing your logic there Kroketje. If the coins have 2 heads how could they possibly com up tails? And even if only the second coin has 2 heads the first coin has already come up heads..

What are you talking about? There’s 2 coins, not 3.

I mean that if coin 1 turns up heads first then coin two has 2 possibilities, and if coin 2 turns up heads first, coin 1 has also 2 possibilities. so there are 4 possibilities not 3.

so the possibilities are:

T-H

H-H

H-H

H-T

Kroketje,

There are 4 possibilities, but one of them can be excluded because of the scenario. He asks: “What is the probability that the other one is also a head?”

That means you can exclude the T-T scenario, because in that case he would not show you a coin. That leaves 3 valid possibilities and only one of them is H-H.

You’re right. It’s 50%

It’s the same case that in this previous puzzle https://richardwiseman.wordpress.com/2012/08/06/answer-to-the-friday-puzzle-166/

Are independent variables, so is equivalent to ask the probability of toss one coin. 50%

-M- 3 possibilities but not with the same chances. HH can be viewwed from two sides so this possibility has 50% or twice 25% if you prefer.

Juan,

No.

A) T (50%) – T (50%) = 25%

B) T (50%) – H (50%) = 25%

C) H (50%) – T (50%) = 25%

D) H (50%) – H (50%) = 25%

T-T excluded leaves three times 25%. 1 out of 3. You can’t count option D double. Take two coins and test it.

It’s just like throwing with two dice gives 7 in most cases.

And you are right, the puzzle you refer to is false.

I think its 50-50

in Richards solution we have:

HH

HT

TH

Right? this means that he counts HH as HH regardless of order, but that also means that HT and TH should be considered the same outcome: one T and one H, regardless of order. If we are going to count HT and TH as two different outcomes, then hH and Hh is also 2 different outcomes. So its either:

Hh

hH

tH

hT

=50%

OR

HH

HT

=50%

Hmm, john_H’s explanation made it clearer, I suppose my reasoning is wrong. 1/3 it is then..

Anders, with your H h and order….. the problem you are making is you aren’t considering all the combos if you are looking at all the possibilities

you list :-

Hh

hH

tH

hT

but using that scheme, the list is actually :-

Hh

Ht

Th

hH

hT

tH

excluding…. tT and Tt

so now you have 6 possibilities, with 2 being hH or Hh, meaning the probability is 1/3

What confused me was whether the order mattered, it doesnt.

Coin 1= ht

Coin 2 = HT

I kept thinking there were 6 outcomes in total, but;

There are 8 possible outcomes:

hH

Hh

hT

tH

Th

Ht

tT

Tt

6 of these has heads, and only 2 of those 6 has 2 heads 2/6=1/3

Heh, you are right keith, I wrote my own correction before seeing yours, haha

That’s indeed the right answer. Well done Richard.

No, it is not the same case at all. Think of it. Conditional probabilities are not always easy to figure out. Many people have big problems with them.

In this coin problem, we start we ZERO coins tossed.

In the Boy-Girl problem, we start with ONE GIRL already guaranteed to be there.

This is something that takes getting used to. In conditional probability, the probabilities change depending on previous situations EVEN IF the variables are independent.

Another classic that has caused big headaches for many people, and has bloodied many a nose including of people who consider themselves experts, is the world-reknowned/feared Monty Hall problem.

Very happy to have got this one without any arguments over the question

I agree with Julie that it’s nice to have an unambiguous question and an answer that was pretty quick to figure out.

Kroketje is konfused. The puzzle doesn’t ask about the “first coin” or the “second coin”; you look at both simultaneously, and look if “at least one” is heads (etc.).

I must admit, I don’t quite get it. We already know that one coin has come up heads, so its probability is 100% heads. Now there are two possibilities, either the other one is heads or it’s tails. So the probability asked is 50%. Can you explain please?

No. If one coin has already come up heads, that eliminates the possibility of both coins being tails. So the only possibilities left are TH, HT, or HH (first letter designates coin 1, second letter coin 2). All three possibilities are equally likely, so the chances of both coins being heads is 1 in 3. If you don’t believe it, you can just go ahead and flip two coins enough times to convince yourself.

I agree

I don’t agree, sorry!

The first coin tossed has no bearing on what the second coin will land on. There are ONLY two options, heads or tails!

You did actually say you ‘tossed two coins’ which could have been done simultaneously, the ONLY possible outcomes being:

HH

TT

HT/TH

Therefore a one in two chance the other is a head!!!!

TT cannot happen “If at least one of them comes up heads, I (Richard) show it to you.” If HH happens, Richard can show either head to you. If HT or TH comes up, Richard can show you the coin that is heads. Out of these three possibilities only HH will allow Richard to show you another head. His chances of doing that is 1 in 3.

Ah, ok, I overlooked the fact that the coin could be chosen from the two after they have been tossed.

i am puzzled. prob seems same as asking “i hide two hands behind my back. one hand (obviously) is a Right hand and the other is a Left hand. i show you the (eg) Right hand. what possibility is there for the Other hand?” so ans here is 100%

Haha wow. I don’t know what it is you smoke between Friday puzzles, but it’s really taking its effect now.

I want some of what he’s taking….

Catherine, the order doesn’t matter – what matters is that after the two coins were flipped, ONE was shown to you with a head on it – which means that out of the four possible combinations for the two coin flips, we can eliminate one. There are therefore three of the original 4 options remaining (HT. TH, HH) – and since we have shown ONE head already, only ONE of those three cominations could be the “other head”, so 1 in 3. The crux of this question rests on the one head being shown before asking what the probability is.

You can eliminate 2 of the 4 options…. If you show the coin witch is head, your only two options are: H-T and H-H.

kroketje: no, that is not true.

1) In case of T-T, Richard shows nothing

2) in case of H-T, Richard shows the H, and still has the T

3) in case of T-H, Richard shows the H, and still has the T

4) in case of H-H, Richard shows the H, and still has another H.

-M-: Thanks for that explanation. I was under the assumption that the remaining coin would have a 50% chance of being a head, since there were only two possibilities left. But, with the explanation you just posted in this reply, I can actually see the logic of the answer being 1 in 3.

-M- the 4 point can be I show the coin 1 or I show the coin 2. Are two possibilities:

You can show:

1H – 2T

2H – 1T

1H – 2H

2H – 1H

I am sure this is wrong but I think the chances are only 50%. Since we already know that one coin is heads, then the options are only HH or HT, the third option of TH is a repetition of HT (it is virtually the same since the coin order was not specified) so it can be eliminated also. Since heads for one coin is predetermined, the chances of the other coin being heads reverts back to 50 50.

exactly what i think

That is a classical mistake. The 50/50 solution would be valid on the condition that you have not seen what happened to the first, i.e. if the last one is the only one you are confronted with.

The Monty Hall problem is another classical example.

It doesn’t matter which coin is first, so in that sense TH is a repetition of HT, but this doesn’t mean that you can simply ignore it. The relevant point is that there’s a 50% chance of getting one H and one T, with only a 25% chance of HH (and 25% TT). We know that TT is out of the question, so now we have either a 2-in-3 chance of ‘one of each’ and a one-in-three chance of HH. Which gives the answer.

Exactly!

If 1H – 2H is the same that 2H – 1H then too the TH is the same that HT

I would submit that the probability is 0, and here is why.

If

at least oneis a head, Richard shows it.Therefore, if the first one is a head, he shows it: H

He now turns over the second one, which is also a head, and he shows it: H

Q: What is the probability that

the other oneis also a head?A: none, there are only two coins and both have been shown. The probability that “the other one” has a head is 0, since there is no other coin available.

Yes……

one COULD interpret it that way…

I have to say that I read the question this way first too; “at least one” means >1, “one or more” means >=1.

However, I thought that as you only show one head, the other one is always going to be a head … trivial!

“At least one” means exactly that: “the very least number is one”.

“At least one” is synonymous with “one or more”

But I think Bart B. Van Bockstaelhe’s point is more that the wording of the question implies that if it is more than one (i.e. two) heads then both are shown. I think the word “it” rules that out.

“If at least one of them comes up heads, I show it to you”. “It” here implies one, not all, even if both are heads.

Thanks. I had a problem with “it” because it sounds so out of place. So, I interpreted “it” not as “one coin”, but as “the situation”, meaning that one coin would be shown if there was one head and two if there were two heads.

Whoops, didn’t mean to post anonymously.

“At least one” means exactly that: “the very least number is one”.

“At least one” is synonymous with “one or more”

But I think Bart B. Van Bockstaelhe’s point is more that the wording of the question implies that if it is more than one (i.e. two) heads then both are shown. I think the word “it” rules that out.

“If at least one of them comes up heads, I show it to you”. “It” here implies one, not all, even if both are heads.

One answer is wrong. This is the same case: https://richardwiseman.wordpress.com/2012/08/06/answer-to-the-friday-puzzle-166/ but with Girl-Boy

And in the other puzzle the answer was 50%

No it is not the same case. It is very different.

In the Boy-Girl problem, you are GUARANTEED to have a girl.

In the coin problem you HOPE for a head.

“If at least one of them comes up heads, I show it to you.”

This guarantees that there is a Head, it’s not a hope.

No it does not. The only guarantee here is that you will be shown IF AND ONLY IF there is a head.

I make two childs = I toss two coins

If at least one is girl = If at least one is head

I show one girl = I show one head

You don’t know if is child #1 or #2 = You don’t know if is coin #1 or #2

What’s the difference?

Oops. I posted my reply in the wrong thread. Sorry!

Here it is:

I understand the confusion, but you really are misunderstanding the problem. That’s not an insult, it takes practice to recognise the subtleties of conditional probability.

In the Boy-Girl case, you start with KNOWING that there is a girl

In the coin case, you start with HOPING/WISHING that there is a head.

The point is this:

Since you already KNOW there is a girl, the probability of getting this girl is no longer relevant. It has already happened. Therefore, you only have to look at the second girl or boy, and that gives you a 1 in 2 probability. You do NOT have to take into account the possibility that the first child will NOT be a girl.

In the case of the coins, you start out BEFORE the first coin has been flipped. As a result, you MUST take into account the possibility that the first coin will NOT be a head.

See? It is quite different.

If it is not clear, I will do my best to post a more graphic explanation, but I will need some time for that.

Thanks for your time Bart. :)

But the text is clear ” If at least one of them comes up heads, I show it to you”

Is not a hope/wish, is a fact. Only in this case we play.

I can assume that the other puzzle have a wrong answer but, in fact are the same.

I suggest you look up the meaning of the word

IF. That should make it clear that both situations are very different from each other.Sorry Juan AR,

Richard got the answer to the boy/girl question wrong and the answer to this one right.

You have to remember that Richard is a human being, just like you and me.

As such he is not infallible. Nor am I. Nor are you.

The only person who is infallible is the Pope (when speaking ex-cathedra) – eat your heart out Richard Dawkins/Phillip Pullman/Christopher Hitchens.

The Contrarian (aka El Twato)

The statement that only the pope is infallible, shows just how infallible and gullible people truly are.

I think it’s one in three too,

The only twist I devised was that we were asked to imagine that two coins were tossed,

I imagined, that when tossed, one turned into a goldfish and the other a peanut so the probability was zero.

I understand the confusion, but you really are misunderstanding the problem. That’s not an insult, it takes practice to recognise the subtleties of conditional probability.

In the Boy-Girl case, you start with KNOWING that there is a girl

In the coin case, you start with HOPING/WISHING that there is a head.

The point is this:

Since you already KNOW there is a girl, the probability of getting this girl is no longer relevant. It has already happened. Therefore, you only have to look at the second girl or boy, and that gives you a 1 in 2 probability. You do NOT have to take into account the possibility that the first child will NOT be a girl.

In the case of the coins, you start out BEFORE the first coin has been flipped. As a result, you MUST take into account the possibility that the first coin will NOT be a head.

See? It is quite different.

If it is not clear, I will do my best to post a more graphic explanation, but I will need some time for that.

Someone shouting on the internet again…….

I think the confusion here is to do with how you imagine the trial is carried out, and the fact that this is not clear from the question!

If you interpret the scenario as “I reveal to you one coin and it is heads, now what it the probabilty that that both coins are heads?” then you would come up with 50%.

However the question is “At least one coin is heads” – so this would require so foreknowledge i.e. me to look at both coins (without showing you) and declare to you that at least one of them is heads. Then it should be obvious that the probability of both being heads is 1 in 3.

“If at least one of them comes up heads, I show it to you. What is the probability that the other one is also a head?”

I interpret “I show it you”, that he shows me the head, so in this case we are in your first scenario.

Do you understand the meaning of the word

IF?Shouting in Bold now……

Just proved Richard’s result to myself on Excel using the random number generator and ‘=IF(E6>=0.5,1,0)’ to model the trials (T=0, H=1) and then using ‘=IF((SUM(G6:H6))>=1,SUM(G6:H6),””)’ and checking the ratio of HH results to all other results.

If you’d like a copy of my spreadsheet please let me know (@edzeteito).

Nah, you’re ok thanks.

I’m still a little confused. How does “If at least one of them comes up heads, I show it to you” eliminate the possibility of T-T? Doesn’t that mean that if the first coin comes up tails, he doesn’t show it to you?

You’re right that it can’t be T-T, Paul. So the remaining 3 options are H-H, T-H, or H-T. In any of those cases, he shows you One coin, a Head – all three options have at least one Head. In only one of those 3 cases is the other coin also a Head – in the other 2 cases, the other coin is Tails, so there is a 1 in 3 chance the other coin is also a Head.

The person with the pre-knowledge (that at least one is heads) would have had to look at one coin. If it was heads he can skip the next step. If it was tails he would have had to look at the 2nd coin in order to establish the fact that at least one is heads. If it was T-T then that is a different scenario from the one given in the question.

The question is misleading because it states “at least one is heads” without explaining how that fact is established, almost implying that one coin has been revealed to be heads and therefore what is the chance the other is heads. For the answer given, the trial in practice would require someone to look at least one coin (as describe above) in order to then tell you that at least one is heads.

Ok, I sort of get it now.

If the words “If at least” were removed, and it started off with “One of the coins is heads, and I show it to you”, it would be clearer.

I think Richard is totally wrong here.

I have two approaches:

1.) The second coin does not know anything about the toss of the first coin, so chances that the second coin is heads, is exactly 50%

2.) There are 4 possibilities:

1H – 2H and the first coin is shown

1H – 2H and the second coin is shown

1H – 2T and the second coin is shown

1T – 2H and the first coin is shown

in two of them head is up, and in two of them tail is up, so again 50%

Sorry, I meant to say:

2.) There are 4 possibilities:

1H – 2H and the first coin is shown

1H – 2H and the second coin is shown

1H – 2T and the first coin is shown

1T – 2H and the second coin is shown

in two of them the other is head up, and in two of them the other tail is up, so again 50%

Important are two facts:

– Its asked for the AVERAGE chance, so try it out to toss the coins one hundred times and you will see what i mean. Yeah just try it out.

– Its only asked for a part of the results (and thats the trick) The TT Combination is excluded

To solve with math (and get the really average random) i recommend to use the binary digit system for this case:

00

01

10

11

After deleting 00, we get “one in three.”

BECAUSE they dont know anything about the others, its possible to get two results HT and TH, whyle HH is only one

you aren’t working out the combinations correctly

you throw the coins…..

First coin will either be H or T, Second will be H or T…

so when first is H the other will be H or T giving

H H

H T

and when the first is T the other will be H or T

T H

TT

so the combinations that occur are :-

HH

HT

TH

TT

this is all before any coin is chosen…

now, given the those Richard needs to choose a H from each combination.

TT – can’t choose anything, so doesn’t end up showing anything, so doesn’t match the stated problem.

HH, he can choose either, and the other will be H

HT, chooses the H the other is T

TH, chooses the H and the other is T

so, given all his choices, the coin is

H

T

T

probability of H being chosen, 1 in 3.

So what are the chances in two roulette tosses, if the first toss is red, that the second toss is red too? 50%, right? Otherwise I’d be rich.

You forgot the zero. The probability to get red is less than 50%. On top of that, there is a betting ceiling.

of course, assuming there is no zero….

according to Richards answer I don’t care about betting ceilings, because then I would always bet black after a red, and would have a 2/3rd chance, always betting the same amount, and being really really rich.

Your example is not comparable. Richard does not say that the first coin is a head. He says that at least one of the coins is a head. That’s a different thing.

This is the same thing. If I ask you in two consecutive roulette tosses, if one of them is red (either the first one, or the second one), what are the chances that the other one is red? 50%. The 2 tosses do not know from each other!

A roulette wheel with no zero? Show me the casino that offers that game. The zero, or double zero, are what give the house an edge. Without it, they are playing a zero sum game. Vegas was not built on the house breaking even.

To address your original comment, no, your roulette example and Richard’s problem are not the same. In Richard’s problem, the coins are tossed in tandem. He then looks at them and reveals if one is heads. He does not toss them in series. Your roulette wheel is spun in series. That changes the probability.

Here is my two step, foolproof guide to dealing with these puzzles:-

1. Read the question

2. Answer the question

Simples

You could also have

3) Read the answer on Monday.

This step can be skipped if

1) You are convinced you have the correct answer and wont be persuaded otherwise (a popular choice)

2) You really don’t care

It’s not that math that makes it true – it’s the clever wording at the beginning. On some cases even with the wording the answer is pure cow poop, but he got me and a lot of us this time, fair and square.

If it makes anyone feel better, just remember that he has a really, REALLY crappy commute a few days a week. =p

I think the logic given above is correct but I disagree with the Boy-Girl logic. Following that logic:

> We know that BB is impossible, so we are left with:

> BG

> GB

> GG

There are 3 possible outcomes:

H T

H H

T H

> If the girl was born first then the possible combinations are

> GB

> GG

If the coin shown was the first then the possible combinations are

HT

HH

> If the girl was born second then the possible combinations are

> BG

> GG

If the coin shown was the second then the possible combinations are

TH

HH

Either way, the answer is 50:50

Where is the flaw in the above when it applies to coins but not to kids? For me both the situations are the same with the 1/3 solution.

I also thought the boy-girl logic was wrong, but there is the difference of the way the B-B (or T-T) was excluded.

There are two kids, giving 4 scenario’s

1)B 2)B

3)B 4)G

5)G 6)B

7)G 8)G

You got to see one of the two kids at random, giving 8 possible scenario’s. So you got to see for example -4-. Then the other is a boy. Since this happened to be at random, you might as well have seen a boy in stead. The chance that you saw a girl was 50%.

You are left with scenario 4, 5, 7 and 8. Two of them are G-G giving it 50%.

…

In the H-T scenario the 1-2 scenario’s are excluded. However, the scenario 3 is replaced by scenario 4. Scenario 6 is replaced by scenario 5.

You are left with scenario 4, 4 (in stead of 3), 5, 5 (in stead of 6), 7 and 8. That leaves 6 scenario’s and 4 of them is H-T.

M

I’m not a racist, but could we have that in English please?

Ta old bean (don’t ya know)

XRayA4T

For what it is worth, I have spent more time thinking about this puzzle – or rather the boy/girl puzzle – than any other posted by RW in the last year (absent the monks/bluespot thing, which is bananas)

For what it is worth, I also agree with you, the answer to both quesrions is 1/3.

i intend to not think about either anymore.

You have to admit, however, this puzzle is really good at winding people up, which it is renowned for doing. ( Which you will see should you be sad enough to Google it, as I am).

Anyroad I peck your liver

TMT

For all of you who are convinced the answer is 1/2, I suggest you toss 2 coins together 100 times (or as many as you can be bothered doing) and make a note of the result. Cross out any TT (no heads). Then count how many HH there is compared to HT/TH. There will always be some exceptions to the rule because it’s random, but most people will get an answer approximately 1/3.

If you dont believe it, try it and see.

Can I suggest (Dave permitting) that you do this on a spreadsheet. Much quicker and easier.

It’s only quicker and easier on a spreadsheet if you know how to generate random coin tosses (or something representative). Perhaps you could teach us?

Easy! Just use the ‘=rand()’ function which generates a random number between 0 and 1. Then use an if statement to get H (for numbers >= 0.5) and T (for numbers =0.5,”H”,”T”)’. Copy rows to make as many trials as you like.

Then use an if statement to get H (for numbers >= 0.5) and T (for numbers =0.5,”H”,”T”)’.

You get the idea…

I’m not much of a coder but here you go: http://codepad.org/M9EkMZZt

Feel free to increase the ‘max_flips’ variable to change the size of the sample.

PS: I not only permit, I enable. ;)

The question should be rephrased as this “Imagine that I toss two coins. If at least one of them comes up heads, I show the coins to you. What is the probability that the other one is also a head?”

The other way heavily insinuates that if a coin comes up heads then that coin is shown to us. This takes that coin out of the equation and leaves a 50-50 chance that the other one is also a head. Rephrasing it allows for the given answer.

Good evening Furie

I have noticed over the last few weeks that you and your chums seem unable to answer the question as presented. You then answer some other question and get into all sorts of scrapes.

I think the question should be rephrased as “What is the capital of England?”

I have to tell you that on this basis Richard’s answer falls far short of adequate…..

PS SPOILER ALERT..it begins with “L”

Actually, it doesn’t talk about order, so, there are two choices: HH andHT.

1) Sorry to post a week late – I was on vacation. But the correct solution hasn’t been stated yet.

2) “Order” isn’t the right word – it’s “distinct.” The two coins are distinct, whether or not there are any words that distinguish them in the problem statement, or even if you can tell the difference from flip to flip. The point is that any time you do flip one, you can call one the “left” coin and one the “right” (and if you try to get pedantic and claim there could be a tie, one is “far” and one is “near”).

If you still doubt this, think about comparing two variations. In one, Richard flips two identical coins. In the other, Richard tells us they are different, but never tells us which coin landed heads. Do you really think the chances are different in the two cases?

3) It is not correct to treat the two coins as independent random variables. They are, but the information we have about them is not about either one, so we must use a joint probabilities.

4) But the question was ambiguous. The answer depends on why Richard told us that one coin came up heads:

4A) If he was pre-disposed to tell us about heads, then in three out of four sets of tosses he would tell us there was a heads, and in one out of four there will actually be two. That makes the answer (1/4)/(3/4) = 1/3. But we don’t know what he would do in 1/4 of the cases, when there are two tails.

4B) If he simply picks a face that is showing, in half of the cases where there is a heads and a tails, he will tell us about the tails. So he only tells us about a heads in two out of four cases, and the answer is (1/4)/(2/4) = 1/2. This is the more realistic case.

4C) This can be easier to see if you change it to dice. If I roll two, what are the chances I rolled doubles if I tell you one is a three? Does it change if I tell you a different number? If you think it does, what determines which number gets which probability? If you think it doesn’t, how is the question different than if I don’t tell you a number?

And if you still doubt, get a friend to play the dices game a hundred times, and count.

5) And in fact, 4B is the solution that is universally accepted by mathematicians in all problems of this type EXCEPT this one, where you can find proponents of both. The characteristic that defines the type is that two properties CAN exist, at least one (either one) ALWAYS exists, and you somehow learn about one. You are asked for the probability that only one property applies. The archetype is Bertrand’s Box Paradox, where a box contains two coins that can be either silver or gold. The two properties are “one is (gold or silver),” which we can compare to “one landed (heads or tails).” Bertrand used three boxes instead of four, but his solution when applied to four boxes is 1/2, not 1/3.

The famous Monty Hall Problem is another example. The two properties are “The good prize is not behind the door to the (right or left) of the contestant’s door.” If we apply Richard’s solution for today’s puzzle, there are 2 possible outcomes that remain: the prize is behind the contestant’s door, or the prize is behind door that was not chosen by either the host or the contestant. That makes the answer 1/2, which is wrong. It is 2/3

Nice explanation, although I don’t agree that this is a matter of opinion among mathematicians. The posted solution is THE correct solution to this problem as it is unambiguously stated (ie your 4A), and 2/3 is THE correct solution to Monty Hall. Both can be verified with computer simulation. If you are saying that applying Richard’s solution to the Monty Hall problem leads to error, all that would mean is that it is a misapplication, and indeed it is.

This is also a nice place to bring up Bayes’ rule. We are looking for the probability of 2 heads, given that we are shown 1 head.

p(2H|shown 1H) = p(shown 1H | 2H) * p(2H) / p(shown 1H)

The first term – the probability of being shown a head given that 2 were thrown, is 1.

The second term – the probability of 2 heads, is 1/4

The third term – the probability of being shown a head, is 3/4

1*(1/4)/(3/4) = 1/3

Laura:

Your answers only can be “verified” by computer simulation if you make different assumptions for the two problems (or it is made clear in the problem statement). For the Two Coin Problem, you must assume that anybody who flips a heads and a tails will always show you the heads, and never the tails. Which means that the answer to “If I show you a Tails, what is the probability that the other one is also a Tails” is 0. Now I agree that Richard’s question is clearer than I originally thought; I reacted too quickly. But that is a common mistake – most people will read it as I re-phrased it, and most mathematicians who answer what I re-phrased will say “1/3,” which is wrong. They often will resort to a computer simulation which imbeds the assumption, or Bayes Rule, both of which can also “verify” that the answer to the coin problem is 1/2, and the answer to the game show problem is 1/2, by switching the assumption.

My point, which I still want to emphasize, is that any good soluiton needs to discuss the importantce of that assumption.