Fog Creek Software
Discussion Board

painfully easy

The answer to painfully easy is incorrect.

Each coin flip is an independent event.  There is always a 50% chance of getting heads.

Knowing that one coin is heads, does not change the odds that the other is also heads.

If you flip one coin and it comes up heads, what are the odds that it will come up heads again? 



Paul Whitehurst
Monday, April 21, 2003

Flip two coins one after another. At least one of them landed heads.


Either the first coin landed heads, so there is a 50% chance that both are heads,


the first coin landed tails, so there is a 0% chance that both coins are heads.

So you see the probability that both are heads if at least one is, is less than 50%.

But there's a more subtle problem with the puzzle.

Say I flip two coins, and I tell you "one of the coins landed heads."

But if both coins came up tails, I must said say something different.

"One of the coins landed tails?" But then you could deduce immediately that they were both tails, since in any other situation I would have said "one of the coins landed heads."

Unless sometimes I get to choose whether to say "one of the coins is heads" or "one of the coins is tails" in the case where there are one of each. But this makes the proposed answer invalid.

The 1/3 answer supposes that I am 100% likely to say "one of the coins is heads" if it is true. But this assumption isn't given in the problem statement nor is it likely if I were playing the game for real.

Peter Meilstrup
Tuesday, April 22, 2003

Paul, the answer is right. The analysis in the 'official' answer gives the case exactly.

Remember that you are not being told which coin is heads. If you were told 'the penny is heads' you would be right about the chance of the dime being 50%. But what you are really being told is 'either the penny is heads, or the dime is heads, or both'. Put like that it's fairly clear that there are three possibilities, in which only one is the 'other' coin heads.

You can try this yourself with two coins. Record all the results, eliminate any where no coin was heads, and work out what fraction of the remainder had two heads.

David Clayworth
Tuesday, April 22, 2003

50% is correct.
Using the notation of the "answer":
TT,TH,HT,HH, we know that H* is true, which implies that
the only remaining possibilities are HT and HH,
both equally likely. Hence, 50%.
The thing to realise is that the T outcomes of the
announced coin have probability zero.

Richard Rodger
Tuesday, April 29, 2003

Richard, you are NOT told which of the coins is heads. There are three results that satisfy the condition 'one of the coins is heads'; HH, HT and TH. All are equally likely. Only one has both coins heads.

David Clayworth
Tuesday, April 29, 2003

Say I throw the first coin today on monday
and the second on tuesday. You can swap dime and penny
as the cases are symmetrical.

On monday I get H.
What is the probability of H on Tuesday?

Conversely, on Monday I get T.
What is the probability of H on Tuesday?

Does the answer change if I reduce the time difference
to an hour?

Does the answer change if I reduce the time difference
to zero?

Richard Rodger
Wednesday, April 30, 2003

David is right - the answer does become 1/3 if you
throw the coins simultaneously (zero time difference),
because the announcer gets to choose any coin.

Richard Rodger
Wednesday, April 30, 2003

To David Clayworth;

Actually, you are told which one is heads. The 'first' one.
When writing out the codes of HH, HT, TH and TT, you are assuming that the first letter is for the dime, and the second letter is for the penny (or vise versa). But the type of the coin is irrelevant. Try categorizing them differently; The first letter is for the coin that is declared (heads), and the second coin is the one that has yet to be determined. You are now stuck with only two options - HH, or HT, as the first coin has been claimed Heads already. A fifty percent chance.

Due theory does not exist. The state of the first coin has already been declared, and isin the past. The second coin does not 'know' that the first coin has flipped heads or tails. And, that second coin, standing alone, has a 50% chance of landing on heads. You may only use the probability of both coins if they are thrown /and/ declared together (as one random instance), in which case, your chances are 25% that they will both be heads.

Try upping the number of coins from 2 to 1000, and declare that 999 of those coins have landed heads. It still will not impact the probability of that 1000th coin being heads or tails, after all 999 coins have been declared heads. Now, the chances of getting 999 heads in a row is really, really small. But since you've already declared it, it's already happened, you've bested that probability. You're really lucky. However, that last final coin has only a 50% chance on it's own, and cares nothing for the other 999 heads.

Lemuel Pew
Wednesday, May 07, 2003

One thing that is confusing about this puzzle is the phrasing. This puzzle isn't asking for the probability of double heads on two uncalled coins. It's only asking for the outcome of one coin - the one that hasn't been shown. The fact that the two coins are different types doesn't matter one bit, because the coin that's already made heads isn't a part of the question. It's a precursor to the actual question, and bears little relevance.

In fact, It's kind of a trick question.

Lemuel Pew
Wednesday, May 07, 2003

Ho Hum

Agreed that when the announcer is identifying a particular coin as the one that came up heads (the dime, the first thrown, the leftmost, the heaviest or whatever) then the probability is 50%. If he means 'one or the other is heads' then it is 33.3%

David Clayworth
Friday, May 09, 2003

i will just reiterate whatever has been said above but in a different manner so as to speak,

two coins are flipped,
Sample Set : HH,HT,TH,TT

one of the coins gives a head,
Reduced Sample Set : HH,HT,TH

case when the other coin is also head,
Favourable set : HH

Probability = 1/3

I think the confusion generally arises in the development of Reduced Sample Set.

Arun Iyer
Sunday, June 01, 2003

Lets assume that probability when event always happens is 1. Never happens - 0.

Then probability for the first coin to come heads:      0.5
And probability for the second coin to come heads:  0.5

To calculate the probability for two non-related events A and B to come together you should multiply their standalone probabilities, so the probability that both coins will come heads is 0.25.

But the puzzle says, that it's announces whenever one of the coins is already heads, so this event is already happened. Because those two events a non-related, probability for the second coin to come heads stays the same: 0.5.

0.3 is not correct answer or questions is not put correctly, because 0.3 is an overall probability for 2 coins to come heads simultaneously when a probability of them both coming tails during trials is discarded.

Announcement "one of the coins came up heads" is part of  the puzzle, and given as a fact and the questions is "What is the chance that the other coin also came up heads?". And it's still 50% or 0.5, because first event has already happened and it's fact and we can't anymore talk about probability of those events happening together, we can only talk about the second coin.

So the overall probability for them to come heads at the same time is 25%, the probability for one of them to come heads, when another is already heads is 50%.

The answer "1/3" can never be correct, because author discards the probability of them both to be tails, but still tries to calculate overall probability of HH outcome.

Although overall probability of both coins to come heads would be possible, for example if:
first coin comes heads: 0.5
second coin comes heads: 2/3, i.e. in 66.(6) cases out of 100.

Anyway, I believe, that the correct answer to the puzzle is 0.5 or 50%, if you tell exactly that it asks and 1/3 is wrong.

Friday, July 18, 2003

The solution is 50%. If one coin is H(eads) than there are 3 possibilities left:


If one coin is already H than TH = HT, therefore the chance is 50%.

Another way of explaining: if you flip one coin, see that it's H, and then flip another coin, what is the chance of it being H? 50%

It's tricky because the puzzle is telling you you already know what the first coin is. That's were the confusion starts.

John Willemse
Thursday, March 24, 2005

*  Recent Topics

*  Fog Creek Home