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Painfully easy solution is wrong

i flip a penny and a dime and hide the result from you. "one of the coins came up heads", i announce. what is the chance that the other coin also came up heads?

When listing the possible outcomes, you should assign a coin to a column. So lets say the first column is the dime, the second the penny.


we know one coin is heads.  50% chance that coin is the dime.  which means there are two possible outcomes, HH or HT.  50% chance of HH, therefore 25% chance of occuring.

likewise, theres 50% chance that the coin that is the penny.  once again, two possible outcomes, HH or TH.  50% chance of HH, therefore 25% chance of occuring.

so overall, the chance of HH occuring is 25+25 = 50%

you can also reason that the chances of the unknown coin being heads is 50% no matter what other conditions that exist.  I can flip 100 coins all landing on heads.  what is chances of the next coin landing on heads, still is 50%.

or maybe I'm wrong

Friday, February 27, 2004

"or maybe I'm wrong"

:)  maybe...

Michael Pryor
Fog Creek Software
Saturday, February 28, 2004

the information about one coin being a penny and another a dime isnt really applicable.

P - the event that one coin turns up head. This can happen in 3 ways- HH, HT and TH. But only
one out of these three - HH - satisfies the condition that the other coin also turned up heads. So the chance of the other coin turning up heads given that one turned up heads is 1/3

Monday, March 1, 2004

Actually, assuming that the flipper/announcer is speaking normal English, the chance that the other coin is also heads is zero, since the normal pragmatic meaning of "One of the coins came up heads" when the speaker knows the state of both coins is "Exactly one of the coins came up heads," not "At least one of the coins came up heads."

Analogy --  if there were two apples in a bag, and your friend looked in, examined them both, and said "One of the apples is rotten", and you replied "Well give me the other one then," if he then gave you a rotten apple and said "The other one is rotten too!" you would be annoyed with him for pulling your chain.

Of course, at technical interviews the language spoken is usually not "Normal English" but rather "Math Problem English," in which case the initial answer as posted is indeed correct as explained elsewhere -- i.e. P(Both coins are heads|At least one is heads) = 1/3.

Jonathan Segal
Tuesday, March 2, 2004

I agree with J.
The second coin is independent of the first. It doesn't matter what the first coin is, the chances of the other being heads is still 50/50

Thursday, March 11, 2004

The coin has already been flipped and they guy is stating a fact about its current state.

See the monty hall problem.

Michael Pryor
Fog Creek Software
Friday, March 12, 2004

Basically the problem corresponds to the following experiment.

Someone has two coins, you can assume coin #1 is a penny and coin #2 is a dime if you want, but it does not really matter.

The person flips coin #1 1000 times, 499 of the time the coin came up heads, 501 of the time tails.

The person then flips coin #2 1000 times, 500 of the time the coin came up heads, the other 500 tails.

The person then flips both coins simultaneously, and keeps track of two counters, called cTotal and cHH.

Initially, both counters are zero.

If both coins came up tails, then
      Both counters stay the same
      The person flips again
      Increase cTotal by one
      If both coins came up heads, then increase cHH by one
      The person flips again

Do this until cTotal reaches 1000.

The ratio (cHH / cTotal) is the experimental approximation of “the chance that the other coin also came up heads.”  For sufficiently large cTotal, the experimental result should approach the theoretical result.  And the theoretical result is 1 / 3.

This experiment is best simulated by a computer program with a random number generator.  However, a real world experiment should work, too, just a bit time-consuming and tedious, but I am sure it is worth the effort if it provides more insight, knowledge, and confirmation to the solution.

Monday, March 15, 2004

So the confusion about this problem comes because it's not immidiately obvious how much information the statement "One of the coins came up heads" reveals.

If you were told "the dime came up heads" or "the penny came up heads" that would be VERY different from "the dime or the penny came up heads".

Since it's "the dime or the penny came up heads" then you know its one of HT HH TH.


P.S. As a side question, Joel, why do you have a picture of a Yale library on the sidebar? 

Ya'ir Aizenman
Wednesday, April 28, 2004

I agree that the real answer was given by Jonathan. However I disagree with the 1/3 answer. This answer would only apply if before flipping the coins the flipper asked the chances both would be heads. But because they have already been flipped and the results of each coin are independent you can only judge each ones results independently. Had the flipper said at least one coin is heads. This would rule out one of the coins probabilities of being tails. This only leaves two possibilities for the other coin. In other words you are using the result HT and TH when in fact these are actually the same result. You can only factor a guaranteed HH or HT and that is it. The result leaves you with a 50% probability of the second coin being heads.

Shane Hayen
Thursday, May 13, 2004

I retract my above reasoning. I didn't factor in the fact that they just say a coin is heads. This does add one more equation into factor. I still do think the real answer is zero though.

Shane Hayen
Thursday, May 13, 2004

I believe the answer is 1/3, although it seems counter-intuitive at first.

Once the event has occurred, the numerical distribution of the alternatives do not change.  But, when information given occurrences, the relative distribution (a.k.a probability) DOES change.

100 flips -- HH 25 - HT 25 - TH 25 - TT 25

When the coins are flipped, the 4 alternatives -- HH, HT, TH, TT -- have an even distribution, and consequently each has a 25% probability. 

But, information eliminates the TT alternative.  So, for the purposes of calculating the probability of a remaining alternative, the 25 TT flips must be eliminated from consideration.

HH 25 - HT 25 - TH 25  -- 75 total flips
--> each alternative has 33% (25 / 75)
The HH alternative is what we are looking for, and it has 33% probability.

The question "the dime is heads, what is the probability that the penny is also heads" would have a 50% answer, because 2 of the 4 alternatives would be eliminated.

The question "what is the probability of both coins being heads" yields a 25% answer.

Tom Jedrzejewicz
Tuesday, May 18, 2004

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