# Probability problem

I have $3$ coins, $1$ coin has $2$ heads (HH), 1 coin has $2$ tails (TT), $1$ coin has $1$ head and $1$ tail (HT). I toss the coin, it fells on my hand, and the side i see is a tail. What's the chance that the other side is also a tail?

I got this as a teaser from a friend, possible from here, as you can see he is insisting on 1/2 as not being the correct answer, I got 1/3 as my answer, am I right?

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I imagine you meant $2/3$. – André Nicolas Apr 16 '12 at 22:44
Hint: how many sides have a tail? What proportion of those have a tail on the opposite side? – Henry Apr 16 '12 at 23:16
According to Sheldon Ross "[this] example has often been used by unscrupulous probability students to win money from their less enlightened friends". – David Mitra Apr 16 '12 at 23:18
The even more unscrupulous players palm a fair coin and replace one of the "doubles" for a while, and then swap the double in again so that the mark does not wonder why the T-T coin (say) has not shown up for a while. Suppose a man with three walnut shells and a pea rolling between them hides the pea and asks you to pick a shell. When you choose one, he lifts another shell and shows you there is nothing underneath that. He then offers you a chance to switch. Should you accept? Should you accept Monty Hall's offer? What is the difference between the two offers? – Dilip Sarwate Apr 16 '12 at 23:26
Here it is again with coins – MJD Aug 28 '12 at 16:45

I will assume you initially choose one of the three coins at random.

There are 6 equally likely possibilities for the side you might see: the "head side" of the HT coin, the tail side of the HT coin, one "head side" of the HH coin, the "other head side" of the HH coin, one tail side of the TT coin, and the other tail side of the TT coin.

Since you observed a tail, three of the six, equally likely, possibilities listed above are ruled out, and you know you either saw one side of the TT coin or you saw the tail side of the TH coin.

The probability that the other side of the coin is also a tail is 2/3.

Here's a different way to do it:

Let $A$ be the event the observed side is a tail, $X_{HH}$ be the event that you picked the HH coin, $X_{TT}$ be the event that you picked TT coin, and $X_{TH}$ be the event that you picked the TH coin.

The desired probability is $P(X_{TT}|A)$. We have: \eqalign{ P(X_{TT}|A)&={P(A|X_{TT})P(X_{TT})\over P(A)}\cr &={P(A|X_{TT})P(X_{TT})\over P(A|X_{HH})P(X_{HH}) + P(A|X_{HT})P(X_{HT}) +P(A|X_{TT})P(X_{TT})}\cr &={ 1\cdot(1/3) \over 0\cdot(1/3)+(1/2)(1/3)+1\cdot(1/3) }\cr &={1 \over (1/2)+1}\cr &=2/3. }

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+1,Thanks David. I like the way you metamorphosed the problem into a Baye's theorem application. – Quixotic Apr 17 '12 at 4:38

Seeing it from another angle,let T1 be the event of the first side to be a Tail and T2 the event of the second side to be a Tail. From the conditional probability we have: $$P(T2|T1)=\frac{P(T2 \cap T1)}{P(T1) }=\frac{\frac{1}{3}}{\frac{3}{6}}=\frac{2}{3}$$

That's because the probability of both sides to be Tails is 1/3 and the probability of the first side to be a tail is 3/6. I believe neither of you were right.

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Another solution to the problem is as follows. Two of the three coins are "doubles" meaning that they have the same face on both sides. So if you always bet that the other face is the same as the one you are seeing, then you will be correct two times out of three.

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strictly, this solution depends on the symmetry between Heads and Tails. For example, it would not be correct in the case where we had coins: HH, HH, HH, TT, HT, HT. – Ronald Apr 17 '12 at 0:47
@Ronald Yes of course. It would also not apply to the situation described in my comment on the main question where a cheater swaps coins in secret, changing HH, HT, TT to HH, HT, HT. – Dilip Sarwate Apr 17 '12 at 0:52

Here's Perl code that you can execute that performs the experiment and reports the results, which do indeed match the predicted result of $\frac23$:

        my @cards = qw(HH HT TT);

my $n_trials = shift || 100; for (1 ..$n_trials) {
my $coin =$coins[ int(rand 3) ];
my @faces = split //, $coin; my$face_choice = int(rand 2);

my ($face,$other_face) = @faces[$face_choice, 1-$face_choice];

if ($face eq "H") { # this trial is spoiled; do it over redo; } else {$count{$other_face} += 1; } } print "In$n_trials trials:\n";
for my $other_face (keys %count) { print " The other face was '$other_face' in $count{$other_face} trials\n";
}


Here's a typical output:

In 100 trials:
The other face was 'H' in 37 trials
The other face was 'T' in 63 trials

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You can find a very intuitive and illustrative explanation here:
http://www.datagenetics.com/blog/june72013/index.html

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Let me try and write the simplest solution.

$\displaystyle\text{Probability}= \frac{\text{Number of favourable events}}{\text{Number of total events}}$

Number of favourable events provided one side is tails$=2$ (there are $2$ coins with both sides tails).

Number of total events provided one side is tails=$3$ (there are $2$ coins with both sides tails and $1$ coin with one side heads and the other tails).

Hence, $\text{Probability}=\frac{2}{3}$.

This solution follows from the very definition of probability. The concept of "conditional probability" follows naturally.

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