Bertrand's box paradox

Question

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 $\frac12$ as not being the correct answer, I got $\frac13$ as my answer, am I right?

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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

    from sys import argv
    from random import choice, randrange
    coins = [ ('H', 'H'), ('H', 'T'), ('T', 'T') ]
    count = { "H": 0, "T": 0 }

    if len(argv) < 2:
        n_trials = 100
    else:
        n_trials = int(argv[1])

    trial = 1
    while trial <= n_trials:
        faces = choice(coins)
        if randrange(2) == 1:       # flip the coin over
            faces = (faces[1], faces[0])
        if faces[0] == "H":
            # this trial is spoiled; do it over
            continue
        else:
            count[faces[1]] += 1
        trial += 1

    print("In {} trials:".format(n_trials))
    for other_face, c in count.items():
        print("The other face was '{}' in {}'\ trials.".format(other_face, c))

Here's a typical output:

    In 10000 trials:
    The other face was 'H' in 3361 trials.
    The other face was 'T' in 6639 trials.

Answer 5

A very basic solution will be as follows. Since we know that a tail has appeared, there can be three outcomes. We can be looking at the 1). Tail side of the HT coin 2). Face 1 of the TT coin 3). Face 2 of the TT coin So 2 of the 3 outcomes possible are in favour of the coin being a 2 tailed coin.