Probability of coin flips conditioned on an assistant lying about the results

Question

During a project researching coin flips, an assistant is asked to flip a fair coin twice. But the assistant is feeling lazy, and instead of following the directions, he does the following: He starts by flipping the coin once. Then, with probability 1/3 he copies the outcome of the first flip as the outcome of the second flip. With probability 2/3 he flips the coin a second a time (as instructed) and writes down that outcome.

(a) What is the probability that the assistant reports Heads for the second toss, given that the first toss was Heads? Explain and/or calculate.

(b) The assistant reports that both tosses were Heads. Given this, what is the probability that he is lying? (You may assume you got the probability from part (a) correct.)

So (a) for me was simple enough(i think) - simple law of total probability problem to calculate (1/3)1 + (2/3)(1/2) = 2/3

But (b) is a different beast. I'm trying to find the probability of P(L | two heads), and I tried using Bayes Law to equate this to

P(lying | two heads) = ( P(two heads | lying) * P(lying) ) / P(two heads)

I calculate the probability of P(two heads | lying) as 1/2, thinking that if he lied, he will give an answer of either two heads or two tails, and the probability of P(L) as 1/3, and the probability of P(HT) as 2/3 (as the only way you could get two heads is by conditioning on the first one being heads anyway... I'm very unsure about this one though.) Giving me a total answer of

(1/2 * 1/3) / (1/4) = 1/4

So, I'm pretty insecure about this answer as it intuitively doesn't make too much sense to me - can anybody point out where I might have made a mistake?

Answer 1

You've calculated everything correctly except for the unconditional probability of getting two heads. Intuitively, this cannot be greater than $1/2$, because the symmetry of the outcomes (heads vs. tails) implies that $\Pr[HH] = \Pr[TT]$, and since there are four possible outcomes, we now can see $\Pr[HH] < 1/2$.

To compute the desired probability, condition on the outcome of the first toss: $$\Pr[HH] = \Pr[HH \mid H]\Pr[H] + \Pr[HH \mid T]\Pr[T].$$ The second term is zero, of course. But what you reasoned was $$\Pr[HH] = \Pr[HH \mid H] = 2/3,$$ which you got from part (a); however, you can now see you have to multiply by the probability that the first toss was heads to begin with, which is $1/2$, giving you $$\Pr[HH] = 1/3.$$

Answer 2

That's a pretty lazy assistant who doesn't want to flip a whole coin!

You have made two errors, one of which is made irrelevant by the other. The first error is in your arithmetic:

$$ \frac{(1/2)(1/3)}{1/4} = \frac{1/6}{1/4} = \frac{2}{3} \not= \frac{1}{4} $$

But that doesn't matter because your denominator is incorrect; the probability of $HH$ is $1/3$: The probability of flipping heads first is $1/2$, and as you determined in part (a), the probability of recording heads the second time, given that heads was flipped first, is $2/3$, so the probability of recording heads both times is $(1/2)(2/3) = 1/3$. So the actual probability for part (b) is

$$ \frac{(1/2)(1/3)}{1/3} = \frac{1}{2} $$

ETA: In words, one might reason as follows: The sequence $HH$ can only happen if the first flip was heads, so assume that that is the case. Then with probability $1/3$, the assistant lied and recorded heads again; with probability $(2/3)(1/2) = 1/3$, the assistant flipped heads and recorded heads; and with the remaining $1/3$ probability, the assistant flipped tails and recorded tails. Two times out of three, the assistant records heads, but one of those two times was the result of a lie, hence the probability $1/2$.

Answer 3

If you find you're having trouble seeing the intuition, you can try stating out your probability space.

You can view your space as being: $$ \mathcal{C}:=\{(c_1, c_2)\,\lvert\,c_1\in\{H,T\}, c_2\in\{H,T,L\}\} $$ where for the second coin, $H$ means that the second coin was tossed and came up Heads, $T$ means that it was tossed and came up tails, and $L$ means the assistant lied and didn't flip the coin at all.

This presentation is convenient, because the two coordinates are independent of eachother: if we describe our (random) outcome as $(C_1, C_2)\in\mathcal{C}$, then $$ P(C_1=H)=P(C_1=T)=\frac{1}{2}, $$ while $$ P(C_2=H)=P(C_2=T)=\frac{2}{3}\cdot\frac{1}{2},\qquad P(C_2=L)=\frac{1}{3}. $$ In particular, every outcome in $\mathcal{C}$ is assigned probability $\frac{1}{6}$.

You needed to compute $P(\text{lying}\,\mid\,\text{2 heads})$. The event "2 heads" consists of $(H, H)$ and $(H, L)$; of these, only $(H, L)$ corresponds to lying. So, , $$ P(\text{lying}\,\mid\,\text{ 2 heads})=\frac{P(\text{lying, 2 heads})}{P(\text{2 heads})}=\frac{P\{(H,L)\}}{P\{(H,H)\text{ or }(H,L)\}}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}. $$