# Probability of two heads given the probability of a head on a Saturday

The probability that a fair-coin lands on either Heads or Tails on a certain day of the week is $1/14$.

Example: (H, Monday), (H, Tuesday) $...$ (T, Monday), (T, Tuesday) $...$
Thus, $(1/2 \cdot 1/7) = 1/14$. There are $14$ such outcomes.

In some arbitrary week, Tom flips two fair-coins. You don't know if they were flipped on the same day, or on different days. After this arbitrary week, Tom tells you that at least one of the flips was a Heads which he flipped on Saturday.

Determine the probability that Tom flipped two heads in that week.

I know that this is a conditional probability problem.

The probability of getting two heads is $(1/2)^2 = 1/4$. Call this event $P$.

I am trying to figure out the probability of Tom flipping at least one head on a Saturday. To get this probability, I know that we must compute the probability of there being no (H, Saturday) which is $1 - 1/14 = 13/14$.

But then to get this "at least", we need to do $1 - 13/14$ which gives us $1/14$ again. Call this event $Q$.

So is the probability of event $Q = 1/14$? It doesn't sound right to me.

Afterwards we must do $Pr(P | Q) = \frac{P(P \cap Q)}{Pr(Q)}$. Now I'm not quite sure what $P \cap Q$ means in this context.

Intuitively, I would think the result would be greater than $\frac{1}{2}$ because of that slight chance we get $2$ heads on Saturday.

Let $P$ denote the event that we flip $2$ heads that week.

Let $Q$ denote the event that we flip at least one head on Saturday.

I find it easier to flip $P(P\mid Q)$ into $P(Q\mid P)$

We have

\begin{align*} P(P\mid Q) &=\frac{P(P\cap Q)}{P(Q)}\\\\ &=\frac{P(Q\mid P)\cdot P(P)}{P(Q)}\\\\ &=\frac{\left({2 \choose 2}\left(\frac{1}{7}\right)^2+{2 \choose 1}\left(\frac{1}{7}\right)\left(\frac{6}{7}\right)\right)\left(\frac{1}{2}\right)^2}{{2 \choose 2}\frac{1}{14}^2+{2 \choose 1}\left(\frac{1}{14}\right)\left(\frac{13}{14}\right)}\\\\ &=\frac{13}{27} \end{align*}

where $P(Q\mid P)$ can be thought of as we're given that we got two heads but what are the chances that at least one was from Saturday with probability $\frac{1}{7}$ for an individual coin.

Note: My answer contradicts my intuition! This serves as further proof that intuition can lead you astray in probability. To see why my intuition was incorrect, see @jgon's answer.

Remy has already given the correct answer, but is not confident because of a missing intuition, and I already more or less answered the question in comments on NewGuy's answer, so I'll just write it up and try to give an intuition for it.

The sample space for a single coin flip is $\Omega=\newcommand{\set}{\left\{#1\right\}}\set{H,T}\times \set{M,Tu,W,Th,F,Sa,S}$, and it has the uniform distribution, with each pair equally likely. We can think of this as flipping a fair coin and rolling a fair 7 sided die labeled with the days of the week together (a d7).

The sample space then for two coin flips is $\Omega \times \Omega$, which again is the same as flipping 2 fair coins and rolling 2 d7s.

If $M$ is the event that both coins are heads, and $N$ is the event that at least one of the coins was flipped on Saturday and was heads. Then $M\cap N$ is the event that both coins were heads and at least one was flipped on Saturday. Now we're interested in $$P(M|N) = \frac{P(M\cap N)}{P(N)}=\frac{|M\cap N|}{|N|},$$
so we just need to compute the sizes of $N$ and $M\cap N$. Let's start with $M\cap N$. Since we know both coins came up heads, we just need to work with the days of the week. The number of ways that at least one of the days of the week can be Saturday is $1+6+6=13$, corresponding to the possibilities $(Sa,Sa)$ or $(Sa,\text{not }Sa)$ or $(\text{not }Sa,Sa)$.

Now we can do a similar thing for $N$. We get $|{N}|=1+13+13=27$ corresponding to the possibilities $(HSa,HSa)$ or $(HSa,\text{not }HSa)$ or $(\text{not }HSa,HSa)$.

Intuition: Why does knowing that one of the coins was a head flipped on a Saturday reduce the probability that the other coin was also a head (13/27) compared to say having a bronze and a silver coin and knowing that the bronze coin was a head flipped on a Saturday (probability that the other coin was also a head 1/2)?

The issue is essentially, for each state in $M\cap N$, $HH(day_1)(day_2)$ if only one of those days is Saturday, say $day_1=Sa$ we get two states in $N$: $HHSa(day_2)$ and $HTSa(day_2)$, but if both days are $Sa$, we get three states in $N$: $HHSaSa$, $HTSaSa$ and $THSaSa$. I.e. in the case when both days are Saturday, we get an extra way to fail to be both heads. Or viewed the other way, the fact that the Saturday flips are interchangeable when they both come up heads means that while $HTSaSa$ and $THSaSa$ are different, they only have one success case associated to them namely: $HHSaSa$.

• Ah okay, that makes sense now! – Remy Mar 20 '18 at 0:25
• Why did you choose $N$ to be the event that exactly one flip was $HSa$ and not at least one $HSa$ like @Remy did it? If you did do that, what would $M \cap N$ mean? – udpcon Mar 20 '18 at 3:55
• Sorry that was a typo, gimme a moment – jgon Mar 20 '18 at 3:56
• $N$ should be the event that at least one was $HSa$ because that is the information we're given. – jgon Mar 20 '18 at 4:00
• Okay, I believe it clicked now. Thank you for taking the time to write this! – udpcon Mar 20 '18 at 4:17

CORRECTED AS REASONS GIVEN BY BY JGON

Sample Space for single throw = {MH,MT,TuH,TuT,...........,SH,ST} = 14

Sample Space for two throws = $14*14 = 196$

M: 2 heads are thrown = $7*7= 49$

N: atleast single head is thrown on saturday = {(SaH,?)(?,SaH)} = $2*14$

But we have counted twice {(SaH,SaH)} therefore one has to be subtracted

= $2*14-1=27$

To Find P(M|N) = $\frac{P(M\cap N)}{P(N)}$ =$\frac{n(M\cap N)}{n(N)}$

$M\cap N$ = One head throw on saturday and other head can be on anyday ={ (SaH,?H)(?H,SaH)}= $2*7$

But double counting also take place here =$2*7-1=13$

P(M|N) = $\frac{13}{27}$

• You're not quite right. There are $2\cdot 14 - 1 =27$ ways to get $N$, the first coin could be $SaH$ or the second could could be $SaH$ – jgon Mar 19 '18 at 23:53
• Similarly $M\cap N$ is also not quite right. It should be $2\cdot 7 -1 = 13$. – jgon Mar 19 '18 at 23:54