# Two fair coins, three unfair, one toss

some more questions about conditional probability:

You have five coins, of which two have heads on both sides, one has tails on both on both sides, and the other two are fair. You flip a randomly chosen coin. a) What is the probability that the lower side is heads? b-i) You open your eyes and see that the upper side is heads. What is the probability that the lower side is heads? b-ii) If you flip the coin again, what is the probability that the lower side is heads?

a) I'd say the phrasing here is a "trick", because in this particular case it doesn't matter if it's the upper or the lower side we're interested in. My answer is $\frac{2}{5} \cdot 1 + \frac{1}{5} \cdot 0 + \frac{2}{5} \frac{1}{2} = 0.6$.

b-i) This is asking for $P(\text{lower side is heads}| \text{upper side is heads})$. This equals $P(\text{lower side is heads AND upper side is heads})/P(\text{upper side is heads})$. The numerator is $2/5$ and the denominator $3/5$ by the argument in exercise (a), whence the answer is $2/3$.

So far so good?

b-ii) I'm a bit lost here because I'm really not sure how to interpret the question. I take it that the coin being tossed here is the one whose upper side previously came up as heads, and that we never actually inspected the lower side of that coin, whence we don't know what coin we're tossing now - well, we know it's the not the coin with tails on both sides, so is the answer $\frac{1}{2} \cdot 1 + \frac{1}{2} \frac{1}{2} = 3/4$?

Your hints and thoughts on all the exercises would be appreciated, because this all went way too easy :P

EDIT: This solution to (2b-ii) is due to user heropup (see comments): Let $A_1$ be the event that the two-heads coin was chosen, $A_2$ the fair coin, $A_3$ the two-tails coin, $L_i$ the event that the lower side is heads in the i:th coin flip, $U_i$ that the upper side is heads in the i:th coin flip. To determine $P(L_2|U_1)$, put $Q(E) =P(E|U_1)$. Then by the law of total probability, $Q(L_2) = \sum_{i=1}^3 Q(A_i)Q(L_2|A_i)$, with the third summand clearly zero. The event $U_1$ doesn't impact $Q(L_2|A_i)$ for either $i$, whence $Q(L_2|A_i)=1/i$ and $Q(L2)= Q(A_1)+(1/2)Q(A_2)$. Finally, apply Bayes' thm to $Q(A_i)=P(A_i|U_1)$ to yield $P(L_2|U_1)=5/6$.

I agree with your interpretation of part c. Assuming that is correct, let $Y_1$ be the upper side of the first flip and $Y_2$ be the lower side of the second flip. So $$P(Y_2=H|Y_1=H)=\frac{P(Y_2=H,Y_1=H)}{P(Y_1 = H)}=\frac{\frac15(1+1+0+1/4+1/4)}{6/10}=\frac{5/10}{6/10}=\frac56.$$

The answer to (b)(i) gives us a mechanism by which to answer (b)(ii). For an equivalent interpretation of the previous part is that the probability the coin we selected is two-headed is $2/3$, and the probability it is fair is $1/3$. So the probability the next coin flip has the lower side heads is $(2/3)(1) + (1/3)(1/2) = 5/6$.
• Thanks. But, why is the probability the coin we selected is two-headed $2/3$, rather than $2/4=1/2$? I'm thinking that since we know that the coin has at least one side that's heads we can exclude the possibility that the coin we're tossing this second time around is the two-tails coin, hence the probability should be $(2/4)(1)+(2/4)(1/2)=3/4$. If from the start there had been only one fair coin then I'd agree on your solution of $5/6$. (This is where I hope you just made a mistake ;) ) – Erik Vesterlund Sep 20 '15 at 14:43
• After part (b)(i), you have already observed the upper face to be heads. Do you agree that the probability that the lower (unseen) side is also heads is $2/3$? That is exactly what you stated as your answer. Do you also agree that the probability of the lower side is tails must then be $1/3$? Then that means that the probability the coin is two-headed is $2/3$ and the probability that the coin is fair is $1/3$. – heropup Sep 20 '15 at 16:41
• The reason why your reasoning is invalid is because you have incorrectly assumed that given you have observed the upper face to be heads, that it was equally likely that this head was observed from a two-headed coin as it was from a fair coin. But a two-headed coin has two heads, and although it was equally likely before you flipped the coin and saw the result to have selected a two-headed coin as it was to have selected a fair coin ($2/5$ versus $2/5$), this is no longer the case after you observed the result. – heropup Sep 20 '15 at 16:43
• I tried working through your solution, see the edit to OP. Now, as you imply, the outcome after the first coin flip impact the subseuent probabilities, but since I did not see that at first, I wanted a formula for $P(L_2)$ that incorporated that (and other possible influencing factors) so that one could do calculations without having to think about the problem (while it's paramount to learn to think a certain way in statistics, I think it's neat if the mathematics can encompass all relevant factors). I tried applying tot. prob. to $P(L_2)$ but it doesn't incorporate influences... – Erik Vesterlund Oct 6 '15 at 9:05