Find the Fake Coin

Let's call the coin I've flipped A , and the other coin B . Based on the results of my flipping so far, the probability that A is the unfair coin is given by Bayes' Theorem (I've "pre-cancelled" all the prior probabilities that A or B is unfair, since they're both $\frac{1}{2}$ ):

$\frac{\frac{3}{4}\cdot\frac{1}{4}}{\frac{3}{4}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{2}}=\frac{3}{7}$

There are four possible "endgame" scenarios:

1. I flip A again and get H . In this case, the probability that A is unfair is

$\frac{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}}{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}}=\frac{9}{17},$

so I would choose A as being the unfair one, with probability $\frac{9}{17}$ .

2. I flip A again and get T . In this case, the probability that A is unfair is

$\frac{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}}{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}}=\frac{3}{11}$

so I would choose B as being the unfair one, with probability $\frac{8}{11}$ .

3. I flip B and get H . In this case, the probability that A is unfair is

$\frac{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{2}}{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{4}}=\frac{1}{3}$

so I would choose B as being the unfair one, with probability $\frac{2}{3}$ .

4. I flip B and get T . In this case, the probability that A is unfair is

$\frac{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{2}}{\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{4}}=\frac{3}{5}$

so I would choose A as being the unfair one, with probability $\frac{3}{5}$ .

Since there is a $\frac{3}{7}$ chance that A is unfair, the probability that it will come up H if I flip it is $\frac{3}{7}\cdot\frac{3}{4}+\frac{4}{7}\cdot\frac{1}{2}=\frac{17}{28}$ . So the probability that I would correctly guess the unfair coin is

$\frac{17}{28}\cdot\frac{9}{17}+\frac{11}{28}\cdot\frac{8}{11}=\frac{17}{28}.$

If I were to flip B instead, the probability that I would get H is $\frac{3}{7}\cdot\frac{1}{2}+\frac{4}{7}\cdot\frac{3}{4}=\frac{9}{14}$ . So the probability that I would correctly guess the unfair coin is

$\frac{9}{14}\cdot\frac{2}{3}+\frac{5}{14}\cdot\frac{3}{5}=\frac{9}{14}.$

Since $\frac{9}{14}>\frac{17}{28}$ , I should flip B for my final flip, and I will have a $\frac{9}{14}$ chance of correctly identifying the unfair coin, so the answer is $9+14=\boxed{23}$ .

We first calculate the probability that the coin flipped in the beginning is the real coin or not.

The probability that the real coin gets the sequence HT is $\dfrac{1}{2}\cdot \dfrac{1}{2}=\dfrac{1}{4}=\dfrac{4}{16}$ .

The probability that the fake coin gets the sequence HT is $\dfrac{3}{4}\cdot\dfrac{1}{4}=\dfrac{3}{16}$ .

Therefore the probability that the coin flipped at the beginning was the real coin is $\dfrac{4}{7}$ , and the probability that it as the fake coin is $\dfrac{3}{7}$ .

We are not faced with a decision. Should we flip the coin we already flipped, or should we switch to the other coin. Well, let's do casework.

Case 1: Choose same coin. We are again faced with a decision. If it comes up heads, should we pick this coin as the unfair coin or the other coin as the unfair coin? Also, if it comes up tails, which coin should we pick?

To help us in this decision, we can use constructive counting. Let $F$ be the case where the first coin flipped is the fair coin, and $U$ the unfair coin. $\begin{array}{l|cr} \text{F: }\dfrac{4}{7}\times & \dfrac{1}{2}(0\text{ or } 1)+ & \dfrac{1}{2}(1\text{ or }0)\\ \text{U: }\dfrac{3}{7}\times & \dfrac{3}{4}(0\text{ or }1)+& \dfrac{1}{4}(1\text{ or }0)\\\end{array}$ where the $0\text{ or }1$ correspond to whether you choose to pick the coin you flipped as the unfair coin or the coin you didn't flip as the unfair coin.

Obviously, maximal probability is achieved in this scenario: $\begin{array}{l|cr} \text{F: }\dfrac{4}{7}\times & \dfrac{1}{2}(0)+ & \dfrac{1}{2}(1)\\ \text{U: }\dfrac{3}{7}\times & \dfrac{3}{4}(1)+& \dfrac{1}{4}(0)\\\end{array}$ which corresponds to choosing the same coin if it comes up heads and the other coin if it comes up tails. This probability is evaluated and we get $\dfrac{17}{28}$ . But is this maximal?

Case 2: Choose other coin. This scenario is similar with the first, except that the probability that the coin you flip is the fair one is now $\dfrac{3}{7}$ and the coin being the unfair one is now $\dfrac{4}{7}$ . $\begin{array}{l|cr} \text{F: }\dfrac{3}{7}\times & \dfrac{1}{2}(0\text{ or } 1)+ & \dfrac{1}{2}(1\text{ or }0)\\ \text{U: }\dfrac{4}{7}\times & \dfrac{3}{4}(0\text{ or }1)+& \dfrac{1}{4}(1\text{ or }0)\\\end{array}$

Clearly, maximum is achieved the same way as last case: $\begin{array}{l|cr} \text{F: }\dfrac{3}{7}\times & \dfrac{1}{2}(0)+ & \dfrac{1}{2}(1)\\ \text{U: }\dfrac{4}{7}\times & \dfrac{3}{4}(1)+& \dfrac{1}{4}(0)\\\end{array}$ which corresponds to picking the same coin if it comes up heads and the other coin if it comes up tails. This probability is evaluated and we get $\dfrac{18}{28}=\dfrac{9}{14}$ . Since this probability is higher than the last case, and there are no more cases, this must be the maximal, so our final answer is $9+14=\boxed{23}$ and we are done. $\Box$

There is a $\frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}$ chance that the fair coin will flip HT, and a $\frac{3}{4}\cdot \frac{1}{4} = \frac{3}{16}$ chance that the unfair coin will flip HT.

If the real coin is chosen, switching gives a $\frac{1}{4}\cdot\frac{3}{4} = \frac{3}{16}$ chance of HTH, and a $\frac{1}{4}\cdot\frac{1}{4} = \frac{1}{16}$ chance of HTT. Staying gives equal chances between HTH and HTT.

If the fake coin is chosen, staying gives a $\frac{3}{16}\cdot\frac{3}{4} = \frac{9}{64}$ chance of HTH, and a $\frac{3}{16}\cdot\frac{1}{4} = \frac{3}{64}$ chance of HTT. Switching gives equal chances between HTH and HTT.

Since $\frac{3}{16} : \frac{1}{16} :: \frac{9}{64} : \frac{3}{64}$ , we choose to switch. (Note that we switch because it is more likely that HT came from the real coin than the fake coin.)

Therefore, our chance of identifying the unfair coin is $\frac{\frac{1}{4}\cdot\frac{3}{4}+\frac{3}{16}\cdot\frac{1}{2}}{\frac{1}{4}+\frac{3}{16}} = \boxed{\frac{9}{14}}$ .

The probability the fair coin gives flips HT is $\frac{1}{4}$ , and the probability the unfair coin gives flips HT is $\frac{3}{16}$ . Thus, the coin we flipped is most likely the fair coin. For our second flip, we don't want to flip the fair coin, as the result gives us no information about the coin. Thus, we want to flip the other coin. If it is heads, we guess it is the unfair coin, and if it is tails, we guess the first coin is the unfair coin. Then, the probability is $\dfrac{4}{7}\cdot\dfrac{3}{4}+\dfrac{3}{7}\cdot\dfrac{1}{2}=\dfrac{9}{14}$ giving the answer of $9+14=\boxed{23}$ .

I cannot rigorously prove something else doesn't work, because there are so many possibilities. Intuition tells me to make the most out of all the information, giving my strategy, but I cannot show everything like "choose one coin at random and only use the result of that flip" is worse.

Optimum strategy: Flip the other coin. If heads, choose the other coin. Otherwise, choose the first coin.

Let XH, XT, FH, FT be unfair heads, unfair tails, fair heads, fair tails

Of all the possible histories, (1/2)(3/4)(1/4)(1/2) = 3/64 of them are { XH, XT, FH } (1/2)(3/4)(1/4)(1/2) = 3/64 of them are { XH, XT, FT } (1/2)(1/2)(1/2)(3/4) = 6/64 of them are { FH, FT, XH } (1/2)(1/2)(1/2)(1/4) = 2/64 of them are { FH, FT, XT }

Combined, these make up 14/64 of all possible outcomes. Out of those, 9/64 will be correctly guessed by optimum strategy. Therefore, probability is 9/14 that the unfair coin will be correctly identified.

Therefore, 9+14= 23