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You play against your friend in a coin flipping game, where the objective is to get the most heads after three coin flips. A player wins if they have more heads than the opponent. If the numbers of heads are equal, then no one wins; it is a tie. You will take turns flipping coins, and your friend flips first. You want to cheat, so you created a fake coin that looks identical to a real, fair coin. The fake coin has a 80% chance of getting heads. Your friend is suspicious, so your friend gets to pick first from the two coins randomly, with equal probability. You are forced to take the other coin. You will both use your own coin for all three coin flips. The game begins, and your friend flips the coin once and gets heads.

(a)Given that your friend’s first coin flip returns heads, what is the probability that your friend got the fake coin, to your disadvantage?

(b)Given that your friend’s first coin flip returns heads, what is the probability that you will win the game?

I don't know what did I do wrong in part b as well as how to use part a's answer.

My Idea:
Me get real coin: Me get 2 head, friend gets 1 or Me get 3 head, friend gets 2;+
Me get fake coin: Me get 2 head, friend gets 1 or Me get 3 head, friend gets 2.

My try:
$0.5*C^3_2*0.5^2*0.5*C^2_2*0.2^2+C^3_3*0.5^3*C^2_1*0.8*0.2+$
$0.5*C^3_2*0.8^2*0.2*C^2_2*0.5^2+C^3_3*0.8^3*C^2_1*0.5^2*0.5=0.2235$

Answer:$\frac{2.88}{13}=0.22154...$

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  • $\begingroup$ For $a)$. There are two ways the friend could have gotten that $H$, either by getting the fake coin and throwing $H$ or by getting the real coin and throwing $H$, thus probability $\frac 12\times .8+\frac 12\times \frac 12$. Of that, $\frac 12\times .8$ is explained by getting the fake coin. $\endgroup$ – lulu Feb 11 at 13:07
  • $\begingroup$ Why are you only considering the winning results $2:1$ and $3:2$? There are lots of others? $\endgroup$ – joriki Feb 11 at 14:22
  • $\begingroup$ @joriki those are the only possibilities I can win. Friend already got one head in the first round, and it starts my turn $\endgroup$ – keanehui Feb 11 at 14:34
  • $\begingroup$ @keanehui: Ah, right, so not lots. But still $3:1$ is missing? $\endgroup$ – joriki Feb 11 at 14:35
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    $\begingroup$ Sorry, what I wrote about the sum being $2$ above was wrong; I hadn't realized that you included an extra factor $0.5$ that doesn't correspond to a coin flip. But this shouldn't be the prior probability $0.5$ but the posterior probability that you calculated in a). You're effectively ignoring the information that the result of your friend's first flip gave you about the coin assignment. $\endgroup$ – joriki Feb 11 at 15:39
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Let us define the event $A$ which will represent the friend choosing the weighted coin and $A^c$ where the friend chooses the fair coin. Note that $\mathbb{P}(A)=\frac{1}{2}=\mathbb{P}(A^c)$. Now, let's define $B$ to be the event that a person flipped a heads.

Part (a): We want to find $\mathbb{P}(A|B)$. In words, given that he friend flipped a heads, what is the probability that the friend chose the weighted coin?

Since $\mathbb{P}(B)>0$, we can use Bayes' Theorem which says $$\mathbb{P}(A|B)=\frac{\mathbb{P}(B|A)\mathbb{P}(A)}{\mathbb{P}(B)}$$ In words, $\mathbb{P}(B|A)$ is "given that the friend chose the weighted coin, what is the probability that he flipped a head?". We know that that is equal to $\frac{4}{5}$. We also know that $\mathbb{P}(A)=\frac{1}{2}$. Now, using the Law of Total Probability, we can calculate $\mathbb{P}(B)$ $$\mathbb{P}(B)=\mathbb{P}(B|A)\mathbb{P}(A)+\mathbb{P}(B|A^c)\mathbb{P}(A^c)=\frac{4}{5}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}=\frac{13}{20}$$ Where $\mathbb{P}(B|A^c)$ is "given that the friend chose the fair coin, what is the probability that he flipped a heads?" Therefore, all together we have $$\mathbb{P}(A|B)=\frac{\frac{1}{2}\cdot\frac{4}{5}}{\frac{13}{20}}=\frac{8}{13}$$

Part (b): There are three (which are really six) possible scenarios where we win the game and the friend loses. They are:

  1. We flip 3 heads and the friend flips 2.
  2. We flip 3 heads and the friend flips 1.
  3. We flip 2 heads and the friend flips 1.

And, each one of these scenarios is really two scenarios: one for the event where the friend has the weighted coin and one for when we do. We already calculated in part (a) what the probability is that our friend has the weighted coin $\left(\frac{8}{13}\right)$, and we also know what the probability is that we have the weighted coin ($1-\frac{8}{13}=\frac{5}{13}$) First, let's calculate the three scenarios in the event that we chose the weighted coin times the probability that we in fact chose the weighted coin: $$\mathbb{P}(A^c |B)\cdot[\mathbb{P}(3heads-2heads)+\mathbb{P}(3heads-1heads)+\mathbb{P}(2heads-1head)]$$ $$\frac{5}{13}\cdot\left[{3\choose 3}(4/5)^3 {2 \choose 1} (1/2)^2 + {3\choose 3}(4/5)^3 {2 \choose 2} (1/2)^2 + {3\choose 2}(4/5)^2 (1/5){2 \choose 2} (1/2)^2\right]=0.18462$$ Now, let's calculate the three scenarios in the event that the friend chose the weighted coin times the probability that he in fact chose the weighted coin: $$\mathbb{P}(A|B)\cdot[\mathbb{P}(3heads-2heads)+\mathbb{P}(3heads-1heads)+\mathbb{P}(2heads-1head)]$$ $$\frac{8}{13}\cdot\left[{3\choose 3}(1/2)^3 {2 \choose 1} (4/5)(1/5) + {3\choose 3}(1/2)^3 {2 \choose 2} (1/5)^2 + {3\choose 2}(1/2)^3 {2 \choose 2} (1/5)^2\right]=0.03692$$ All together we have $0.18462+0.03692=0.22154$

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