Smith is offered the following gamble: he is to choose a coin at random from a large collection of coins and toss it randomly.The proportion of the coins in the collection that are loaded towards a head is $p$. If a coin is loaded towards a head, then when the coin is tossed randomly, there is $\frac{3}{4}$ probability that a head will turn up and a $\frac{1}{4}$ probability that a tail will turn up. Similarly, if the coin is loaded towards tails, then there is a $\frac{3}{4}$ probability that a tail will turn up and a $\frac{1}{4}$ probability that a head will turn up. If Smith tosses a head, he loses \$$100$, if he tosses a tail, he wins \$$200$. Find the probability value $p$ for which Smith's gain is $0$ when taking the gamble.
[Math] Probability and Stats (loaded coin)
probability
Related Solutions
The friend says yes if he got one head or two.
When a coin is tossed twice, the following $4$ outcomes are equally likely: HH, HT, TH, TT. For $3$ of these outcomes, the friend says yes. In $2$ of these outcomes, there is a tail. So the probability that there is a tail given that there is at least one head is $\dfrac{2}{3}$.
This sort of semi-formal argument can be treacherous. So let's do it more formally. Let $B$ be the event there is at least $1$ head, and let $A$ be the event there is a tail. We want $\Pr(A|B)$ (the probability of $A$ given $B$).
By a standard formula, $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
The probability of $B$ is $\dfrac{3}{4}$.
The probability of $A\cap B$ is $\dfrac{2}{4}$. Divide.
Added: A second part has been added. This one requires interpretation. Answers to similar problems can be quite interpretation-dependent.
Suppose that with probability $\frac{1}{2}$ we get to have a peek at the result of toss $1$, and with probability $\frac{1}{2}$ we get to see the result of toss $2$. Let $S$ be the event we see a head, and let $T$ be the event there is a tail.
We want $\Pr(T|S)$. The computation is much like the one in the answer to the first question. After examination of cases, we find that $\Pr(S)=\frac{1}{2}$ and $\Pr(S\cap T)=\frac{1}{4}$. From that we conclude that $\Pr(T|S)=\frac{1}{2}$.
It is worth going through the effort of calculating the probability via the definition of conditional probability in early examples.
$$Pr(A\mid B):=\frac{Pr(A\cap B)}{Pr(B)}$$
Let $B$ be the event that the first coin flipped is not a head (i.e. the first coin flipped turned up tails).
Let $A$ be the event that the coin is flipped exactly three times.
We are tasked with calculating $Pr(A\mid B)$, the probability that the coin is flipped exactly three times given that the first flip did not turn up heads.
We can draw ourselves a tree diagram or however else we like to arrive at the following table of outcomes and respective probabilities:
$$\begin{array}{|c|c|}\hline\text{Outcome}&\text{Probability}\\\hline H&\frac{1}{2}\\\hline TH&\frac{1}{4}\\\hline TTH&\frac{1}{8}\\\hline TTT&\frac{1}{8}\\\hline\end{array}$$
It is worth taking a moment to check that this does in fact make sense as a probability distribution by verifying that the probabilities add up to exactly one. Indeed $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}$ does equal $1$.
The event that the first flip is not heads corresponds to all of the above listed outcomes except the first and so occurs with probability $\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$ so we learn that $Pr(B)=\frac{1}{2}$.
The event that the first flip is not heads and it takes three flips in total corresponds to the last two outcomes in the above table and so occurs with probability $\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$ so we learn that $Pr(A\cap B)=\frac{1}{4}$.
Putting this information together, we get:
$$Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{1/4}{1/2}=\frac{1}{2}$$
Best Answer
I'm assuming the coins that are not loaded towards heads are loaded towards tails. Write out the expected gain in terms of p, set it to 0, and solve.
Compute the expected gain in terms of P(heads) and P(tails), the overall probability of flipping heads, or tails, respectively.
We can then compute P(heads) and P(tails) in terms of p, the percentage of coins biased towards heads.
P(tails):
Now just solve for p: