# Probability of X being a trick coin (heads every time) after heads is flipped k amount of times

A magician has 24 fair coins, and 1 trick coin that flips heads every time.

Someone robs the magician of one of his coins, and flips it $$k$$ times to check if it's the trick coin.

A) What is the probability that the coin the robber has is the trick coin, given that it flips heads all $$k$$ times?

B) What is the smallest number of times they need to flip the coin to believe there is at least a 90% chance they have the trick coin, given that it flips heads on each of the flips?

Here is my approach:

Let $$T$$ be the event that the robber has the trick coin

Let $$H$$ be the event where the robber flips a heads k times in a row

$$Pr(T) = 1/25$$

$$Pr(H|T) = 1$$

$$Pr(T') = 24/25$$

$$Pr(H|T') = 1/2$$ when $$k=1$$, $$1/4$$ when $$k=2$$, $$1/8$$ when $$k=3$$... etc

$$Pr(T|H) = (1 * 1/2) / (1 * 1/2 + Pr(H|T') * 24/25) = 1/13, 1/7, 1/4,...$$ etc

So the Pr(T|H) answer changes for every k, do I answer with the formula? How can I answer A? How do I make a probability distribution when k can be infinite?

Also is B 8 flips? Since when k = 8, Pr(T|H) = 1/256.

Thanks for any help.

• @Arthur fixed, cheers – Jeremy May 14 at 8:35

It might be more handsome to let $$H_k$$ denote the event that the stolen coin will give $$k$$ heads by the first $$k$$ flips.

A) To be found is $$P(T\mid H_k)$$ where:$$P(T\mid H_k)P(H_k)=P(T\cap H_k)=P(H_k\mid T)P(T)\tag1$$

You already found values for $$P(H_k\mid T)$$ and $$P(T)$$ so $$(1)$$ allows you to find $$P(T\mid H_k)$$ if you can find $$P(H_k)$$. This can be done on base of:$$P(H_k)=P(H_k\mid T)P(T)+P(H_k\mid T^{\complement})P(T^{\complement})$$ Again $$P(H_k\mid T)$$ and $$P(T)$$ are well known and of course $$P(T^{\complement})=1-P(T)$$. Finally it is quite evident that $$P(H_k\mid T^{\complement})=2^{-k}$$.

B) So you end up with an expression in $$k$$ for $$P(T\mid H_k)$$ and to be found is the smallest $$k$$ that satisfies:$$P(T\mid H_k)\geq0.9$$

$$P(trick|H_k)=\frac {P(trick \cap H_k)} {P(H_k)}=\frac {P(H_k|trick).P(trick)} {P(H_k)}.$$ Now,

\begin{align}P(trick)&= \frac{1}{24}\\ P(H_k|trick)&=1\\ P(H_k)&=P(H_k|trick)\cdot P(trick)+P(H_k|fair)\cdot P(fair)\\ &=1\cdot\frac{1}{24}+\frac{1}{2^k}\cdot\frac{23}{24}\end{align}

Hope this helps