# Probability of all coins tails-up.

A magician has $$5$$ coins. He initially places $$3$$ of the coins with heads-up and the rest with tails-up. Then he performs a process in which he flips one coin every second. The process stops when all the coins are tails-up. What is the probability that the process ends in $$3$$ seconds.

I found two methods two solve this but both are giving different answers.

Method 1: The process will end in exactly $$3$$ seconds when at each step the the coin with heads-up is flipped.

The probability of choosing one heads-up coin in first step is $$\frac{3}{5}$$.

Now we have flipped one heads-up coin. So, probability of choosing one heads-up coin in second step is $$\frac {2}{5}$$.

Similarly in third step it is $$\frac{1}{5}$$.

So, the probability that the process ends in three seconds is $$(\frac{1}{5} \cdot \frac{2}{5} \cdot \frac{3}{5})=\frac{6}{125}$$

Method 2: The process will end in $$3$$ seconds if we flip heads-up coin in all the steps.

We wi ll find all the possible sequences of steps:

$$HHH$$

$$HHT$$

$$HTH$$

$$THH$$

$$HTT$$

$$TTH$$

$$THT$$

Where $$H$$ or $$T$$ at the $$i_{th}$$ position represents the heads-up or tails-up coin respectively flipped at $$i_{th}$$ second.

So, the required probability is $$\frac{1}{7}$$

Why am I getting different answers? Which one is wrong?

Method $$1$$ is correct. Method $$2$$ is completely wrong: the probability of picking a heads-up coin to flip changes with every step.
• @AmitShah There's also the fact that not all the sequences are equally likely. You can't just list all the sequences and say "oh, the probability of $HHH$ occurring is $\frac17$". It's like saying that the Large Hadron Collider will destroy the world with a 50% chance because either it will happen or it will not (and this claim was actually made to the press I think in 2008). – Parcly Taxel Apr 4 '20 at 9:14
• @AmitShah Yes, a correct solution based on method $2$ would be far longer and excessive than one using method $1$. – Parcly Taxel Apr 4 '20 at 9:16