Time for a somewhat different post… yesterday I found an interesting interview question while browsing:

*“Let’s say you were given a biased coin, but you don’t know how it is** biased. How can you make a fair decision using the coin?”*

I assume the author meant a 50/50 decision when he/she said ** fair**. After thinking for a few minutes, I realized this could have been meant as an interesting pre/post-probability question. Let me explain. First of all, note that it doesn’t say how much the coin is biased. Therefore, you are allowed to make the assumption that it is completely biased – this always helps in thinking and calculation. Let’s say you ask someone to choose heads or tails before the coin is thrown. These are the possible outcomes:

- The person chooses heads, the coin is biased to heads
- The person chooses heads, the coin is biased to tails
- The person chooses tails, the coin is biased to heads
- The person chooses tails, the coin is biased to tails

Obviously, in two of these cases, the person would choose well, in two cases, he/she would lose. So, in fact, **it is a 50/50 chance**, no matter if the coin is biased, if you consider the person could choose what they wanted! I think, however, that the author meant a different kind of solution – one that makes the coin “fair”. The toss is considered “unfair” if you choose heads, but have no (or small) chance of getting it. This can also be done – **flip the coin two times**. If it ends up HT or TH, the result is the first one of the sequence. Note that both of these have the same probability. If it ends up HH or TT, continue flipping it again two times until you get HT or TH. This won’t work only in the case the coin is 100% biased.

There is an interesting way to solve this as a post-probability problem – you can **first throw the coin and then ask** the person – what do you think landed? If the coin is 100% biased, this is the same as asking “which way is the coin biased”. Obviously, you have a 50/50 chance of guessing, since there are only 2 options.

This reminds me of a more famous problem of this kind – the Monty Hall problem. The **Monty Hall problem** is a probability puzzle loosely based on the American television game show *Let’s Make a Deal* and named after the show’s original host, Monty Hall.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?”

Let’s analyze a bit. Obviously, before anything happens, you have **1/3 chance** of picking the right door. This is your pre-probability and is something that can not change no matter what happens. Hence, the probability the prize is behind some of the **other doors is 2/3** (since probabilities always sum up to 1). This, also, always stays the same. Now, after the host opens one of the other doors, what happens? Let me sum up:

- The probability that you picked the right door is 1/3. If you stick to it, you will win this show one in three times.
- The probability that the prize is behind some other door is 2/3. After the host does his thing, there is only
**one**other**door**left (but the probability remains**2/3**). If you pick that other door, you will win 2 out of three times.

So, you should pick the other door! People often don’t believe this result. For those who do not, here is a small generalization. Let’s say you have a 100 doors and one holds the prize (the host know where it is). You pick a door (with a chance of 1/100). The chance it is behind some other door is 99/100. The host opens 98 doors which don’t hold a prize. Will you switch? Put this way, everyone would switch – but put with three doors, and it is hard to believe you should…