Here is an interesting but misleading probability question: A drunk man reaches his home and tries the door key from a bunch of 10 keys. If the first key doesn’t work, he returns the key to the bunch and randomly selects another key repeatedly until he opens the door. The question is: which precise trial has the highest probability of opening the door?
The question is misleading because it does not ask you to guess by which try the person will find the right key and open the door. We’ll come to that topic a little later. The task is to find the probability of finding the right key at each try and show which one is the highest.
The probability of finding the right key in his first attempt = 1/10; ten keys and finding one at random is one out of 10 possible options.
The probability of finding the right key in his second attempt = (9/10)x(1/10); it is the joint probability of not getting the right key in the first attempt (9/10) AND the chance of hitting the right on the second try.
The probability of finding the right key in his third attempt = (9/10)x(9/10)x(1/10). Here is the summary:
| Attempt | Probability |
| 1 | 0.1 |
| 2 | 0.09 |
| 3 | 0.081 |
| 4 | 0.0729 |
| 5 | 0.06561 |
It doesn’t mean that person will open the door in the first attempt or never. We need to estimate something different to find how the probability of opening the door changes with attempts. The chance he opened the door in his second attempt is:
Probability he opened in the first OR probability he opened in the second = 1/10 + (9/10)*(1/10). Here is how that develops.
| Attempt | Probability | Probability of Right key by Attempt |
| 1 | 0.1 | 0.1 |
| 2 | 0.09 | 0.19 |
| 3 | 0.081 | 0.27 |
| 4 | 0.0729 | 0.34 |
| 5 | 0.06561 | 0.41 |
| 6 | 0.059 | 0.47 |
There is about a 50% chance he will open the door by his sixth attempt.
