Before the holidays I posted a question to you all to see what you thought. The question was the famous Monty Hall problem where there are 3 doors and a contestant is told 1 has a fabulous prize while the other 2 have goats. The contestant selects a random door and the host reveals one of the doors that have a goat and gives you a chance to switch doors if you want or not. The question was: is it better to keep the original door, switch doors, or are the odds the same? Here's the link if you want a refresher.
So most people will look at this problem and say the chances are the same whether you switch or keep the same door. The logic usually follows like this: There are 2 doors left. One has a present, the other a goat, so there is a 50% chance that either of them will have a goat. Thus each door has the same probability of having the prize.
Although this is sound logic, there is only one big mistake, the assumption that the 3rd door doesn't matter. Let's start from the beginning and work our way up in order to see what the answer is. Since we first choose a random door let's pretend it's door A. We now have 3 options: the present is behind door A, the present is behind door B, the present is behind door C.
Since those are the only doors available those are only ways of playing. Now let's look at the winning ratio. If we stayed the same we would have won 1 game. If we switched then we would have won 2 games. So by switching doors you have increased your chances of winning to 66%! So remember to always switch when given a choice!