1. Cars, Goats, and Sample Spaces

这道题的精髓在于：(Relevant) information improves your chance of winning.
1. 不换：You are essentially not using the extra information to improve your chances. Thus your likelihood to win is the same as before Monty opens the other door.
2. 换：在这里很多人摔跤，因为觉得现在只剩下两扇门了，随机选一个的胜率就肯定是1/2。但是这两扇门是不一样的。对比一下下面这两种情况：
- 有A、B两扇门，其中一扇门背后有大奖，you do not know anything else about A or B and if you make random choice, your chance to win is indeed 1/2.
- 有A、B两扇门，其中一扇门背后有大奖。你选了A，但是现在你知道B是知情人Monty没有选的。这样的话，A和B两扇门的“信息量”就不同了。
Better Explained上的解释说的比我好：
Assuming that “two choices means 50-50 chances” is our biggest hurdle.

Yes, two choices are equally likely when you know nothing about either choice. If I picked two random Japanese pitchers and asked “Who is ranked higher?” you’d have no guess. You pick the name that sounds cooler, and 50-50 is the best you can do. You know nothing about the situation.

Now, let’s say Pitcher A is a rookie, never been tested, and Pitcher B won the “Most Valuable Player” award the last 10 years in a row. Would this change your guess? Sure thing: you’ll pick Pitcher B (with near-certainty). Your uninformed friend would still call it a 50-50 situation.

在这种情况下把Car = A、Car = B两个事件model成independent random events就不行了。

利用Bayes theorem的解法：
不换的胜算：1/3
换的胜算：假设有A、B、C三门，你选了A，大奖在B，Monty打开了C门。换的话你就得选B。
P(car = B | Monty opens C) = P(Monty opens C | Car = B) * P(car = B) / P(Monty opens C)
- P(Monty Opens C | Car = B) = 1
- P(car = B) = 1/3
- P(Monty opens C) = 1/2
- 换的胜算是2/3
我们也可以用Bayes theorem来写一下stick with A的胜率：
P(car = A | Monty opens C) = P(Monty opens C | Car = A) * P(car = A) / P(Monty opens C)
- P(Monty opens C | Car = A) = 1/2
- P(Car=A) = 1/3
- P(Monty opens C) = 1/2
- 不换的胜算是1/3。