I was struggling to understand the Monte Hall problem. This explanation helped.

I think the confusion is between dependent and independent probability. Ironically, I dropped my first course in Statistics because the prof was confusing the two.

Here’s the explanation that makes sense to me:

• If you choose the correct door and you switch, you lose.
• If you choose incorrect door-a and you switch, you win.
• If you choose incorrect door-b and you switch, you win.

Ergo, in 2 of 3 cases you win by switching.

Now consider:

• If you choose the correct door and you don’t switch, you win.
• If you choose incorrect door-a and you don’t switch, you lose.
• If you choose incorrect door-b and you don’t switch, you lose.

Ergo, in 2 of 3 cases you lose by not switching. (That’s logically equivalent to saying that in 1 of 3 cases you win by not switching.)

The thing is, this is **dependent **probability, but it is dependent upon whether you picked correctly on the FIRST guess, not on where the win is.

There are two possibilities for your first choice: You picked a winner or you picked a loser.

• Given that you picked a winner, what is the probability that switching will win? Zero
• Given that you picked one loser, what is the probability that switching will win? 50%
• Given that you picked the other loser, what is the probability that switching will win? 50%
• What is the probability that your initial pick was a loser? 2 in 3.
• What is the probability that your initial pick was a winner? 1 in 3.

By not switching, you are betting that your initial pick was right, which has only a 1 in 3 probability.