Monty Hall Proof – The Formula is here. My two previous posts described the Monty Hall Problem – Can You Solve This Maths Puzzle? and Monty Hall Solution – Advanced! Well, this is the next installment of the trilogy, a simple mathematical proof.

If you don’t like Maths (Mathematics, Math) then, well, you have serious problems – get some help :^) This isn’t difficult at all, it’s just a bit of simple probability and algebra, yep ALGEBRA ♥

The Probability that you will Win is the quotient of the Number of Cars, and (divided by) the Number of Doors. To represent that symbolically using algebra is simple:

\(P(W) = \frac{NC}{NDtot} \) … Equation (1)

The Probability that you will Lose is a little more interesting, it is the quotient of the Number of Doors less the Number of Cars, and (divided by) the Number of Doors, in symbolic notation this is:

\(P(L) = \frac {NDtot – NC}{NDtot} \) … Equation (2)

There’s one last equation we want, and it says the Probability that we either Win or Lose is 1 – since these are the only two possible events. In other words, we have to either win or lose – there are no other possible events (see my earlier post re the philosophical and physics debates on that general point). Anyway, to represent this symbolically:

\(P(W) + P(L) = 1 \) … Equation (3)

(Equation (3) is based on Kolmogorov’s second axiom i.e. \(P(\Omega) = 1\))