# Monty Hall Proof – The Formula

Get the free App – Monty Hall Game with Monty Hall Proof.

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. $$P(\Omega) = 1$$)

# Monty Hall Solution – Advanced!

Get the free App – Monty Hall Game

This post “Monty Hall Solution” continues on from my previous post Monty Hall Problem – Can You Solve This Maths Puzzle? If you haven’t read that post, then read it now before reading this. Because I will now show you even more Monty Hall Solution coolness! We saw that you could increase (double) your chances of winning a car by understanding some maths, so let’s delve further into it and who knows, you might win something big (then again you might not, but hey!).

So what happens, if there are 4 doors instead of 3 doors? And what happens if there are 5 doors, 6 doors, hmmm, more code required – cool!

Checkout the code here Alan Cowap GitHub.
Get my free Monty Hall Game on Google Play.

# Does The Monty Hall Solution Hold True For 4 or More Doors?

Door 2 now has a 2/3 chance

For 4 doors we would expect that the odds would be 1/4 (25%) for not changing Vs 1.5/4 (37.5%) for changing. “How did you get those figures?” I hear you ask. Well, with 4 doors, each door has a 25% chance of being correct. Our 1st chosen door has a 25% chance – the other 3 doors have a combined 75% chance. When Monty removes one of those 3 doors by opening it -the remaining 2 doors still have a combined 75% chance – which is now divided by the 2 remaining doors i.e. 37.50% chance each. Once again the figures from the 70 million simulations are very precise.

Doors in Game: 4

Unchanged Wins: 2498993, Changed Wins: 3750566
Unchanged Wins: 24.99%, Changed Wins: 37.51%

Unchanged Wins: 2498720, Changed Wins: 3750419
Unchanged Wins: 24.99%, Changed Wins: 37.50%

Unchanged Wins: 2500812, Changed Wins: 3748980
Unchanged Wins: 25.01%, Changed Wins: 37.49%

Unchanged Wins: 2499171, Changed Wins: 3747829
Unchanged Wins: 24.99%, Changed Wins: 37.48%

Unchanged Wins: 2499602, Changed Wins: 3749207
Unchanged Wins: 25.00%, Changed Wins: 37.49%

Unchanged Wins: 2497548, Changed Wins: 3748332
Unchanged Wins: 24.98%, Changed Wins: 37.48%

Unchanged Wins: 2499206, Changed Wins: 3749860
Unchanged Wins: 24.99%, Changed Wins: 37.50%