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 Proof – The Formula, Here Comes the Proof!

Those 3 Equations give us what we need to check the proof. If the Equations are correct then when we combine the equations the result should give us an equality – that’s why *equations* are also known as e*qualities*!

We start with Equation (3), into which we will substitute P(W) from Equation (1), and P(L) from Equation (2); as follows:

\(P(W) + P(L) = 1 \) … Equation (3)

\(\frac{NC}{NDtot} + \frac {NDtot – NC}{NDtot} = 1 \) … Substituting for P(W) and P(L) … Equation (4)

Now we want to simplify, an easy simplification is to combine the two terms of the left-hand-side of the equation since they have the same denominator (NDtot), which gives us:

\(\frac {NC + NDtot – NC}{NDtot} = 1 \) … Combining the left hand terms

Can you spot the next simplification? Take a look and see. Did you get it? That’s right, the ND and -ND will cancel each other out, giving us:

\(\frac {NDtot}{NDtot} = 1 \) … The NC terms cancel each other out

Can you finish the Monty Hall Proof? Yes, any term divided by itself equals 1, giving us:

\(\frac {1}{1} = 1 \) … we get 1 = 1 which is a proper equality, we did it!

**Q.E.D.**

# Monty Hall Proof – Test it, try some actual numbers.

So it seems we have this all wrapped up, but let’s try it with the actual numbers from the App. We have 3 doors, and 1 car, Recall Equation 4:

\(\frac{NC}{NDtot} + \frac {NDtot – NC}{NDtot} = 1 \)

Now let’s put in our values for total number of doors: NDtot =3, and number of cars NC = 1, giving us:

\(\frac{1}{3} + \frac {3 – 1}{3} = 1 \) … using our actual numbers from the App

\(\frac{1}{3} + \frac {2}{3} = 1 \) …do the math 🙂

\(\frac {3}{3} = 1 \) … great, it’s correct!

**Q.E.D.**

Go ahead and try with 2 cars and 3 doors; or try with 3 cars 3 doors, or 4 cars and 20 doors, it even works with 4 cars and 3 doors 😀

I hope you enjoyed the Monty Hall Proof, including the previous posts, and found them informative. Feel free to leave a comment, or if you see an error let me know, finally a nod to Andrey Kolmogorov and his pioneering work on Probability, he was born 115 years and 1 week ago.

Checkout the code on GitHub.

Get the free Monty Hall Game on Google Play.