In a Monty Hall Problem with four doors, in which only one door has the price and the rest are goats.
(a) There are four different possible
strategies.
There were only two solutions in the original Monty Hall problem
the sticking strategy where you'd stick with your original pick
after Monty opens the first door, or the switching strategy where
you'd change your original pick once Monty opens the first
door.
Now, there are two stages where a decision is required, and there
are two choices at each stage. These are
This gives you 2*2 = 4 different strategies as shown in the table below.
Stage | 1 | 2 | 3 | Probability of winning |
Strategy 1 | Pick | Stick | Stick | ?? |
Strategy 2 | Pick | Switch | Stick | ?? |
Strategy 3 | Pick | Stick | Switch | ?? |
Strategy 4 | Pick | Switch | Switch | ?? |
(b) Now let's complete the table by calculating the Probability of winning for each strategy
I will use the following notation
So, just to be clear, WLW will correspond to Win Lose Win
Note that we only Win, when we choose the
Winning door in the last time.
Strategy 1: Stick in both
Here we can have the following cases:
Since you only won in the 1st case i.e. WWW, so P(Winning | Strategy 1) = 1/4 = 0.25
Strategy 2: Switch and Stick
Here we can have the following cases:
Since you only won in the 3rd case i.e. LWW, so P(Winning | Strategy 2) = 0.375
Strategy 3: Stick and Switch
Here we can have the following cases:
Since you only won in
the 2nd case i.e. LLW, so P(Winning |
Strategy 3) = 0.75
Strategy 4: Switch and Switch
Here we can have the following cases:
Since you won in the 1st and 3rd case i.e. WLW and LLW, so P(Winning | Strategy 4) = 0.25 + 0.375 = 0.625
So now finally we can complete our table as follows:
Stage | 1 | 2 | 3 | Probability of winning |
Strategy 1 | Pick | Stick | Stick | 0.25 |
Strategy 2 | Pick | Switch | Stick | 0.375 |
Strategy 3 | Pick | Stick | Switch | 0.75 |
Strategy 4 | Pick | Switch | Switch | 0.625 |
So Strategy 3 i.e Stick and Switch is the best strategy for winning.
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