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In the three-door Monty Hall problem, there are two stages to the decision, the initial pick followed by the decision to stic

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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

  • Either you stick with your current door, or
  • you can switch to another door (stage 1 has two possible doors to choose from, choose one of them randomly)

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

  • WWW = to denote picking the Winning door in each of the 3 times
  • LLL = to denote picking the Losing door in each of the 3 times

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:

  • Pick the Winning door first time and stick to it, so WWW
    Probability(WWW) = Pr(picking the wining door out of 4 doors in first time) = 1/4 => YOU WIN
  • Pick any of the three wrong doors in the first time and stick to it
    Probability(WWW) = Pr(picking 3 wrong door out of 4 doors in first time) = 3/4 => YOU LOSE

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:

  • Pick the Winning door first time, Switch to Losing door, Stick to it, so WLL
    P(WLL) = 1/4 = 0.25
    => YOU LOSE
  • Pick the Losing door first time, Switch to Another Losing door, Stick to it, so LLL
    P(LLL) = 3/4*1/2 = 0.375
    => YOU LOSE
  • Pick the Losing door first time, Switch to Winning door, Stick to it, so LWW
    P(LWW) = 3/4*1/2 = 0.375
    => YOU WIN

​​​​​​​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:

  • Pick the Winning door first time, Stick to it, and Switch to losing door, so WWL
    P(WWL) = 1/4 = 0.25
    => YOU LOSE
  • Pick the Losing door first time, Stick to Losing door, and Switch to Winning door, so LLW
    P(LLW) = 3/4 = 0.75
    => YOU WIN

​​​​​​​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:

  • Pick the Winning door first time, Switch to Losing door, and Switch to Winning door, so WLW
    P(WLW) = 1/4 = 0.25
    => YOU WIN
  • Pick the Losing door first time, Switch to Winning door, and Switch to Losing door, so LWL
    P(LWL) = 3/4*1/2 = 0.375
    => YOU LOSE
  • Pick the Losing door first time, Switch to Losing door, and Switch to Winning door, so LLW
    P(LLW) = 3/4*1/2 = 0.375 => YOU WIN

​​​​​​​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.

Please upvote and provide feedback if this answer helped you. This would help me improve and better my solutions.
I will be happy to answer your doubts, if any in the comment section below. Thanks! :)

​​​​​

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