# Variation of Monty Hall game. Homework. Need checking.

Assume that you are playing the following variation of the Monty Hall game: A valuable prize is randomly placed behind one of the four doors, numbered from 1 to 4, and nothing behind the remaining three doors. You are allowed to select a door. After your selection, the game show host will open one of the remaining doors, but he will always open a door without the prize. However, the host will always open the door with the smallest number if no prize is behind the remaining doors; or smaller number if the prize is behind one of the remaining doors. After opening a door, the host will offer you the possibility to switch to one of the two doors. We will also assume that your aim is to win the prize and you know all the above information (such as host always opens a door without prize which has the smaller or smallest number).

Consider the following strategies: (I) Initially you select door 4 and you stick with it.

(II) Initially you select door 4 and you will always switch if you are offered the opportunity to switch. You’ll choose the door to switch to at random from the available possibilities.

(III) Initially you select door 4. You will switch to door 1 if the host opens door 2; you will stick with your initial choice of door 4 if the host opens door 1.

Choose one

My solution

There are 4 possibilities for the arrangement of the prizes, namely

X X X P / X X P X / X P X X / P X X X

Consider Strategy (I). Chance of winning is a flat $1/4$.

Consider Strategy (2). He will open the smallest door. For my cases, I end up with OXXP\OXPX\OPXX\POXX. Chance of winning if I switch randomly is $1/4*1/2+1/4*1/2+1/4*1/2+0=0.375$

Consider Strategy (3). Cases are OXXP\OXPX\OPXX\POXX. Probability is $1/4+1/4=1/2$. Because there is only 2 situations in which this can win!

Am I right to say that strategy 3 yields me the highest chance of winning?

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Why would you switch to a random door if he opens door 2 and door one is up for grab? – Jean-Sébastien Oct 16 '12 at 0:05

I agree with your calculation. For a check of your value for 2, you could think this way: If the prize is in 4 (probability $\frac 14$) you lose. Otherwise, you know it is neither $4$ nor the door opened, so you have $\frac 12$ chance now to win. Overall, this gives $\frac 34 \cdot \frac 12=\frac 38$. I give you strategy 4) Initially select door 4. If host opens 2, change to 1. If host opens 1, change to a random door. What chance do you have now? Jean-Sébastien has given a good intuitive reason why strategy 3 has to beat strategy 2. Can you do the same for this one?