# Modified monty hall problem

Hello how to show the following

You are given the choice of 3 doors. Behind one is a car and the other two are goats. You pick a door uniformly at random say 1, and Monty opens another door, say 3 which has a goat. Monty asks "do you want to switch your choice to door 2?" If we have the following modified assumptions:

*The car was put behind a uniformly chosen random door.

*Monty knows where the car is and always opens a door with a goat behind it.

*When Monty has a choice(i.e. the contestant picked the door with the car behind it), he chooses the door with the larger number with probability 3/4.

Show that if the contestant switches, she wins with probability 4/7 not 2/3.

Thanks a lot!

• Are you sure that you stated the problem in its entirety? As it stands, Monty’s choice strategy appears to be irrelevant. – Brian M. Scott Feb 18 '13 at 10:55
• Hi Prof. Scott yes i am sure the setting is this way... – Salih Ucan Feb 18 '13 at 10:57
• I would assume that the contestant's strategy should be something like "Switch if the lower numbered of the two non-chosen doors is opened". As @BrianM.Scott says, which goat Monty chooses is irrelevant if the contestant has already decided to switch. – Arthur Feb 18 '13 at 10:59
• As the problem is stated, she wins with probability $\frac23$ if she switches, since if she switches she wins if and only if her initial choice was wrong. – Brian M. Scott Feb 18 '13 at 11:00
• I have clarified the problem i hope.. i edited the problem.. – Salih Ucan Feb 18 '13 at 11:13

If she always switches, she wins exactly when her original choice was wrong, which means that she wins with probability $\frac23$.

Suppose that her strategy is to switch if and only if Monty opens the lower-numbered of the two doors that she did not choose initially. She wins if she chose right initially and Monty opened the higher-numbered of the other two doors, or if she chose wrong initially and Monty opened the lower-numbered of the other two doors.

The probability of the first of these alternatives is $\frac13\cdot\frac34=\frac14$.

Now suppose that she guessed wrong, which of course occurs with probability $\frac23$. Consider the $3$ equally likely arrangements, CGG, GCG, and GGC. If she guessed wrong in the first one, Monty didn’t open the lower-numbered of the other two doors, since the car is behind it. If she guessed wrong in the second one, Monty opened the lower-numbered of the other two doors with probability $\frac12$. And if she guessed wrong in the third one, Monty had to open the lower-numbered of the other two doors. In short, if she guessed wrong, the probability that Monty opened the lower-numbered of the other two doors is $\frac12$. The probability of this alternative is therefore $\frac23\cdot\frac12=\frac13$.

Her probability of winning if she does switch according to this strategy is then $\frac14+\frac13=\frac7{12}$.

• I think "her probability of winning if she does switch" should just say "her probability of winning"? – joriki Feb 18 '13 at 11:19
• @joriki: Yep, either that or the version that I think was in the back of my mind as I wrote; thanks! – Brian M. Scott Feb 18 '13 at 11:26

For your question to make sense, I assume you want to show that $\frac{4}{7}$ is the probability of winning if the contestant switches given that the contestant chooses door 1 and Monty opens door 3. Let $A, C, M$ be the random variables representing the number of the door which the car is behind, which the contestant chooses, and which Monty opens respectively. \begin{align} & \text{P(car is in door 2|contestant chooses door 1 and Monty opens door 3)}\\ =&P(A=2|C=1, M=3) \\ =& \frac{P(A=2,C=1, M=3)}{P(C=1, M=3)} \\ =&\frac{P(A=2,C=1, M=3)}{P(A=1,C=1, M=3)+P(A=2, C=1, M=3)} \\ =&\frac{\frac{1}{9}}{(\frac{1}{9})(\frac{3}{4})+\frac{1}{9}} \\ =&\frac{4}{7} \end{align}

Because which door Monty opens makes a difference, the probability of winning if the contestant switches depends on the doors chosen by the contestant and opened by Monty.

Edit: I suppose you can write it as follows:

$P(A \not =C|M \text{is the larger of the 2 numbers other than } C)=\frac{4}{7}$

Similarly we would have

$P(A \not =C|M \text{is the smaller of the 2 numbers other than } C)=\frac{4}{5}$

• Yes, so you need to know both the door chosen by the contestant and the door opened by Monty to perform the calculation. (And it was specified that the contestant chooses the door uniformly at random) – Ivan Loh Feb 18 '13 at 11:44
• Sorry, I'd already deleted the comment because I noticed I'd overlooked that specification. No, you don't need to know the door chosen by the contestant; the number of that door is never used, and the entire problem can be phrased in terms of just the two numbers on the doors not chosen by the contestant. – joriki Feb 18 '13 at 11:46