Problem 5.21 Easy Free 3 variations

Monty Hall

There are three doors: one hides a car and the other two hide goats, placed independently of your choice. You choose a door uniformly at random. The host, who knows what is behind every door, then opens one of the remaining two doors and it always shows a goat (he never opens your door or reveals the car). You are offered the chance to switch to the other, still-closed door. If you switch, what is the probability that you win the car?

Work it out, then compare with the answer.

SolutionShow the worked solution

The host always opens a goat door from the two you did not choose. So the closed door you could switch to is determined by your first pick:

If your first pick hides the car, the other two doors both hide goats. Whichever one the host opens, the remaining door is a goat, so switching loses.
If your first pick hides a goat, the host cannot open the car door. He must open the only other goat door. The remaining closed door is then the car, so switching wins.

Switching therefore wins exactly when the first pick was wrong. The car is placed independently of your uniform choice, so this happens with probability $\frac23$ :

\mathbb{P}(\text{win by switching}) = \mathbb{P}(\text{first pick is a goat}) = \frac{2}{3}.

\boxed{\tfrac{2}{3}}

Key idea. An informed host who never shows the car reveals nothing about your own door, which keeps probability $\frac13$ . The remaining $\frac23$ moves entirely onto the one other closed door.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 5.21a

There are three doors: one hides a car and the other two hide goats. You choose a door uniformly at random. Instead of the host, an audience member who has no knowledge of what is behind any door is picked to open one of the two doors you did not choose, selected uniformly at random between them. This door happens to reveal a goat. You are then offered the chance to switch to the other closed door. If you switch, what is the probability that you win the car?

Work it out, then compare with the answer.

Solution

Let $C$ be the event that your first pick hides the car, and $G$ the event that the opened door shows a goat. The audience member picks one of your two unchosen doors uniformly, with no knowledge of the prizes.

If $C$ holds, both unchosen doors hide goats, so $\mathbb{P}(G\mid C)=1$ .
If $C$ fails, exactly one of the two unchosen doors is a goat, so $\mathbb{P}(G\mid C^c)=\frac12$ .

Hence

\mathbb{P}(G)=\tfrac13\cdot 1+\tfrac23\cdot\tfrac12=\tfrac23 .

Switching wins exactly on $C^c\cap G$ , which has probability $\tfrac23\cdot\tfrac12=\tfrac13$ . Therefore

\mathbb{P}(\text{switch wins}\mid G)=\frac{1/3}{2/3}=\frac12 .

Unlike the informed host, this goat reveal is evidence: it is twice as likely when your first pick was the car.

\boxed{\tfrac12}

Variation 5.21b

There are ten doors: one hides $1000 and the other nine hide goats (worth $0). You choose a door uniformly at random. The host, who knows what is behind every door, opens three of the remaining nine doors, all of which reveal goats. You then switch to one of the six still-closed doors other than your original choice, chosen uniformly at random. What is the expected value, in dollars, of this strategy?

Work it out, then compare with the answer.

Solution

Your first pick is wrong with probability $\frac{9}{10}$ . The host only ever opens goat doors, so in that case the money is behind one of the six closed doors you might switch to.

You choose among those six uniformly at random. So given that the money is among them, you find it with probability $\frac16$ , whatever rule the host used to choose his three doors.

If your first pick was right (probability $\frac1{10}$ ), switching always loses. Hence

\mathbb{E}[\text{winnings}] = 1000\cdot\frac{9}{10}\cdot\frac16 = 150 .

This beats staying, which is worth $1000\cdot\frac1{10}=100$ .

\boxed{\$150}

Variation 5.21c

There are five doors: one hides $10,000 and the other four hide nothing. If you pick a door uniformly at random without any further information, what is your expected winnings?

Before picking, you have the option to buy a pair of x-ray goggles that let you see behind exactly one door of your choice before the battery dies. How much should you be willing to pay for this option (in terms of the increase in expected winnings it provides, assuming you play optimally afterward)?

Suppose you use the goggles on one door and see that it is empty. You then pick one of the remaining four doors uniformly at random. The host, who knows where the money is, opens one of the three doors you did not pick (other than the one you already saw was empty), always revealing an empty door. You are then offered the chance to switch to either of the two remaining closed doors. Playing optimally, what is your probability of winning the $10,000?

Work it out, then compare with the answer.

Solution

Without goggles. A uniformly random pick wins with probability $\frac15$ , so the expected winnings are $\frac15\cdot 10{,}000=\$2000$ .

Value of the goggles. Look behind any door.

With probability $\frac15$ you see the money and take that door.
Otherwise you pick uniformly among the other four doors and win with probability $\frac14$ .

Hence

\mathbb{P}(\text{win})=\tfrac15+\tfrac45\cdot\tfrac14=\tfrac25,

so the expected winnings are $\$4000$ . The goggles are worth an increase of $\$2000$ .

After the host’s reveal. Call your pick $A$ and the other unseen doors $B,C,D$ . Each has prior probability $\frac14$ . Suppose the host opens $D$ , choosing uniformly among the eligible empty doors. Then

\begin{aligned} \mathbb{P}(D\text{ opened}\mid A)&=\tfrac13,\\ \mathbb{P}(D\text{ opened}\mid B)=\mathbb{P}(D\text{ opened}\mid C)&=\tfrac12,\\ \mathbb{P}(D\text{ opened}\mid D)&=0 . \end{aligned}

The posterior probabilities are proportional to $\frac13,\frac12,\frac12$ , which sum to $\frac43$ . This gives

\mathbb{P}(A)=\tfrac14,\qquad \mathbb{P}(B)=\mathbb{P}(C)=\tfrac38 .

Your door keeps its $\frac14$ , and the remaining $\frac34$ is split equally between the two other closed doors. Switching to either of them is optimal.

\boxed{\$2000;\ \ \$2000;\ \ \tfrac38}

Spotted an error? Email us