# A variation on the Monty Hall Problem

I was watching Lets Make a Deal this morning and this game was played:

There are 7 numbers with hidden items underneath. 4 have zonks (no prize), 1 has \$500, 1 has \$1000, 1 has a car. You can get rid of 3 of the zonks by playing a game.

Which struck me that this could be a variation of the Monty Hall problem.

So lets say this happens. I pick item #4. I play the game and knock out three of the zonks. Now Wayne (Monty) gives me the option of picking a different number. Should I do it?

And another variation:
In the actual game played, you didn’t get to pick a different number if you wanted. but you didn’t have to pick until after the zonks were knocked out. So lets say when the numbers are first presented, I think to myself “I’m going to pick number 4”. I knock out 3, and number 4 happens to still be there. Should I stick with 4, or change my pick?

To review the basic 3-door Monty Hall game (1 car 2 zonks), make the usual assumed rules:

1. Monty must always open a door not chosen by the contestant;
2. Monty must always reveal a zonk;
3. Monty must always offer the option to switch.

Under these rules, I think the strongest intuition is gained by the following perspective:
After you pick a door, Monty revealing a zonk and giving you the option to switch is logically equivalent to Monty offering you the best of the other two doors. The best of two doors has a superior 2/3 chance of being the car, so you should switch.

Similarly, in the current game (if the rules about what the host must do are similarly rigid), in opening three zonk doors the host is offering you the best three of the other six doors, so just as in regular Monty Hall you should certainly switch.

In the variation, the “mental pick” does not have the same effect as an actual pick, since it does not restrict the host’s actions. Any unopened door has an equal chance.

I think you should swap and I rationalise it thus.

Your initial free choice sees you with a 1 in 7 chance of hitting the car and a 4 in 7 chance of hitting a zero… Switching gives you a 1 in 4 chance of hitting the car and a 1 in 4 chance of hitting a zero. Which set of odds would you prefer to play with?

I’m sure there’s some fiendish probability calculation that lays it out in more precise terms but that was my immediate back-of-a-fag-packet feel for it.

Those odds are not quite right. Odds of getting a car if you switch are the odds that the car was among the 6 you didn’t originally pick (6/7) times the 1 in 3 chance of switching to the correct one among the 3 remaining unopened doors.

6/7 x 1/3 = 2/7, slightly better than 1/4

Since the problem is so similar to Monty Hall, let’s just look at Monty Hall One.More.Time. :smack:

The simplest way to understand Monty Hall puzzle using the rules Riemann posted in #2, is that
(1) Before the Reveal your chance is 33.3333% exactly.
(2) Monty will always Reveal Zonk, so this Reveal gives you no new information about your original selection.
(3) Therefore your win chance with that door is unchanged; it is still 33.3333%.
(4) Exclusive probabilities always sum to 1, so the win-chance with a switch is 100% - 33.33333% = 66.66667%.
Focusing on changes to your information is the way to make short work of many problems of this type.

Intuitively, I figured that like the Monty Hall problem it would be better to switch. I wasn’t sure if have two other lesser prizes had an impact.

In the game, you had a chance to remove up to three of the four zonks. How do the odds change if I only removed one or two?

There’s a 6/7 chance that the car is among the 6 doors you didn’t initially pick.
If host removes 1 zonk to leave 5 doors to switch to, p = 6/7 x 1/5
If host removes 2 zonk to leave 4 doors to switch to, p = 6/7 x 1/4
If host removes 3 zonk to leave 3 doors to switch to, p = 6/7 x 1/3

All are better than the 1/7 chance if you don’t switch

But I don’t think this is the clearest intuitive way to grasp the basic Monty Hall problem. Even people who are good at probabilities often get it wrong when they first encounter it.

Reframe the (3-door) situation as follows:

You initially pick door #1.

Rather than opening another door, suppose Monty now says: you can stick with door #1, or you can switch and I will give you the best of either door #2 or door #3 - in other words, if you switch you get the car if it’s behind either #2 or #3. Obviously you can see that you must switch, two doors is better than one.

If you think about it carefully, you can see that the above is exactly equivalent to the actual game where Monty opens a bad door to eliminate it and offers the switch to the third remaining door. He is telling you which is the best of the two doors that you didn’t originally pick (if they are different), and letting you switch to it.

The above perspective makes it transparently clear what information you have about which doors.

The rules of the game are effectively that Monty will provide you with an additional piece of information: a comparison between two doors to tell you which of those two is better. In your initial pick (say door #1) you are specifying which two doors you want compared - the other two. In opening one of the other two doors, Monty is doing the comparison and telling you which of door #2 and door #3 is better. Let’s say Monty opens door #2 to reaveal a zonk. You can stick with your original choice door #1 or switch to door #3, and the asymmetric information you now have is:

door #1 - no information
door #3 - is better than or equal to door #2

Thus, sticking with door #1 gives you the best of one door.
Switching to door #3 gives you the best of two doors.

Yes, but those are also the chances if you don’t switch, after the three zonks have been removed.

No, the chance of getting the car if you don’t switch is 1/7, just as it was before the zonks were removed.

With three zonks removed and three other doors available to switch to, switching to one of the other three doors gives you a 2/7 chance of getting the car.

1/7 + 2/7 + 2/7 + 2/7 = 1

To complete the picture - switching doubles your chance of winning, just as in the original Monty Hall. That’s because in both games Monty eliminates half of the remaining doors (in the original game, one out of two; in this game, three out of six).

I think the best way of grasping the Monty hall problem is if you imagine 50 doors. After your choice Monty opens all doors except yours and one other, do you now switch?

Of course, with only three doors (or seven choices as in the OP) the odds improve with switching less impressively but they still improve by switching.

Just as in the original Monty Hall problem, we have to be clear about how the elimination works.

If you pick one of the seven first, and then the three zonks that are eliminated are chosen from among the six numbers that you did not pick, then your analysis applies, and you should switch.

If you wait until after the three zonks are eliminated to pick, then you have 4 to choose from, so your chances of picking the car are obviously 1 in 4 (and your chance of picking the forth zonk is also 1 in 4).

If you pick first (either just mentally, or specifying your choice out loud), and the three zonks that are eliminated are chosen at random from all four zonks (including your choice if it happened to be a zonk), one of two things will happen:

(1) your choice might get eliminated. In that case, you’d have to choose again, and your choice would have a 1 in 4 chance of being the car. Or

(2) your choice does not get eliminated. That’s the tricky case. It’s the one I was mainly thinking of when I said “those are also the chances if you don’t switch.” It seems to me to be essentially equivalent to the situation where the three zonks were eliminated before you picked, since your initial pick has an equal chance of being any of the four that remain after the zonks were eliminated.

Ok, so now you’re looking at the “variation” in the last paragraph of the OP. Nothing in my above discussion relates to that.

There’s nothing tricky about it really. If your initial “choice” (whether secret or otherwise) does not restrict the additional information that the host will provide, then it cannot have any effect on the relative probabilities among the doors that remain for your final choice. All remaining doors (including your initial “choice”) must have equal probability - this must trivially be true by symmetry, since two potential players could simultaneously make two different initial choices.

I refer you again to my posts #8 and #9 which I think give the clearest framework for understanding when there is asymmetry of information. The host is effectively offering to provide additional information: a comparison between some (sub)set of doors to tell which of that set of doors is better. If your initial choice restricts the host to exclude the initial door from inclusion in that set, then your information about the initial door is different from the information that you have about the other doors that are included in the comparison set. But if your initial choice does not exclude the initial door from inclusion in the comparison set, then the information you have about the initial door is the same as the information that you have about the other doors.

I wasn’t entirely sure from the OP just what was the “actual game played” and what was the “variation.”

I meant both as a “variation on the original Monty Hall problem.”

So lets say Variation A: You pick first, knock out 3, then you can switch.
Variation B: You don’t pick first (except in your head), knock out 3, then you choose.

To be clear, the difference between A and B is whether your initially chosen door is available for the host to include as one of the 3 zonk doors that he opens. In A your initial choice may not be opened; in B it is available to the host and may be opened.

Then, to summarize clearly the chance of getting the car:

Variation A:
Stick with initial choice, p=1/7
Switch to an other of 3 remaining doors, p=2/7

Variation B:
Stick with initial choice if still available, p=1/4
Switch to another of the 3 (or 4) remaining doors, p=1/4

Once you know the correct answer to that particular puzzle, yours may be the easiest way to convince someone else of that answer.

I was trying to present a general framework that leads to the correct answer in a wider variety of probability puzzles.