TYI38 Lior Kalai: Monty Hall Meets Survivor

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Lior Kalai: Survivor Meets the Monty Hall Puzzle

We start with the classical question and go on with a new version contributed by my son Lior.

Update: A few brief comments on the original problem and Lior’s survivor variant are added to the end of the post. (Spoiler warning) A comment of Vincent describes another lovely variant.

Monty Hall – three doors, two goats and one car

The Monty Hall problem is a famous brain teaser, first posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975. It became famous as a question from a reader’s letter quoted in Marilyn vos Savant’s “Ask Marilyn” column in Parade magazine in 1990.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

 

Survivor – A golden coin in a stone in a case

In the current season of the Isreali Survivor television game each participant receives a case that contains a stone. One of the stones contains a golden coin. Every week one participant is voted off  the game. But, if after being voted off, the participant discovers the golden coin inside his stone then he stays in the game. In our puzzle, the host gives the best player of the round an opportunity to switch his stone with another stone.

Suppose you are in the survivor game and there are three remaining participants. One of the other participants is voted off. He opens his case, smashes the stone, and does not find the gold coin.  Therefore he is out. The host gives you an opportunity: “Do you want to replace your stone with the stone of the other remaining participant?” Is it to your advantage to switch the stones?

 

 

Lior, who contributed the question is our youngest son. Here in the picture from left to right: Hagai (our second child), my wife Mazi, Ilan, Lior, Eran (Neta’s husband), Yoav, and Neta (our first child.) Ilan and Yoav are Neta and Eran’s children. Picture taken by Felix S. Rettberg.  ©Felix S. Rettberg

 

Before moving to the solution and a discussion of our riddle

Breaking news: (added March 20, 2019)

I am very happy to report that yesterday, March 19, 2019, Karen Uhlenbeck was awarded the 2019 Abel prize “for her pioneering achievements in geometric partial differential equations, gauge theory and integrable systems, and for the fundamental impact of her work on analysis, geometry and mathematical physics.” The full citation for the prize and additional material may be found in the Abel Prize page  (Additional sources: I, II, III, IV, V,VI,VII, VIII, IX, X, XI, XII , XIII.) (More links  on the prize & 2019 committee: i, ii, iii, iv)

Warm congratulations to  Karen Uhlenbeck and to the mathematical community as a whole!

 

After you answered to the poll consider this: If your answers to the two polls are different can you put your finger on the difference between the two scenarios? 

An earlier survivor post: Impossibility result for “survivor”.

Another picture of us taken by Felix S. Rettberg.  ©Felix S. Rettberg

Further Discussion. (Spoiler warning)

An overwhelming number of people – 96 percent – gave the right answer to the first problem that it is beneficial to switch. (The probability to win the car goes up from 1/3 to 2/3.)

This answer is based on various  assumptions about the strategy of the host. The problem (as given in Wikipedia) only says: “the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat.” The precise strategy of the host is not entirely clear from this formulation of the problem. If, for example, the host’s choice to open another door depends on whether there is a car behind door number No. 1, switching may not be a good idea.  (For example, if the host opens another door only when there is a car behind door number 1, you had better not switch.)  If the host randomly chooses either door No. 2 or door No. 3, then it makes no difference whether to switch or not.

The problem gives an impression that this is a repeated situation. (And various people have made it explicit.) The game show is played every week, the host always opens one of the other doors, and the door he opens always has a goat behind it, and the host always gives the opportunity to switch.  In this case it is indeed beneficial to switch.

Also 80 % of participants gave the right answer to Lior’s variant. In this case switching does not make a difference.

When you change one word in the original problem: “the host, who does not know what’s behind the doors, opens another door, say No. 3, which has a goat,” then for this version it makes no difference whether or not to switch. However, this version does not specify what would happen if  door number 3 had a car.  (The host will be fired?) This difficulty does not arise in Lior’s survivor version.

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3 Responses to TYI38 Lior Kalai: Monty Hall Meets Survivor

  1. Vincent says:

    Your variation of Monty Hall inspired me to finally write up a different variation I found a few years ago and where the same question in orange (in your post) can be asked. Perhaps you enjoy this version as well, it is here: https://puzzling.stackexchange.com/questions/81569/monty-hall-variation

  2. Martin Fabian says:

    Is not the whole thing with the Monty Hall problem that the host opens the door with the car with probability zero? This is what it means that the host “knows what’s behind the doors”. In that case, which is the way I have always known it, the probability that the first chosen door has the car is 1/3. The probability that the host opens the door with the car is zero. Thus, the probability that the third door hides the car must be 2/3.
    In the survivor game, the “host” does not have any knowledge about where the coin is, and so opens the coin-box with probability 1/3. Therefore, it does not matter if the remaining boxes are switched or not, the probability for each of them holding the coin remains 1/3.

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