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The Magic Cafe Forum Index » » Magical equations » » Monty Hall Dilemma (0 Likes) Printer Friendly Version

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FrankFindley
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Quote:
On May 19, 2022, glowball wrote:
Let's say we add a new rule: The contestant can switch or take the shown prize. Then in situations where the car is shown do you just take it and forget about the $100,000?

A bird in the hand is worth two in the bush?


Looking at the scenarios from above, it affects the two possibilities in bold:

1/3rd of the time you will choose a Donkey, he will show a car, switching will get you $100K Good Switch
1/3rd of the time you will choose a Car, he will show a donkey, switching will get you $100K Good Switch
1/6th of the time you will choose $100K, he will show a donkey, switching will get you a Car Bad Switch but still a winner
1/6th of the time you will choose $100K, he will show a car, switching will get you a Donkey Bad Switch and a loser

So, two out of three times that happens, you would be keeping the car and foregoing the $100K. But one out of three times you would be keeping the car when you otherwise would be getting a donkey.

Therefore, if you are a risk adverse person, you should always keep the car when it is shown as this precludes you from walking away with the donkey. But if you are a gambler, you should forgo the car as 2/3rds of the time you will walk away with $100k, very good betting odds!

It is more complicated if you are neutral to risk. Then it depends on the price of the car. Assuming the donkey is worth nothing, the expected value of switching when you are shown the car is $66.7K. That is: ($100K + $100k + $0)/3 = $66.7K. So if the car is worth that or more, you should take the car. If the car is worth less, you should switch.

All this assumes that there are no taxes involved on winning. Of course we all know in reality, Uncle Sam is always the real winner of these games!
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