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Suppose you're on a game show, and you're given the choice of one hundred 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 switch your door?" Is it to your advantage to switch your choice?

Assumptions

  1. The host must always open a door that was not picked by the contestant.

  2. The host must always open a door to reveal a goat and never the car.

  3. The host must always offer the chance to switch between the originally chosen door and the remaining closed doors.

Edit: Not a duplicate, host is not leaving only two doors remaining, he is only opening one door, leaving 99 remaining

Mike Pierce
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lucy
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  • Look at the second answer. – Zach466920 Jun 23 '15 at 15:24
  • @Zach466920 Isn't this different? I interpret this to mean only $1$ other door is opened out of the $100$ original doors, in that answer all but $2$ doors are opened. – CSCFCEM Jun 23 '15 at 15:28
  • Yeah that is correct, I already saw that version, I am wondering if the host only opens 1. – lucy Jun 23 '15 at 15:31
  • To lucy, its always to your advantage to switch, and there is an answer that hammers that in. That's why I think this is a duplicate. Merely replace the appropriate numbers. – Zach466920 Jun 23 '15 at 15:40

2 Answers2

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The explanation in the question linked above hinges on the fact that you will always switch from goat to car and vice-versa if you do switch, which isn't true if there are more than two doors left unopened as in this question. That being said, the general reasoning behind why switching is always better still applies. We can run through the math to verify this.

If we always keep our original door, then our probability of getting the car is the probability we choose the car door first try, which is clearly $\frac{1}{100}$.

If we choose to always switch, then our probability of ultimately getting the car is the probability we choose a goat times the probability we switch from a goat to a car (we can ignore the case where we start with a car since going from a car to a car is impossible). Our probability of picking a goat initially is clearly $\frac{99}{100}$. Then, once we pick a goat and one goat door is opened, there are 98 other doors, of which one has a car. So our chance of switching from a goat door to a car door is $\frac{1}{98}$. Thus our chance of getting a car if we always switch is $\frac{99}{100}*\frac{1}{98}=\frac{\frac{99}{98}}{100}$ which is slightly greater than $\frac{1}{100}$. So, the optimal strategy is still always switching.

Arun G
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  • Excellent! That's what I was looking for, very clear and helpful! I had thought switching would still be better in this case, but couldn't justify it. – lucy Jun 23 '15 at 15:59
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You might like this short and sweet explanation.

P(win by not switching) = $\dfrac{1}{100} =0.01$

P(win by switching) = P(original choice has a goat)$\times$P(guess correctly after switching)

= $\dfrac{99}{100}\times\dfrac{1}{98} \approx 0.0101, > 0.01$

true blue anil
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