Door 4? No, no, door 7

Probability Level 2

You find yourself on a game show and there are 10 doors. Behind nine of them are assorted, random farm animals including a variety of pigs, goats, and geese, but behind one is a brand-new, shiny, gold car made of real gold!

The game proceeds as follows:

You pick a door.
The game show host then opens a door you didn't choose that he knows has only farm animals behind it.
You are then given the option to switch to a different, unopened door, after which he opens another door with farm animals behind it.

This process continues until there are only two doors left, one being your current choice. (Every time the host opens a door, you are given another option to switch or stick with the one you currently have selected.)

If you play optimally, your chances of winning the car are of the form $\frac ab$ , where $a$ and $b$ are coprime positive integers.

Find $a+b$ .

Details and Assumptions:

You know ahead of time that this process will continue until you are down to only two doors.
Assume that you prefer the car over any of the farm animals!

Photo credit: http://iloboyou.com/

More probability questions:
Check out my other problems .

The answer is 19.

3 solutions

Geoff Pilling
May 6, 2016

When you pick the first door, you know you have a probability of $1/10$ that it has the car behind it. Also, switching from this first guess will only increase the probability that you have chosen the car. At the very end, you will be presented with two choices, the one you have chosen and another one. The one you have chosen will have probability $p$ of having the car behind it, and the other one will have probability $1-p$ . So, you really want to minimize the probability that the one you currently have has the car, and then switch right at the end. This means you should stick with your initial choice until the very end, (keeping it's probability of being the car as low as possible) so the probability of the one you have picked is $1/10$ . Then, when you switch right at the end, the probability will by $1-1/10 = 9/10$ , and $9+10=\boxed{19}$ .

Make sense? Feel free to comment if you can think of a clearer explanation! :)

Great variation of Monty Hall problem!

Mateo Matijasevick - 5 years, 1 month ago

Hey thanks, Mateo! :)

Geoff Pilling - 5 years, 1 month ago

How is 9 + 1 = 19? I put 10 and it didn't work?

Thanks, Stronak the Vulcan

Stronak The Vuclan - 5 years, 1 month ago

@Stronak The Vuclan – The probability is 9 over 10, thus the sum is 19. I think the 9+1 is a mistake and it was meant to say 9+10

Chiara Piombi - 5 years, 1 month ago

@Chiara Piombi – Ooops... You were right... Fixed the typo... Glad it was in the solution and not the problem!

Geoff Pilling - 5 years, 1 month ago

@Chiara Piombi – Thanks for answering my query

Stronak The Vuclan - 5 years, 1 month ago

Little typo at the end ;)

Mateo Matijasevick - 5 years, 1 month ago

I can not see how such a HIGH probability will develop. I agree to stick the selection till 3 doors are leaving and then switch to another one to remain there:

Say door "1" (sticked) together with doors "2" and "3" leave and last you switch to door "2". We define events: w(i)=win is behind door "i", m(k)=host opens door "k"

Then the probability for win behind "2" is:

P(w(2) | m(1)) * P(m(1)) + P(w(2) | m(3)) * P(m(3)) = 1/2 * 1/2 + 1/2 * 1/2 =1/2

The probability for win behind "1" (resp. "3") is:

P(w(1) | m(1)) * P(m(1)) + P(w(1) | m(3)) * P(m(3)) = 0 + 1/2 * 1/2 = 1/4

Therefore the optimal tactic is to choose a door randomly, stick it till 3 doors leave, change the door and hold this one finally. The probability then is 1/2 and the solution should be $\boxed{3}$ .

Andreas Wendler - 5 years, 1 month ago

This is how I understood it. The probability of you not having chosen the car at the start is 9/10, therefore if you stick with your choice until then, and then switch, you will have that same probability of not having the car at the end. When you have two doors left, you can switch and win as long as the door you have now doesn't have the car. It's probably more complex than this but this is how I would explain it... it's a variation of the Monty Hall problem, which was very confusing at first for me.

Chiara Piombi - 5 years, 1 month ago

Good answer, Chiara!

Geoff Pilling - 5 years, 1 month ago

@Geoff Pilling – Thanks! Also, I liked the clarification about choosing the car. XD I wouldn't have been so sad with getting a cuddly farm animal either

Chiara Piombi - 5 years, 1 month ago

I know it's surprising, but the probability does indeed go up to 9/10. Think of it this way. You pick a door, and it has 1/10 probability of being right. Then, effectively, if you are wrong, the host, "reveals" the correct answer by opening all the other doors except for that one. So, unless you picked correctly, you will get the right answer once you switch with a 9/10 probability.

Geoff Pilling - 5 years, 1 month ago

It seems to me very curiously! Can you explain this by formula?

Andreas Wendler - 5 years, 1 month ago

@Andreas Wendler – Ah, good idea! Lemme think about how to explain this via a formula, and then I'll post it... Stay tuned...

Geoff Pilling - 5 years, 1 month ago

This would mean the more doors the higher the probability to pick up the car which would be equal 1 for an infinite count of doors ... but then it is NOT a game anyway! => Something goes wrong here!!!

Andreas Wendler - 5 years, 1 month ago

@Andreas Wendler – Yeah, its definitely counter intuitive. This is an extension of the Monty Hall problem which has confused people for a long time!

So, let me try to explain a bit better with more of a formula... So you start off and pick one of ten. So your probability of being right so far is 1/10. Then, the host reveals eight doors without the cars, and then you finally make the switch.

So, lets look at all the possibilities:

1) You already had picked the car and you switch. Clearly you lose.

2) You didn't pick the car. So, now you know its behind the one that is not yet opened with certainty.

Since there were originally 1 way to do the first one, and 9 ways to do the 2nd one (i.e. 9 ways you could have chosen a door with no car behind it), your chances of winning the car are

P(win car) = $(1/10)*0 + (9/10)*1 = 9/10$ .

Make sense?

Lemme know if I should clarify anything...

And thanks for trying out the problem!

Geoff Pilling - 5 years, 1 month ago

@Geoff Pilling – Please see my comment before regarding the lost game character! I think you don't perform the right thing of probability calculation.

Andreas Wendler - 5 years, 1 month ago

@Andreas Wendler – Its similar to buying a lottery ticket. Its almost certainly not the winner. But if you assume there is one winner of the millions of entrants, if someone (the host in this case) goes around and throws out all the non winners, then you can trade with the only person left he hasn't invalidated, you are almost certain to win!

Geoff Pilling - 5 years, 1 month ago

@Geoff Pilling – Thank you - I understand. The thing I miss is the sense! It has absolutely nothing to do with probability calculation but only with an excluding procedure! Sad!!!

Andreas Wendler - 5 years, 1 month ago

@Andreas Wendler – Well, yes... Actually as the doors approaches infinity, the probability of winning does approach one. Since your first guess has almost no chance of being correct, when the host reveals all but one, the one he doesn't reveal is almost certainly the car! So, the more doors, the better chances you have of winning the car by switching right at the end.

Geoff Pilling - 5 years, 1 month ago

Picked a door $P(H \mid E) = \frac {1 \times \frac{1}{10}} {1 \times \frac{1}{10} + 1\times \frac{9}{10}} = \frac {\hspace{1mm} \frac{1}{10}\hspace{1mm} }{1} = \frac{1}{10}.$

The probability that the car is behind door 1 is completely unchanged by the evidence. However, since the car can only either be behind door 1 or behind the other doors not revealed, the probability it is behind the doors not revealed is $\frac{9}{10}$ .

selected a door $P(H \mid E) = \frac {1 \times \frac{1}{9}} {1 \times \frac{1}{9} + 1\times \frac{8}{9}} = \frac {\hspace{1mm} \frac{1}{9}\hspace{1mm} }{1} = \frac{1}{9}.$

The probability that the car is behind selected door is completely unchanged by the evidence. However, since the car can only either be behind selected door or behind the other doors not revealed, the probability it is behind the doors not revealed is $\frac{8}{9}$ .

$\boldsymbol{\cdot}\boldsymbol{\cdot}\boldsymbol{\cdot}\boldsymbol{\cdot}\boldsymbol{\cdot}$

selected final door $P(H \mid E) = \frac {1 \times \frac{1}{3}} {1 \times \frac{1}{3} + 1\times \frac{2}{3}} = \frac {\hspace{1mm} \frac{1}{3}\hspace{1mm} }{1} = \frac{1}{3}.$

The probability that the car is behind selected door is completely unchanged by the evidence. However, since the car can only either be behind selected door or behind the other door not revealed, the probability it is behind the door that is not revealed is $\frac{2}{3}$

$\therefore \frac{9}{10} \times \frac{8}{9} \times\cdots \times \frac{2}{3} = \frac{2}{10} = \frac{1}{5}$

Getting in the the form $\frac ab$ , where $a$ and $b$ are coprime positive integers. $\therefore 5 + 1 = 6$

Michael Fitzgerald - 3 years, 4 months ago

Benjamin Katz-Crowther
May 7, 2016

I see that 9/10 is a solution. (Very nice btw) but could you prove that there is no better? It may be trivial but I was hoping to find something here and I didn't. Thanks.

Good point... Lemme think about how to prove that you can't do better.

Geoff Pilling - 5 years, 1 month ago

You can do at most 9 door switches and at least 0 (that is, don't change the door you have chosen at the beginning). What we want to prove is that the highest probability of winning the car is achieved when you do a single switch when there's only 2 doors left.

Let's suppose that there exists another strategy of doors switches that let you achieve a higher probability than 9/10. As at the end there will be only 2 doors left and the probability of each door must sum 1, the probability of the door you haven't chosen is less than 1/10. But, this is impossible, because the initial probability of each door is exactly 1/10 and the switching process does not decrease the probability of each door, but it increases it or doesn't affect it. Hence, this is a contradiction, therefore, the highest probability you can achieve is 9/10.

@Geoff Pilling feel free to add this to your solution ;)

Mateo Matijasevick - 5 years, 1 month ago

Nice explanation, Mateo! :)

Geoff Pilling - 5 years, 1 month ago

Elisa Abarquez
Feb 17, 2018

The regular Monty Hall problem with n= 10 doors gives P=9/10( which CAN be applied here with down to 2 doors, 1st and last switch strategy (here)) You can't do any better because switching beforehand will increase the chance of the car being behind 1st door picked. (The host will NOT open door picked in optimal play, thus reducing the chances of winning the car. Exceptions: NONE.)

Door 4? No, no, door 7

The answer is 19.

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