[Misc] Probability problem

Arthritic Toe · Apr 14, 2020

Following on from the success of the bodmas thread. Here's a probability problem for the maths geniuses. The solution is on the face of it, surprising, and when it first came out, baffled many a statistical brain even though its very simple. It's quite famous so if you know it, please don't ruin it - and please don't google it.

The problem is based on the American game show hosted by Monty Hall called "Let's make a deal" - presumably along the same lines as "Deal or no deal" that we have here. Here's the problem:

Monty Hall asks you to choose one of three doors. One of the doors hides a prize and the other two doors have no prize. You state out loud which door you pick, but you don’t open it right away.
Monty opens one of the other two doors, and there is no prize behind it.

At this moment, there are two closed doors, one of which you picked.
The prize is behind one of the closed doors, but you don’t know which one.
Monty asks you, “Do you want to switch doors?”

In order to maximise your chances of winning the prize, should you:
(a) Stick with your original choice
(b) Switch to the other closed door
(c) Do either - both outcomes are equally likely

Is it PotG? · Apr 14, 2020

There are 2 doors, one has a prize.

57.3% to 42.7% chance I would say.

Choose the bigger door.

Jesus Gul · Apr 14, 2020

c) Shirley

Brok · Apr 14, 2020

Why does maths, in such simple forms, baffle me?

sparkie · Apr 14, 2020

Does Monty know where the prize is or does he open one of the 2 other doors randomly ?

Berty23 · Apr 14, 2020

This is a great puzzle. I know it but I am a data analyst and used to like a gamble so have a good understanding of odds and implied probability. I will read the comments with interest but not write the answer here.

Arthritic Toe · Apr 14, 2020

sparkie said:
Does Monty know where the prize is or does he open one of the 2 other doors randomly ?

Yes he does know where the prize is.

sparkie · Apr 14, 2020

Arthritic Toe said:
Yes he does know where the prize is.

In that case, I reckon you switch doors.

Diablo · Apr 14, 2020

(c) Do either - both outcomes are equally likely.

There are two closed doors one of which has a prize so each door has 50% chance of being the one with prize.

There was a one in three chance before first door opened but because that had no prize the odds are now one in two.

pb21 · Apr 14, 2020

There is a 1/3 probability it is the box you chose and a 2/3 probability it is in one of the other two boxes.

:shurg:

Bulldog · Apr 14, 2020

When you chose the first door, you had a 1/3 chance of picking the prize.

You now have a 1/2 chance of picking the prize so switch from the 1/3 door to the 1/2 door.

Lincolnshire Seagull · Apr 14, 2020

Diablo said:
(c) Do either - both outcomes are equally likely.

There are two closed doors one of which has a prize so each door has 50% chance of being the one with prize.

There was a one in three chance before first door opened but because that had no prize the odds are now one in two.

This sounds like the sensible answer (to a non-mathematician).

But I bet we are wrong

Diablo · Apr 14, 2020

If Monty does know where prize is you should switch.
As If you had chosen the one with prize outcome would be as said 50/50.Monty would choose one of the other empties so one with prize one without.
But if you had chosen empty one Monty would have had to have chosen the other empty one so swapping would mean you had the prize.

So if you had first chosen empty door you def get prize and if you chose prize door outcome 50/50 !

Berty23 · Apr 14, 2020

pb21 said:
There is a 1/3 probability it is the box you chose and a 2/3 probability it is in one of the other two boxes.

:shurg:

But when one of the other boxes is removed (which will always be possible no matter which is picked)

studio150 · Apr 14, 2020

sparkie said:
In that case, I reckon you switch doors.

I will go with this also

Badger · Apr 14, 2020

Bulldog said:
When you chose the first door, you had a 1/3 chance of picking the prize.

You now have a 1/2 chance of picking the prize so switch from the 1/3 door to the 1/2 door.

That's correct. This is a classic probability question. I think my lecturer posed this to us in the first lecture of one of my computer science undergrad courses.

PILTDOWN MAN · Apr 14, 2020

The probability remains 1 in 3 as this was the first choice you made, so it remains with the first option.

father_and_son · Apr 14, 2020

Arthritic Toe said:
Following on from the success of the bodmas thread. Here's a probability problem for the maths geniuses. The solution is on the face of it, surprising, and when it first came out, baffled many a statistical brain even though its very simple. It's quite famous so if you know it, please don't ruin it - and please don't google it.

The problem is based on the American game show hosted by Monty Hall called "Let's make a deal" - presumably along the same lines as "Deal or no deal" that we have here. Here's the problem:

Monty Hall asks you to choose one of three doors. One of the doors hides a prize and the other two doors have no prize. You state out loud which door you pick, but you don’t open it right away.
Monty opens one of the other two doors, and there is no prize behind it.

At this moment, there are two closed doors, one of which you picked.
The prize is behind one of the closed doors, but you don’t know which one.
Monty asks you, “Do you want to switch doors?”

In order to maximise your chances of winning the prize, should you:
(a) Stick with your original choice
(b) Switch to the other closed door
(c) Do either - both outcomes are equally likely

Always swap. Monty can only ever remove an incorrect door and so you chances increase from 1in3 to 50/50.

Unless you are from Birmingham... In which case don't worry. The prize will be a speedboat and useless for you so drop your darts on the floor and leave.

Berty23 · Apr 14, 2020

PILTDOWN MAN said:
The probability remains 1 in 3 as this was the first choice you made, so it remains with the first option.

This is the classic error.

The way I explain this to people is to ask whether they would rather take a choice about a one in three chance or a one in two. Obviously people would rather a 50/50 shot but this puzzle is interesting because it kids you into thinking the odds are 50/50 for both boxes left (which is true) BUT when you picked it you had a 33% chance. So you switch.

The removal of one box and the relatively small change in implied odds makes it tricky to change your mind. No law imagine there were 100 boxes and you picked one. 98 incorrect boxes were removed. Would you swap? Of course you would because it is obviously now a 50/50 shot whereas on first guess you had 1% chance of being right. It is all about the information available when you make the decision.

Arthritic Toe · Apr 14, 2020

sparkie said:
In that case, I reckon you switch doors.

father_and_son said:
Always swap. Monty can only ever remove an incorrect door and so you chances increase from 1in3 to 50/50.

Unless you are from Birmingham... In which case don't worry. The prize will be a speedboat and useless for you so drop your darts on the floor and leave.

In actual fact it doesn't matter whether Monty knows or not. We are where we are, with one door having been revealed not to have the prize. Monty may have opened the door with the prize and the game would have been over, but he didn't, and here we are with two closed doors, one of which hides the prize

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[Misc] Probability problem

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