[Misc] Probability problem

[Billy the Fish]

Kevin Spacey told me the answer to this in 21.

Great film if anyone hasn't seen it.

 

[pastafarian]

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

In essence then the last gamble becomes irrelevant and you NOW have a brand new gamble with two doors and you can choose either door.
No need to overthink this its C 50-50

 

[Herr Tubthumper]

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 Monty an Eastern European type positioned on Oxford Street with a couple of look outs? If so, there’s no prize, walk away.

 

[Goldstone1976]

Is Monty an Eastern European type positioned on Oxford Street with a couple of look outs? If so, there’s no prize, walk away.

With the current lockdown in London? Wouldn’t happen in England.

Swap, btw.

 

[Perfidious Albion]

Interesting..... it still appears 50/50 to me.. but what do I know?
I have seen long and complicated explanations of this on the net and am none the wiser.
I am sure Tony Bloom would understand the calculations of the odds!

 

[withdeanwombat]

In all probability I’d be watching a documentary on another channel.

 

[Uh_huh_him]

Interesting..... it still appears 50/50 to me.. but what do I know?
I have seen long and complicated explanations of this on the net and am none the wiser.
I am sure Tony Bloom would understand the calculations of the odds!

It took me a while to get my head around it.

The easiest way for me to think about it is that if you did this experiment 900 times your first pick hold the prize 300 times.
One of the other two would hold the prize 600 times.

Monty eliminating one of the losers provides the illusion of leaving you with a 50/50 choice.
In reality 600 times out of 900 he has to remove the other loser, leaving the prize unopened.

Therefore you are twice as likely to find the prize if you switch.

 

[Fran Sandaza 8 my hamster]

https://www.bbc.co.uk/programmes/p00wj3bd


Sent from my BLA-L09 using Tapatalk

 

[Baldseagull]

Or look at it this way.

When the empty door is gone - saying stik or swap is akin to saying pik any of the 2 doors you like - a 50/50 bet. The 1/3 bet is washed away by new info that now presents a 1/2 bet.

No, sticking is still a one in 3 chance, changing is a 2 in 3 chance, not 50/50.

 

[Fran Sandaza 8 my hamster]

Great Video demo of the problem.
Only spoiled by all the Arsenal!!

Sent from my BLA-L09 using Tapatalk

 

[Tom Bombadil]

It helps if you list the outcomes.

So at the first stage you have three possible outcomes, youve picked the correct door or you've picked wrong door 1 or wrong door 2
2nd stage you can either stick or swap.
If you picked the correct door originally then there are two wrong doors left, one of which is removed and whichever door is removed you are left with a wrong door if you swap
But if you picked one of the two wrong doors at the first stage Monty will discard the other wrong door and you're left with the prize door if you swap.

So the list of possible outcomes is

1st stage correct, Stick Win
1st stage correct, Swap Lose

1st stage wrong door 1, Stick Lose
1st stage wrong door 1, Swap Win

1st stage wrong door 2, Stick Lose
1st stage wrong door 2, Swap Win

You can see from this that If you stick there is a 1 in 3 chance of winning and if you swap there is a 2 in 3 chance of winning

 

[Baldseagull]

In essence then the last gamble becomes irrelevant and you NOW have a brand new gamble with two doors and you can choose either door.
No need to overthink this its C 50-50

No, it isn't.

 

[Bakero]

Switch.

 

[father_and_son]

Imagine no one opens any door, they just say, do you want to stick with what is behind your one door, or do you want to take what is behind the other two doors? You know without anyone opening anything that at least one of the other 2 doors is going to have nothing behind it, but you have 2 chances v 1 chance of getting the winning door if you switch.

Brilliant explanation and one I've not actually seen used before.
There are many ways to walk someone through why switching is the best outcome but this is the most succinct I've seen. Well done sir.

 

[Baldseagull]

Brilliant explanation and one I've not actually seen used before.
There are many ways to walk someone through why switching is the best outcome but this is the most succinct I've seen. Well done sir.

Why, thank you.

 

[Mtoto]

An extension of this, lifted from 538's (excellent) puzzle column. Making my brain hurt already but that's the idea.


https://fivethirtyeight.com/features/can-you-beat-the-goat-monty-hall-problem/

Riddler Classic

The Monty Hall problem is a classic case of conditional probability. In the original problem, there are three doors, two of which have goats behind them, while the third has a prize. You pick one of the doors, and then Monty (who knows in advance which door has the prize) will always open another door, revealing a goat behind it. It’s then up to you to choose whether to stay with your initial guess or to switch to the remaining door. Your best bet is to switch doors, in which case you will win the prize two-thirds of the time.

Now suppose Monty changes the rules. First, he will randomly pick a number of goats to put behind the doors: zero, one, two or three, each with a 25 percent chance. After the number of goats is chosen, they are assigned to the doors at random, and each door has at most one goat. Any doors that don’t have a goat behind them have an identical prize behind them.

At this point, you choose a door. If Monty is able to open another door, revealing a goat, he will do so. But if no other doors have goats behind them, he will tell you that is the case.

It just so happens that when you play, Monty is able to open another door, revealing a goat behind it. Should you stay with your original selection or switch? And what are your chances of winning the prize?