The Monty Hall Problem

[Chookes 965]

No cheating.

 

Imagine Deal or No Deal with only three sealed red boxes.

 

The three cash prizes, one randomly inserted into each box, are 50p, £1 and £10,000. You pick a box, let's say box two, and the dreaded telephone rings.

 

The Banker tempts you with an offer but this one is unusual. Box three is opened in front of you revealing the £1 prize, and he offers you the chance to change your mind and choose box one. Does switching improve your chances of winning the £10,000?

[tronic44 3,632]

How would you cheat?

 

Do you want to answer here or shall i put it in a spoiler, i'm even sure it's right to be honest but makes the most sense to me?

[Chookes 965]

As in don't Google it. I suppose these questions only really work if you've never heard it before. Answer it anyway Rich, I got it wrong.

[tronic44 3,632]

You start off with a chance of 1/3 of getting it right but then after the first door is opened, if you switch, then you double your chances? 2/3? Now i've written it down it doesn't seem right at all but it must be because it can't be a 50/50 chance? 

 

If it's a 50/50 chance then screw you, damn if it's that easy!

 

I should point out i seem to remember seeing something like this before (might not be this at all) i didn't get it to begin with but kinda do now.

 

*hovers over the post button for 5mins* 

[tronic44 3,632]

Also i enjoyed Maths at school lol

[Chookes 965]

Yeah that's right, I went for 50/50, adamant at the point that when the box is open its effectively a new choice so you're 50/50.

A famous Hungarian mathematician got this wrong so it's not easy!

I have another I don't know the answer too that I'll post tonight, hopefully someone can answer that one for me as I'd like to know!

[tronic44 3,632]

I so thought you were going to say it's 50/50 lol Even though i could see it was 2/3 my mind was like "but Rich, this is Euan, it's never straight sailing with this man"  In fact i kinda expected the answer to be apple or something.

[burnfitbillyboy 505]

I had the same mindset as Euan with this question... Damn you for making me think! My brain hurts.

[Jason 1,245]

Well the odds shift in your favor, so you start with a 1 in 3 chance to win.

That's goes up to a 50/50 chance, right?

So switching would mean you have a 2/3 chance. Does that count though because you still had a 50/50 shot you already picked the right case.

Confused. Lol

[Docwagon 856]

Intuition seems to say it wouldn't matter if you switched or not.  You know one of the boxes you didn't pick isn't the winner, so it should be 50/50.

 

Figuring that's too simple, I drew it out.  I drew 12 sets of boxes, with the prize in a different box each time (It appears in each position 4 times total) then always selected the middle box.  I'm assuming the game show host will know which box has the prize and won't open that one, or he'll ruin the game.  So, you eliminate one box as the one the host opens, and then switch every time.  I "won" 8 times out of 12 by switching.  The experiment seems to show you have an increased chance of winning by switching.

 

P = Prize  X=other box.

 

P  X  X

X  P  X

X  X  P

P  X  X

X  P  X

X  X  P

P  X  X

X  P  X

X  X  P

 

Your initial selection is always the middle box (to simulate a random guess, since you are making your own experiment obviously you know which is the prize).

 

So now P = Prize, S = Selection, E = Elimination  x= No prize  p = prize (Elimination will always be no prize)

 

#1:  P  Sx  E.  By switching, you win.

#2:  X  Sp  E.  By switching, you lose.

#3  E   Sx  P.  By switching, you win.

 

and this pattern repeats, resulting in a 2/3 win ratio if you always switch vs a 1/3 win ratio if you never switch.

 

I don't really get why, it still seems intuitively wrong, but that's what I came up with.

 

My actual results from always switching with the above 12:

 

Win

Loss

Win

Win

Loss

Win

Win

Loss

Win

Win

Loss

Win

 

**on edit**

I'm going to play around with it and see what the result are if the host doesn't know which box holds the prize.

 

** final edit**

 

Ok, that's why.  If the host doesn't know which box has the prize in it, the intuitive answer works.  1/3 of the time he'll eliminate the prize box, resulting in a loss for the player.  2/3 of the time he will reveal a non-prize box, and its 50/50 on if you win by switching or not.  Assume you don't switch and the host always reveals the first box.

 

HL = Lost because host reveals prize

W = Win by not switching

L = Lose by not switching

 

HL

W

L

HL

W

L

HL

W

L

HL

W

L

 

Total odds of winning 1/3

 

 

Now eliminate the times the host causes you to lose by taking the correct box out of play:

 

 

W

L

 

W

 

L

W

L

 

W

L

 

You have 8 chances to win if the host reveals a non-prize box.

 

Not switching will result in 4 wins (as shown above), so of course switching would have the same number of wins, just opposite times.

 

So switching when the host doesn't know the correct box has no affect on your odds.

[tronic44 3,632]

I can tell you all i did not put in that kinda work ^

 

Damn Doc! Interested in your findings

[Plumbers Crack 4,042]

That's one smart cop!

[OTC Mike 53]

Definitive answer = yes. changing does improve your odds. I actually came up with a simple program  to run the experiment 10,000 times for a college class

In it's simplist terms:

 

Staticistically speaking the first box you pick will be wrong 2 out 3 times.

 

This means that the correct box is one that you didn't choose 2 out of 3 times

 

Since the host will never reveal the winning box, when he reveals one box, the other un-opened box now has a 2 out of 3 chance of being right.

.

.

.

And Doc, it makes a huge difference whether or not the host knows which box to open or is randomly choosing one of the 2 boxes because you can't eliminate when the host reveals the winning box. You've lost the game right there.

[JsinOwl 644]

Yaaaaaaaaaaaaaaay, Mike is here! Figures a math problem brought you out of the woodwork. lol 

[Chookes 965]

Ok, the second BBC question:

 

 

Imagine that 1% of people have a certain disease.

 

A diagnostic test has been developed which performs as follows - if you have the disease, the test has a 99% chance of giving the result "positive", while if you do not have the disease, the test has 2% chance of (falsely) giving the result "positive".

 

A randomly chosen person takes the test. If they get the result "positive", what is the probability that they actually have the disease? The answer, 1/3, is perhaps surprisingly low.

 

Answers on a postcard please.

[OTC Mike 53]

Yaaaaaaaaaaaaaaay, Mike is here! Figures a math problem brought you out of the woodwork. lol 

HAHAH. I've been lurking for a while. Just needed the right topic to respond to.

 

Kinda wondered what happened to the old FG forum though.

[Docwagon 856]

Definitive answer = yes. changing does improve your odds. I actually came up with a simple program  to run the experiment 10,000 times for a college class

In it's simplist terms:

 

Staticistically speaking the first box you pick will be wrong 2 out 3 times.

 

This means that the correct box is one that you didn't choose 2 out of 3 times

 

Since the host will never reveal the winning box, when he reveals one box, the other un-opened box now has a 2 out of 3 chance of being right.

.

.

.

And Doc, it makes a huge difference whether or not the host knows which box to open or is randomly choosing one of the 2 boxes because you can't eliminate when the host reveals the winning box. You've lost the game right there.

 

That actually makes it make sense.  I figured out the ratios but didn't really understand WHY those were the ratios.  Thanks.