permalink for this thread :
bororeddaz Posted on 7/3 11:00
mathmatical question

This one will cause a fight.
I have 3 boxes, whilst your back is turned, I place a grand in one of the boxes. I know which one it goes in but you don't.
You have to pick a box, if you choose the right box the money is yours.
You pick a box but before confirming your final answer I open one of the other boxes to show you it is empty.
I now offer you the chance to stick with your box or switch to the other remaining box.
should you stick or switch? Why?
There is an actual answer, let the fighting commence

Lefty3668 Posted on 7/3 11:03
re: mathmatical question


Moses_Kiptanui Posted on 7/3 11:03
re: mathmatical question

What Lefty said.

zoec Posted on 7/3 11:03
re: mathmatical question

What Lefty and Moses said.

trodbitch Posted on 7/3 11:04
re: mathmatical question

Stick. Switching makes no difference.

bororeddaz Posted on 7/3 11:05
re: mathmatical question

here we go we have a contender

SmogOverOrmesby Posted on 7/3 11:06
re: mathmatical question

Stick, the boxes are open.

MA_07 Posted on 7/3 11:07
re: mathmatical question


UndercoverElephant Posted on 7/3 11:11
re: mathmatical question

Only a fool switches. Your odds are better if you stick.

mattfj Posted on 7/3 11:12
re: mathmatical question

Its a 50/50 split. You will have a 1 in 2 chance of choosing the right one.

Lefty3668 Posted on 7/3 11:12
re: mathmatical question

Are you having a laugh UE?

bororeddaz Posted on 7/3 11:12
re: mathmatical question

"Only a fool switches. Your odds are better if you stick."

hmmm most interesting

trodbitch Posted on 7/3 11:12
re: mathmatical question

All switching does is give you another roll of the dice in effect.

cumbria_boro Posted on 7/3 11:12
re: mathmatical question

Mathematically you should stick, but in practice it doesn't matter. Possibly.

UndercoverElephant Posted on 7/3 11:12
re: mathmatical question

Not true at all Matt. It's 2/3 to 1/3 in your favour if you stick.

mattfj Posted on 7/3 11:14
re: mathmatical question

Undercover Elephant-but one of the boxes is shown to be empty-surely therefore its 50/50????

cumbria_boro Posted on 7/3 11:14
re: mathmatical question

No hang on, mathematically you should change. Yes. That's right.

john_lillie Posted on 7/3 11:16
re: mathmatical question


Your box has a 1/3 chance, the other has a 1/2 chance of been a grand.

boksic Posted on 7/3 11:16
re: mathmatical question

Ah, the old Bill Murray Problem.

The answer is, counter intuitively, you always have a better chance when you switch.

See the link and let the debate run.

Congratulations reddaz, will this be your first ton?

Link: For Once I Agree WIth Wikipedia

bororeddaz Posted on 7/3 11:16
re: mathmatical question

the clue is in the thread title

UndercoverElephant Posted on 7/3 11:16
re: mathmatical question

OK, the Maths are as follows:

The chances of you picking the right one in the first place are 1 in 3, right? As are the chances of any of the other three having the money. The chances of you having picked the wrong one are 2 in 3.

If one of the other boxes is opened, and doesn't contain the money, then the chances of you picking the wrong box is 1 in three, thus the chances of you picking the right box in the first place are 2 in 3.

Only a fool switches.

andybarca Posted on 7/3 11:17
re: mathmatical question

no you should stick
it's 2/3 or 66% that boxes you haven't got the money in if one is empty then the probability that the last box is empty is 1/3 or 33% thus the odds that your box has the money in is now 66% and you are a 2-1 fav

ayrshire_red Posted on 7/3 11:18
re: mathmatical question

You never had odds of 2/3 as you would need to have been given 2 choices from the start. HE chose the first box and knew nothing was in it so effectively you have 1 in 2 chances to pick the correct box regardless of changing your mind. Its the same as a toss of a coin - only 2 possabilities and changing your mind before you toss the coin will have no change on your odds.

mattfj Posted on 7/3 11:19
re: mathmatical question

Undercover Elephant-I now agree with you!

cumbria_boro Posted on 7/3 11:20
re: mathmatical question

Ok. It's 1/3 when you first pick that you'll have the money so 2/3 that the other boxes have the money. If you know which of the other boxes don't have the money then you go with the 2/3 chance of having the money, i.e switch boxes.

trodbitch Posted on 7/3 11:21
re: mathmatical question

Jesus Christ, my head is bleeding.

Juventus2 Posted on 7/3 11:21
re: mathmatical question

I've been away from this board for 6 months. Today I come back and it was like I was never away. Good old Monty Hall.

mattfj Posted on 7/3 11:21
re: mathmatical question

Hang on, no I dont as when you are given the option to choose again, its a 1 in 2 chance of getting the cash, as opposed to a 1 in 3 at the beginning.

UndercoverElephant Posted on 7/3 11:22
re: mathmatical question

But Cumbria, you know that one of the boxes you didn't choose doesn't contain the money, so your odds get better.

boksic Posted on 7/3 11:22
re: mathmatical question

Changing always gives you the better chance to win, look at the decision tree on the wiki article.

Juventus2 Posted on 7/3 11:23
re: mathmatical question

It always amazes me how no matter how many times this comes up I know you should always switch because of the maths, but then the maths confuses me because even people who think you should stick have maths to back it up. But yeah, switch. The key is that the guy KNOWS which one is wrong, and you didn't when you picked.

trodbitch Posted on 7/3 11:23
re: mathmatical question

"Its the same as a toss of a coin"

It's not. You have a 50/50 chance of each outcome of a coin flip - heads or tails. Here you have a 1/3 chance of cash or no cash.

ayrshire_red Posted on 7/3 11:26
re: mathmatical question

The fact of the matter is you never had 3 boxes to choose from as one was opend before you did anything so you have only got 2 boxes left = 50% chance regardless of switching.

Forget about the 3 rd box because you did not get to choose it - it could not have contained the cash or the game would be pointless.

Lefty3668 Posted on 7/3 11:26
re: mathmatical question

When you first picked you had a 1 in 3 chance of being right.

Then the box is opened. The other box now has a 1 in 2 chance of being right.

trodbitch Posted on 7/3 11:27
re: mathmatical question

A nice summary from another site:

"The easiest route to the truth is to notice that resolving never to switch is equivalent to not having the option to switch, in which case, I'm sure you'll agree, the odds of winning remain one in three. Switching, therefore, has a two-thirds chance at the prize."

andybarca Posted on 7/3 11:27
re: mathmatical question

look it there is a 66% chance that both the boxes you didn't pick have no money in if one is empty then it is half those odds, 33% that the other box is empty so the implied odds of you box having the money in goes up to 66%

Azedarac Posted on 7/3 11:27
re: mathmatical question

We've been through this one before, and I never got my head around it. So I'll have to restore my bruised ego by pointing out that the word is spelt "mathematical".

zoec Posted on 7/3 11:34
re: mathmatical question

"It's not. You have a 50/50 chance of each outcome of a coin flip - heads or tails. Here you have a 1/3 chance of cash or no cash."

It doesn't matter how many boxes you started with. You're left with 2, so it's 50/50. The past doesn't matter.

UndercoverElephant Posted on 7/3 11:37
re: mathmatical question

That would be true, if you didn't know what had gone on and were given a free choice of two boxes. But you do know what happened, so you would be wrong to not use past knowledge to your advantage.

Say there had been 100 boxes to start, and Boreddaz had gone through 98 boxes, opening them all up (empty), and you were down to the last two. Would you still think it was 50/50?

--- Post edited by UndercoverElephant on 7/3 11:38 ---

Azedarac Posted on 7/3 11:39
re: mathmatical question

Isn't there a Flash web page that demonstrates this problem? I can't get past the logic that tells me there is no difference between sticking and switching, but I'm sure I've seen a demo that proves me wrong.

trodbitch Posted on 7/3 11:40
re: mathmatical question

The past doesn't *always* matter.

Souter Posted on 7/3 11:40
re: mathmatical question

There are 3 boxes to start with and the Dealer knows the location of the grand, when you select a box the box you select has a 1/3 chance of containing the grand. You therefore have a 1/3 chance of having the grand in your box and the dealer has a 2/3 chance of having the grand in either of his boxes. When the dealer opens one of his boxes and proves that it is empty his remaining box still has a 2/3 chance of holding the grand therefore you should swap!

zoec Posted on 7/3 11:42
re: mathmatical question

My head hurts. I think I'll go back to one of LL's threads where you don't have to think.

ayrshire_red Posted on 7/3 11:44
re: mathmatical question

As stated before - you can only take the past into consideration if you had some input to it, such as with the 3 box situation YOU only have 66% chance if YOU get 2 pics. The person that picks the 1st box in this situation KNOWS there is nothing there so this can not be taken in to consideration. There are only 2 boxes to choose from and thus giving 50% chance


smoggieboy Posted on 7/3 11:46
re: mathmatical question

surely it depends whether u have guessed right
if u guess box 1 and when u show me if theres money in that box that determines whether i confirm my answer.
if there is money obviousily stick, if not switch

--- Post edited by smoggieboy on 7/3 11:47 ---

ayrshire_red Posted on 7/3 11:48
re: mathmatical question

The box is chosen by the "dealer" not you - so he will allways pick an empity box - leaving 2 - one with the cash in.

SmogOverOrmesby Posted on 7/3 11:51
re: mathmatical question

Think of it this way, lets say the money is in box 3.
If you choose box 1, box 2 is shown as empty, if you switch you get the money.
If you choose box 2, box 1 is shown as empty, if you switch u get the money.
If you choose box 3, box 1 or 2 is opened and is empty, if you switch you dont get the money.

Therefore, if you switch you have a 2/3 chance of gettin the money. If you stick you still only have a 1/3 chance of getting it.

bororeddaz Posted on 7/3 11:52
re: mathmatical question

At this point can I add that my eastern european accomplices have been pickpocketing your wallets and purses for the last 50 mins whilst you stand around bickering

UndercoverElephant Posted on 7/3 11:59
re: mathmatical question

SmogOverOrmesby - If there's only two boxes left, your argument falls down. For example:

I picked box 1.
Box two is out of the equation, all the fools on BetFair who thought the market was overvalue have lost thousands.
If the money is in box three, and I switch I win.
If the money is in box one, and I switch, I loose.

50/50, assuming that I don't use past knowledge, which I would, obviously.

--- Post edited by UndercoverElephant on 7/3 12:00 ---

Juventus2 Posted on 7/3 12:01
re: mathmatical question

It's not 50/50 when the third box/door whatever is removed. Your original choice was placed when it was a 1/3 probability. So it is still a 1/3 probability that it was right, no matter what happens.

= a 2/3 probability you were wrong


YodaTheCoder Posted on 7/3 12:02
re: mathmatical question

It's not 50/50.

It's always better to switch.

Only a Mac user would stick.

--- Post edited by YodaTheCoder on 7/3 12:03 ---

ayrshire_red Posted on 7/3 12:06
re: mathmatical question

Its the way which you ask the question that can also confuse poeple..

Another old one is ......

3 old ladied decide to buy a TV set - they see an old one in shop for £30 - they each pay £10 and take it away. The sales assistant then sees they charged the wrong amount - it should have been £25 so promptly runs after the said old ladies, but then thinks about pocketing some of the £5 herself. She decides to give the old ladies £1 each back (and pocket the spare £2 for her self) - so they have now paid £9 each - now £9 x 3 =£27 + £2 for the corrup shop assistant = £29 so where did the other £1 go?

T4Tomo Posted on 7/3 12:17
re: mathmatical question

Everyone who thinks its 50 50 stick or switch hasn't read the question properly - if the dealer opens a box and if that contains the cash you've lost etc then it is 50 50 stick or switch,

but as he is always going to open an empty box your odds become 1/3 sticking 2/3 switching

Azedarac Posted on 7/3 12:26
re: mathmatical question

If we're dragging out old puzzles, there's Zeno's paradox:

Achilles is in a race against a tortoise. Achilles travels 100 times faster than the tortoise, so he gives it a 100 foot headstart. By the time he catches the tortoise up, it has moved on 1/100th of the distance he has covered (1 foot). By the time he's made up this one foot, it has moved on another 100th the distance. Each time he catches up it will have moved on 100th the distance he has just covered. No matter how small the gap, he will never catch up.

billeeee Posted on 7/3 12:57
re: mathmatical question


Some of the logic on previous posts is absolute pisch.

If you need to be convinced, have a go with a mate. Not with boxes and money, just get 3 squares of paper and put an X on one and use this as your 'money box'. You'll find that switching is much better, and you won't have to think too hard about the logic!!

YodaTheCoder Posted on 7/3 15:21
re: mathmatical question


Entertain me mathematically challenged monkeys!

TheYak87 Posted on 7/3 15:40
re: mathmatical question

Here's another problem along similar lines,
Prisoner's Dilemma:
Three prisoners A, B and C are under sentence of death and kept in solitary confinement. The governer decides to execute two of them but pardon the third. He randomly selects the lucky one. Prisoner A hears of this and tries to persuade the governer to tell him of his fate. The governer refuses to discuss A's personal case but does offer to tell A the name of one prisoner, other than A, who will be executed. "Tell me" says A, and the governer says that B will be executed.
A now feels happier, since he believes that his chance of being executed has fallen from 2/3 to 1/2. Is this justified? (answers on a postcard please)

lionrock31 Posted on 7/3 16:02
re: mathmatical question

dead easy Yak.
Prisoner A should 'switch' into prisoner C's clothes, wear his fake moustache and copy all his love/hate tattoos

boksic Posted on 7/3 16:20
re: mathmatical question

but only a Tuesday and if the Governor is wearing beige (and is an ex-member of the Coldstream Guards).

ThePrisoner Posted on 7/3 16:24
re: mathmatical question

No-one will convince me that switching makes the odds better. There are two independent events here. All that showing one empty box has is to increase the odds to 50:50. The previous 1/3 odds are irrelevant as these are independednt trials.

boroborob Posted on 7/3 16:31
re: mathmatical question

You're wrong, theprisoner.

Imagine if there were 100 boxes in a row instead of 3. You turn around and I place a grand in a box.

You then get to pick a box.

I then walk down this line of 100 boxes, opening every single one of them except for one box which I ignore and the box you picked.

Now, there is a 1/100 chance you got the right box. Alternatively, and far more likely, you picked the wrong box I've left the box with the money in. Wouldn't you switch at that point?

riverboat_captain Posted on 7/3 16:44
re: mathmatical question

There are three possible outcomes.
A. you chose empty box 1 and the dealer shows you empty box 2.
B. you chose empty box 2 and the dealer shows you empty box 1.
C. the you chose the money and the dealer shows you an empty box.

In A and B you gain by switching. In C you lose by switching.
Therefore you have a 2/3 chance of winning by switching.

Another explanation:
If you pick box A there is a 1/3 chance that the money is in it.

Therefore there is a 2/3 chance of the money being in B or C

If the dealer opens box C there is still a 2/3 chance that the money is in B or C because the money doesn't move. However, you know that box C is empty so the 2/3 chance all moves to box B.


stewboss Posted on 7/3 16:58
re: mathmatical question

"Another explanation:
If you pick box A there is a 1/3 chance that the money is in it.

Therefore there is a 2/3 chance of the money being in B or C

If the dealer opens box C there is still a 2/3 chance that the money is in B or C because the money doesn't move. However, you know that box C is empty so the 2/3 chance all moves to box B.


This goes against the basic laws of mathematical probabilities.

Probability is obtained by eliminating the impossibilities i.e. events which have ALREADY occured (in this case the opening of one box) and focusing on all possible remaining outcomes.

In this case, AFTER a box has been opened and shown to be empty, there are only two remaining boxes (one of which contains the money).

You cannot apply hindsight and pretend that 3 boxes are still in the equation and that the probibility is 2 in 3. There are only 2 possible outcomes.

50-50 - It doesn't matter what you do.

riverboat_captain Posted on 7/3 17:15
re: mathmatical question

You are wrong and I refer you to my first explanation.

boksic Posted on 7/3 17:19
re: mathmatical question

Stew - you are wrong, you are confusing this with coin flipping adn you are struggling becasue this is counter intuitive, ie your conscious mind is sying this can't be right but here is the proof. Look again:

A. you chose empty box 1 and the dealer shows you empty box 2. If you switch you win.
B. you chose empty box 2 and the dealer shows you empty box 1. If you switch you win.
C. the you chose the money and the dealer shows you an empty box. If you switch you lose.

In A and B by switching you win. In C you lose by switching.
Therefore switching you win 2 out of 3, by sticking you win 1 out of 3.

YodaTheCoder Posted on 7/3 17:24
re: mathmatical question

The people who are STILL claiming it's 50/50 are missing the vital piece of information.


At this point, you are choosing to keep your box (which was chosen with a 1 in 3 chance) OR choose the other box which legitimately has a 50/50 chance of containing the prize.

If the game started with 1,000,000 boxes, only one of which contained the prize. When you choose a box you have a 1 in 1,000,000 chance of winning.

If I then take away 999,998 boxes. The box you chose still only has a 1 in 1,000,000 chance of being a winner.

Yes there are 2 boxes, so if you give someone a choice between them there is a 50/50 chance of winning, but you are not given that option. Your choice is stick with the 1 in 1,000,000 chance, or take the other box which has a 50/50 chance.

The best option is to switch.

--- Post edited by YodaTheCoder on 7/3 17:24 ---

TheYak87 Posted on 7/3 17:26
re: mathmatical question

the trick is that you chose your box BEFORE you are shown one of the empty boxes. if you had to choose your box after being shown an empty box it would be 1 in 2. you should test this problem with a friend stew using a ball and 3 upturned cups. you are twice as likely to win if you switch. do it 20 or so times and i'm sure you'll see the difference.

clag01 Posted on 7/3 17:58
re: mathmatical question

Rubbish! You are making a choice between two boxes. The final choice is two boxes choose one. Stay or switch. 50/50...Its not rocket science is it.

trodbitch Posted on 7/3 17:59
re: mathmatical question

Thank god you don't design rockets then.

NSB19 Posted on 7/3 18:03
re: mathmatical question

Take both boxes off him for being an annoying **** ! Then you win !

--- Post edited by NSB19 on 7/3 18:13 ---

clag01 Posted on 7/3 18:08
re: mathmatical question

Haha! Come on. Its a straight choice between two boxes. What does it matter if you stay or switch,you still have a 50/50 chance....why would switching make any odds? `Oh Ive been shown one of the empty boxes so I must be wrong....switch it is then`????

Lefty3668 Posted on 7/3 18:11
re: mathmatical question

The chances are you got your first choice wrong. If you were offered the choice to switch from the one you chose to BOTH of the others, would you take that choice?

Because that's what you've been given.

TheYak87 Posted on 7/3 18:11
re: mathmatical question

erm, just because there are 2 options to choose between doesn't mean there is a 1 in 2 chance of being correct. There are 20 teams in the premiership if i asked you to pick the winner before the season kicked off would you have a 1 in 20 chance of being right? The two boxes would have to have equal probability of containing the money for it to be 1 in 2 and they DON'T.

YodaTheCoder Posted on 7/3 18:13
re: mathmatical question

The box you are offered to switch to has a 50/50 chance of containing the prize, the box you chose first still has it's initial probability whether that is 1/3 or 1/1,000,000.

clag01 Posted on 7/3 18:22
re: mathmatical question

Yeah but the goalposts have been moved. It was a 3/1 chance initially but you didnt know he was going to open an empty box. So now its a straight 2/1 the hell would switching thro the odds your way??

YodaTheCoder Posted on 7/3 18:24
re: mathmatical question

You did know he was going to open an empty box.

That information is in the original question.

(nearly there bororeddaz, 100 here we come)

trodbitch Posted on 7/3 18:26
re: mathmatical question

It's a play on words. If I have 1/3 chance of picking right and then one is taken away and switch I then have a 1/2 chance, right? OK, what if I choose again from the 2 that are left and pick the same one (ie stick). The chances are 1/2 again so I'm neither improving or worsening my odds.

The answer given on wiki says change but there are caveats that aren't answered.

clag01 Posted on 7/3 18:31
re: mathmatical question

Bugger!! Am still right tho!! One of the empty boxes doesnt come into the equation as you were told it was going to be shown. So therefore its a choice between two boxes....yes/no.....50/50.

Arctic_Mongoose Posted on 7/3 18:33
re: mathmatical question

This must be something in how the question is read out and I'm misunderstanding.
I think at time of confirming, you know money is in either two boxes, therefore 50/50.

If you have eliminated 999,998 boxes you are left with two.

Therefore your information has changed. Therefore the odds change.

You know that 999,998 boxes are empty.
You have two left, money could be in either. Equal chance now.

You are betting in running on a horse race, with three horses, horse A, horse B and horse C, you can't decide but are thinking about horse A even though all horses are level and looking good.....Suddenly either horse A or horse B falls over and dies. Lets say horse A - Are you trying to say that the odds of the horse B winning and the odds of horse C winning are now different?

It would be equal chance for either horse.

NSB19 Posted on 7/3 18:34
re: mathmatical question

Having mulled this over - I now realise a switch is in order. Original choice 1/3 chance of getting it and 2/3 chance of it being in the other 2 boxes. Opening the box does not alter the fact that its still 2/3 chance of being in the other two boxes ? Its obvious if you have intimate knowledge of Bayesian probability !!

Suppose you have three cards:

a black card that is black on both sides,
a white card that is white on both sides, and
a mixed card that is black on one side and white on the other.
You put all of the cards in a hat, pull one out at random, and place it on a table at random. The side facing up is black. What are the odds that the other side is also black?

--- Post edited by NSB19 on 7/3 18:38 ---

YodaTheCoder Posted on 7/3 18:40
re: mathmatical question

Here's a flash game that represents the problem.

Link: Monty Hall

YodaTheCoder Posted on 7/3 18:41
re: mathmatical question

And here's a screen grab after I'd played 100 games. 50 stick, 50 switch.

Link: 100 games

clag01 Posted on 7/3 18:43
re: mathmatical question

It does alter the know its a 1/2 you were told that an empty box was gonna be open and you could change your two boxes, choose one...heads/tails????

YodaTheCoder Posted on 7/3 18:44
re: mathmatical question

OK, someone, anyone, answer this.

There are 1 million boxes, one contains 1000 pounds. You pick up a box. What are the chances that the box you are holding in your hands contains the money?

--- Post edited by YodaTheCoder on 7/3 18:45 ---

clag01 Posted on 7/3 18:49
re: mathmatical question

1/1000,000...but if you are told that 999,998 empty boxes are gonna be open and you have the choice to switch...then its 1/2!

YodaTheCoder Posted on 7/3 18:51
re: mathmatical question

Slow down there sparky.

You have a box in your hands that has a 1 in 1,000,000 chance of containing the prize.

Do you agree?

trodbitch Posted on 7/3 18:58
re: mathmatical question

I agree, go on.

clag01 Posted on 7/3 18:58
re: mathmatical question

Yes...go on.

riverboat_captain Posted on 7/3 19:00
re: mathmatical question

Here's my 100 games.

edit: the link didn't work (how do you do screen grabs) but my scores were

Not switching - 32%
Switching - 70%

--- Post edited by riverboat_captain on 7/3 19:03 ---

Link: here

grocerygirl Posted on 7/3 19:01
re: mathmatical question

i'm presuming its a trick question and it says "if you choose the right box the money is yours" is that the box on the right hand side. if you have picked that box then don't swap. :(

clag01 Posted on 7/3 19:04
re: mathmatical question

Oh bugger!! Good answer.

clag01 Posted on 7/3 19:07
re: mathmatical question

But Im still right about the odds...if it wasnt a trick question...oh I best let it going mad..hearing voices on the FMTTM homepage and now this!!

YodaTheCoder Posted on 7/3 19:09
re: mathmatical question

OK, you keep a hold of that box.

I'll throw away 999,998 empty boxes, bear with me a moment...

Right. I've got 1 box, if you chose right the first time and made you 1,000,000/1 shot then I have the remaining empty box. however, if you chose wrong with your 1,000,000/1 show, then I have a box with the money in.

Do you want to keep your box, your 1,000,000/1 shot?

YodaTheCoder Posted on 7/3 19:11
re: mathmatical question

It's not a trick question.

ThePrisoner Posted on 7/3 19:20
re: mathmatical question

"OK, you keep a hold of that box.

I'll throw away 999,998 empty boxes, bear with me a moment...

Right. I've got 1 box, if you chose right the first time and made you 1,000,000/1 shot then I have the remaining empty box. however, if you chose wrong with your 1,000,000/1 show, then I have a box with the money in.

Do you want to keep your box, your 1,000,000/1 shot?"

And I come along and say "What's going on here then?" and you say "One of these boxes has money in and the other has none in. Have a go at choosing which one the money is in."
So I say "I'll flip a coin to choose because it's 50:50"

According to you Yoda one box has a million to one odds on it so the other must have 999999 to one it but you see that that is bollox because any previous trials are irrelevant to the current situation.

clag01 Posted on 7/3 19:23
re: mathmatical question

But its not a 1,000,000/1 shot coz you have been told that all will be left is 2 money....what does it matter how many boxes in the first place....would have replied quicker but was busy.

YodaTheCoder Posted on 7/3 19:25
re: mathmatical question

"According to you Yoda one box has a million to one odds on it so the other must have 999999 to one it but you see that that is bollox because any previous trials are irrelevant to the current situation."

You come along and see two people holding a box each. I agree, with no prior information if you are allowed one box YOU have a 50/50 chance of the prize.

But clag01 hasn't just come along, he was here when there was 1,000,000 boxes, and the one he is holding has a 1 in 1,000,000 chance of containing the prize.

He isn't choosing between two boxes, he is choosing to keep his 1 in a million chance, or switch.

That's what everyone who is saying "50/50" is failing to grasp.

If you are certain you are right, I will gladly play this game with you 100 times before the Man Utd game if you like. You'll pay for my ticket.

YodaTheCoder Posted on 7/3 19:27
re: mathmatical question

Take 5 minutes to play this game 100 times (link), stick 50 times, switch 50 times.

Then come back and answer honestly which had the most success.

Link: Monty again

boroborob Posted on 7/3 19:28
re: mathmatical question

For those who thinks it makes no difference....

I, and a number of other people, have provided proof mathematically, via links and using 'real world' scenarois that prove that switching is the thing to do.

If you still don't believe, or 'get it', I don't think you ever will. Just accept you don't understand and move on because there is absolutely no way you can prove otherwise.

YodaTheCoder Posted on 7/3 19:29
re: mathmatical question

Ton up.


ThePrisoner Posted on 7/3 19:37
re: mathmatical question

Can't arsed playing it 100 times though I do see that as the number of boxes increases the calculated probabilities converge to 1.1 % with only a few dozen boxes. So that's your theory down the pan.

I think I'll stick with the laws of probability rather than a parlour trick.

boroborob Posted on 7/3 19:42
re: mathmatical question

ThePrisoner, you're dense.

clag01 Posted on 7/3 19:45
re: mathmatical question

Bugger off Yoda :-) Ok will try game later...but it may be fixed...thought of that? Tho cant really comment without seeing it first...but even then what would I know? Anyway I am choosing between two boxes coz there is only two left....oh god gonna watch tv...laters :-)

riverboat_captain Posted on 7/3 19:45
re: mathmatical question

Prisoner, you are not sticking to the laws of probability, you are ignoring them.

There are only three possible outcomes. With two of them you win by switching, with the third, you lose. It's that simple.

Scrote Posted on 7/3 19:48
re: mathmatical question

bloody hell - the stick debate has gone downhill loads since this was last on (and we won a ferrari last time i think (bigger boxes probably))

the problem with switching is that one in three times you will switch and be wrong

so whilst you may improve your odds over a random number of attempts, in any one given situation switching does not guarantee a win and therefore it comes down to how lucky you feel on a 1/3 chance

you don't get another go

ThePrisoner Posted on 7/3 19:51
re: mathmatical question

"ThePrisoner, you're dense."

And you sir are not a gentleman, but I shall be sober in the morning.

Somebody please post the maths up. P(z) and all that. That would convince me.

riverboat_captain Posted on 7/3 19:53
re: mathmatical question

The only way you can lose when switching, is if you have guessed the correct box to start with, but the chances of doing that are only 1 in 3.

It really can't be easier than that.

boroborob Posted on 7/3 19:53
re: mathmatical question

There's been at least three links already, as well as all the explanations.

Apologies for the insulting comment, but given that it's been 100+ posts and you're still trying to argue this with 'laws of probability' I'm confident in saying it.

clag01 Posted on 7/3 20:00
re: mathmatical question

But you do get another can change your mind... in a 50/50 game...heads or tails...coin is sure?...You can say tails if you want...what does it matter if you change or not its 50/ there!

Scrote Posted on 7/3 20:03
re: mathmatical question

in the original problem the box was the prize on a gameshow

you only got one chance to open a box

therefore you only get one go - if you can do it over and over again then it only matters in terms of how much money you win rather than whether you win or lose completely

boroborob Posted on 7/3 20:12
re: mathmatical question



Arctic_Mongoose Posted on 7/3 20:21
re: mathmatical question

Riverboat Captain...

"The only way you can lose when switching, is if you have guessed the correct box to start with, but the chances of doing that are only 1 in 3."

Yes I agree with this. But what happens if I turn up now, I see you have two closed boxes and that you are deciding whether or not to switch.....

There are now two closed boxes rather than three, one contains the money and one doesn't.

In fact, before I arrive, you have thrown away and hidden the open box as it is not involved - I only see two boxes. One with money in, one without. I will have a 50/50 chance of choosing the correct box. At the time of two boxes left on the table if you pick the same box as me you will have a 50/50 chance.

--- Post edited by Arctic_Mongoose on 7/3 20:22 ---

--- Post edited by Arctic_Mongoose on 7/3 20:25 ---

john_b Posted on 7/3 20:24
re: mathmatical question

Can not be faffed to read all 112 odd posts, but it's correct to take the offer of switching boxes. I read Derren Browns entertaining book and he explains it in simpleton terms which almost made sense to me.

jimmyace Posted on 7/3 20:24
re: mathmatical question

you copied that from a curious incident and the dog in the night

Arctic_Mongoose Posted on 7/3 20:53
re: mathmatical question

I get it. I was thinking of it separate events, but you effect what box is opened.

The link explains with pictures.

Link: Monty Hall

clag01 Posted on 7/3 20:58
re: mathmatical question

Tail tucked firmly between my legs...

andyd763 Posted on 7/3 21:01
re: mathmatical question

i've read the monty hall thing and i still don't get it.

Arctic_Mongoose Posted on 7/3 21:12
re: mathmatical question

Try think of it like this....

It's unlikely you will be right with your initial choice - it's 1/3 odds. With 2/3 odds that it's one of the other boxes.

Your choice to switch is a choice to pick one of the other boxes. It's 2/3 that it's one of the other boxes. But because the boxman opens an empty one, he forces your switch to be the closed one.

So the boxman HAS to remove one of the other two boxes and has to be an empty one, so perhaps think of the other two boxes as one big box, i.e one choice - a switch. It's still 2/3.

Hopefully not confusing but the 50/50 thing would only apply (I think), if boxman randomly chooses a box, that's it different to yours and that box was empty.

The boxman HAS to remove an empty box.

clag01 Posted on 7/3 21:13
re: mathmatical question

Right have been converted. You have a 1/3 chance of picking the right one. If you switch you lose. But you have a 2/3 chance of picking the wrong one so when you switch you win. God And I thought I was quite clever..dear me.

clag01 Posted on 7/3 21:20
re: mathmatical question

Bit like them 3D picture things that were on the go a while back. You even get them or you dont..but then you do.....sometimes...better go before I make a bigger fool out of myself...goodnight all.

squiffypants Posted on 7/3 21:31
re: mathmatical question

I'm not reading the 100 odd posts here I haven't got time, but it is correct that you are more likely to be correct if you switch choices after an empty box is taken away.

It stands to reason doesn't it? When you have three boxes your chance of being correct is 1 in 3. Therefore statistically speaking you are more likely to have picked the wrong box initially.

It makes perfect sense for you pick the other box. What we are effectively doing is saying "Which box is more likely to be the right one? Is it this one here (your first choice) or one of those two (second choice)?"

If you were asked are you more likely to get it right if you get to choose two boxes instead of one then I think we would all say that having two attempts doubles our chance of being right from 1 in 3 to 2 in 3. This is exactly what you do on the second choice if you switch.

If it makes anyone feel better some of the worlds greatest mathmeticians struggle to get their heads round this, as on first look it just seems so obvious that it shouldn't make any difference. But it does

clag01 Posted on 7/3 21:34
re: mathmatical question

Smart arse squiffy :-)

minkey_brain Posted on 7/3 21:50
re: mathmatical question

this is simple and savant and etc are wrong

its a simple 33 percent probability in the beggining, but as soon as you take away a box its 50 percent, eitherway so it can make no difference no matter how many silly long division style formulas you choose, for the simple reason that the equation is always 50/50 anyway because the first box is always chosen for you and is ultimately empty, so the equation is never actually a 1 in 3 chance, its always a one in 2

Durham_Red Posted on 7/3 23:36
re: mathmatical question

What is the quiz show host, opens one of the two remaining boxes at random and it is empty. What are the odds now?

To think about the orignal problem use Yodas 1 million box example.

squiffypants Posted on 7/3 23:36
re: mathmatical question

Minkey - The first box isn't chosen for you, you choose it youself. The way it works is that you get to choose a box. You are then asked if you want to keep that box or you can (effectively) have the other two boxes instead.

What throws people is that a box is removed after the original choice, but as we know the box is empty that removal is irrelevant.

The choice you are making is simple it is either your original choice (1 in 3 chance) or an effective switch to the two remaining boxes - one of which has been removed (2 in 3 chance).

If there were ten boxes and you picked one, this would leave nine boxes. I would say the correct box is (nine times) more likely to be one of the remaining nine.

When eight of the nine are removed the guy removing them knows they are empty. This does not involve chance. Your argument would be that at this point it becomes 50/50 as their are two boxes left. If this is true then you will be right 50% of the time if you don't change your original choice which means your original pick will need to be correct 50% of the time even though it was a one in ten chance. This I'm sure you will agree would not be the case.

Scrote Posted on 8/3 1:36
re: mathmatical question

you can't keep bringing 10, 100 or 1,000,000 boxes into it to explain the maths whilst maintaining the premise that the solution applies in the real world

either its a pure maths exercise - in which case always swap due to probability giving better results over time

or it is a real world game in which you get one chance and one chance only with 3 - and only 3 - boxes

in the real world it doesn't matter that the chances are higher 'cos 30% of the time you'll swap and get it wrong - maths goes out the window and it comes down to gut feeling which isn't particularly scientific but has far more application in a real world scenario

NeddySeagoon Posted on 8/3 5:29
re: mathmatical question

Forget about the guy opening empty boxes.
First choice - pick a box.
Second choice - keep the box or discard it and take all the other boxes?
Simple innnit?

Iowapete Posted on 8/3 11:22
re: mathmatical question

This is a psychological, sociological tendencies problem.

First the notion of the probability: as far as a statistical matter--in a proper controlled environment, where one box is filled with the 1000 by one person, then after all are closed, another randomly resorts the 3 boxes--your odds are 1 in 3 to pick the correct one. That is a statistical FACT.

If 2 boxes are at play, and you are to pick one, your chances are 1 in 2 of being correct. Statistics apply to the total SUM of possibilities in play. You will have a 50-50 shot at now being correct. That is a statistical fact also.

If you flip a coin, 50-50 chance, it doesn't mean the second flip will be the opposite of the first because there is theoretically an infinite number of times that coin may be flipped--meaning it doesn't have a terminal value such as the case spelled out here: ONE of TWO.

There might be a tendency for people to pick the so-called middle box, an anti-statistical anomaly in itself; secondly, they may in fact be "hearing" pick the "right" box--a controlled environment, subconsciously suggesting the choice for them. But does that mean the experimenter's right, or your right. Played out 100 times without influencing the person, one will correctly pick the money box roughly 33 times give or take a few. Do the same with 2 boxes as the variables and you will be correct roughly 50 times out of a 100.

There is no statistical basis for switching. By offering the person, after the fact, that he or she may change their mind, places more doubt about the choice. If you tell the experimental subject BEFORE hand that one incorrect choice will be opened, at which time they may change their mind--they will more likely not waffle, because why should they? Tell them: you will have a 50-50 chance of picking correctly. Three boxes were never at issue.

Odds posted by gambling sites and over-unders, and the like, are calculated to make you lose because of studied tendencies, otherwise there would a lot more wealthy gamblers. This is a fixed statistical probabily--and the variable in this instance is the experimenter playing this trick out. Stand pat with your choice.

Lefty3668 Posted on 8/3 11:25
re: mathmatical question

Congrats on the ton, reddaz.

I didn't think you'd get beyond about 20.

YodaTheCoder Posted on 8/3 11:29
re: mathmatical question

"There is no statistical basis for switching"


Link: see if this helps

speckyget Posted on 8/3 11:36
re: mathmatical question

*sniff* Hello again old friend....*splutter*

It's like threadular Old English Spangles.

bororeddaz Posted on 8/3 11:41
re: mathmatical question

There's some mileage left in this one Lefty.
You should see the emails and hear the phonecalls I'm getting from the gay lurker.
Cue. bororedbaz

Lefty3668 Posted on 8/3 11:53
re: mathmatical question


Iowapete Posted on 8/3 12:03
re: mathmatical question

Yoda, all it did is reinforce my point (The Monty Game) I got 35% correct when picking from 3 (apx 1 of three right); and when you have the choice of ONE OF TWOOOOOO! you'll pick correctly about half the time (I got 55%)--these are two separate statistical problems--in the second instance you're merely re-randomizing your selection; you're still picking to see if you'll be right, and that will be roughly 50-50.

YodaTheCoder Posted on 8/3 12:10
re: mathmatical question

"these are two separate statistical problems"


It seems you can't prove a point so someone who doesn't understand either the problem or the maths, so...


trodbitch Posted on 8/3 12:24
re: mathmatical question

The mistake people are making is treating them as two separate problems ie "if I walk in at this point and pick then this it's 50-50". But you don't that's not the problem. The problem described is when you do one thing and THEN do the other, you increase your statistical odds which is a mathematical fact and cannot be disputed. The fact that when you implement the problem by actually playing the game and it works should convince people.

So if when you show someone the mathematical proof and then show how that proof works in the real world and they still don't believe you, stop explaining.

Iowapete Posted on 8/3 12:25
re: mathmatical question

Sigh back at you, Yoda. You're guessing in the first instance to be correct 1 of 3 times, right? In the second instance there are only one of two select-----you re-randomize to assure yourself of being... right half the time....OH I get it. You have to re-randomize to in fact GIVE yourself a 50-50 chance, otherwise you're stuck with 1 out of 3! CHEERS, and my hats off to you.

billeeee Posted on 8/3 12:46
re: mathmatical question

I posted my 'switch' vote yesterday and have just now revisited this debate.

I haven't laughed so much for ages.
I've forwarded the link to this post to a few people who have no interest in the Boro but like a good giggle.

Particularly like the banter between boroborob and Prisoner...."You're thick"!!.... that one.

I was gonna post saying "stick to football talk" but this is just classic. keep it up, especially you're special!

billeeee Posted on 8/3 13:30
re: mathmatical question

please don't stop....come on....more debate!!

billeeee Posted on 8/3 13:43
re: mathmatical question

BTW I think it's a "Mathematical question"

Iowapete Posted on 8/3 13:51
re: mathmatical question

Mathewhat? Well thats jus disturbingly probl'matical. Hung jury, lets start this ova.

billeeee Posted on 8/3 14:20
re: mathmatical question

Pete, If you are in Iowa, then this is a simple "Math" problem.

Iowapete Posted on 8/3 14:54
re: mathmatical question

Just being a goofball, billeee; understanding the rationale to the separate probabilities is one thing; but decontructing the theoretics to realizing you must re-establish a random probability to GET the benefit of a higher probability, as opposed to being left with 1 out of 3... I took a look at the problem last night, mulled it over, and thot geronimo(!) this a.m... I'd done this without having looked at the Monty Hall puzzle... and the initial conclusion I drew was false... PS: Hey, Calculus was not my strong suit--and I'm embarrassed to admit I aced statistics--so what excuse am I going to hang on? Re: Iowans--they're not that dense; their school aptitude scores are some of the highest in the US... but having said that, am just passing through on a self-imposed writing project. Cheers!

billeeee Posted on 8/3 15:36
re: mathmatical question

Is that a 'stick' or a 'switch' then?

wayvvee_dayvee Posted on 8/3 15:52
re: mathmatical question

statistics are crap........and i have the statistics to prove it!!

trodbitch Posted on 8/3 16:14
re: mathmatical question

Is everyone agreed that it's switch and it's switch every time without a shadow of any doubt whatsoever? Because having thought stick, I'm now 100% certain it is switch - FACT.

Iowapete Posted on 8/3 16:14
re: mathmatical question

Switch, definitely, Bill, but there's still this tiny patch of brain not yet in on it (tho I do undertand it now, empirically): like the logic of being told to *steer* yer car INTO a slide on ice or snow...

steve0 Posted on 8/3 16:33
re: mathmatical question

Stuff all that!
What about the £1 the corrupt shopkeeper's nicked off the old dears?
Where has that gone?
Bring back the birch I say! can't trust no-one

Lefty3668 Posted on 8/3 17:06
re: mathmatical question

'Is everyone agreed that it's switch and it's switch every time without a shadow of any doubt whatsoever? Because having thought stick, I'm now 100% certain it is switch - FACT.'

No, I believe Brushy68 is still tenaciously clinging to his right to be a dim-wit. He just isnít doing it on this board as that would require a bit of courage I suppose.

harry_x Posted on 8/3 17:53
re: mathmatical question

nah... undercoverelephant nailed it mid-day yesterday.

Once you remove a box then it becomes a non-runner and all previous bets are off for that box.

All thats left is a two box race and it's 50/50

...think betfair carnage ...

Scrote Posted on 8/3 18:01
re: mathmatical question

i can't actually remember who i argued with on this one the first time it was on (and its been on a few times since) but i do remember asking for someone to put their money where their mouth was so to speak

you put down 20 quid and i'll put down a tenner - get three receptacles and stuff the cash in one of them without me looking - then perform the test - i'm willing to lose my tenner by not switching if you are willing to gamble with 20 quid to prove your "always switch" theory - once and once only

surprisingly enough no-one went for it despite the protestations that always switching was the thing to do - go figure...

i fully understand the maths and i agree that over x number of attempts switching is the best option - but again i will point out that in the original game you ONLY GET ONE ATTEMPT

something as subtle as a wry knowing smile from monty when you pick your first box and he whips away the second could be enough to convince me to switch but then again he might be double bluffing - i have to put up with that once a week in the pub quiz and its far more relevant than whether the statistical analysis of 100 attempts says stick or switch

on a one off chance at 1/3 or 2/3, gut feeling and perceptions of how lucky a guesser you are come into play - statistics don't matter in the real world for single instance events

Iowapete Posted on 8/3 18:27
re: mathmatical question

Good point Scrote. I was trying (though agree with switching) to explain that intangibles come into play, in fact, tho I mentioned getting 35% correct in the pick one of three (in the Monty game), it was actually 38%--that seemed a bit braggadocious--and for most of the 50 tries the percentage was about 42-45%--then began dropping precipitously for the final 10-12.... Insofar as that 10 v. 20 bet, even at 1/3 odds, after three tries you'd have won 20 and lost 20 (when you think about it--and quite possibly WON 40 versus lost 10... a sucker bet). :-)

minkey_brain Posted on 8/3 18:34
re: mathmatical question

yeh yeh, everybody is forgetting , what happens if the first box you choose is the right one, the main problem with this equation is that it involves 2 seperate spaces in time , the first one is the original question, the second part of the question totally removes any furtherpart of the original question void as it is part of a totally different mathematical probability,thus rendering the whole incalulatable, no matter how many silly long winded formulas you give, thats the problem with some of these genious types, to busy making formulae, bet most of em cant even boil an egg

YodaTheCoder Posted on 8/3 18:40
re: mathmatical question

I'll assume minkey_brain is a typo and you were going for either "monkey" or "brian".

"the second part of the question totally removes any furtherpart of the original question void"

And your English is as poor as your maths.

It's not a 50/50 shot.
You win 2/3 of the time if you switch and only 1/3 of the time if you stick.

--- Post edited by YodaTheCoder on 8/3 18:43 ---

Scrote Posted on 8/3 18:46
re: mathmatical question

iowapete - as i said - it was a once only thing - as it would be for an individual on the show

minkey - the fact that it is two seperate choices doesn't matter as far as the maths goes because you directly influence the available choices in the second part with your choice in the first

if i was given a hundred attempts and was allowed to bet stick or switch on each one then the best strategy would be to go switch every time - my point is that with just one attempt the actual decision made has far more to do with factors that the maths doesn't allow for - such as subtle body language clues and the like...

Iowapete Posted on 8/3 19:09
re: mathmatical question

On Let's Make a Deal: Monty opens one curtain (or what's behind the box)--yay!, a year's supply of frozen dinners. Behind the other two: 1. a gaggle of chickens with a bag of feed and 2. a little box with the keys to a sports car... Yoda, you're saying I have a 66% chance of getting the car IF I change my mind? (2/3?). This is not addition. I can live with the 50-50 logic, but what (this hasn't been explained) would compel me to decide I hadn't made the correct choice to begin with?

boroborob Posted on 8/3 19:19
re: mathmatical question

scrote - i wouldn't put down £20 to do it once.

However, if you're willing to do that 100 or even 10 times then i'd be more than happy. That doesn't mean I'm buying into your guff about body language - that has nothing to do with the question posed - it just means I lack balls and that £20 could buy me a lot of beer.

i can't believe there are still morons who don't get it :(

--- Post edited by boroborob on 8/3 19:25 ---

YodaTheCoder Posted on 8/3 19:20
re: mathmatical question


Link: watch this

Iowapete Posted on 8/3 19:51
re: mathmatical question

Now we're arguing apples and pears, Yoda. Already said much earlier that I aggreed with switching, appreciated getting set straight on that, understood the theoryosophy...etc, etc... This is beggining to deteriorate into something akin to Wittgensteinian existentialism. I'm out. Thanks all.

NeddySeagoon Posted on 9/3 7:09
re: mathmatical question

In the long term Scrotes bet works out exactly 50/50 so is not worth taking. As a one off you are betting 20 to win 10, so for me you can keep that one as well!

zaphod Posted on 9/3 7:21
re: mathmatical question

It hinges on the fact that the opener of the box knows what's in the boxes and chooses to open an empty one. So initially you have a 1 in 3 chance and a 2 in 3 chance that it's in the other 2 boxes. When the opener chooses to open one of the boxes because he knows what's in it, it means he doesn't change the initial odds (2 in 3). He has deliberately chosen not to open the other box.

YodaTheCoder Posted on 11/6 22:18
re: mathmatical question


Northsea_Kitten Posted on 11/6 23:04
re: mathmatical question

you should stick
if he knows which one has the money in and you have it then he also knows that the other two are empty so it doesnt matter which one he chooses to open before giving you the option to change.

Mattyk50 Posted on 11/6 23:06
re: mathmatical question

switch FFS.

Saldamuz Posted on 11/6 23:09
re: mathmatical question

Surely it doesnt matter.

If there were 10,000 boxes, regardless of which you choose, you are left with 2, and you can pick one of them = 50/50 for me.

trodbitch Posted on 11/6 23:15
re: mathmatical question

Having written a program to do the picking and switching, after 5 million goes (!) it was 66.66% for the SWITCH team and 33.33% for the STICK team.

If you pick a box that is empty (2 out of 3 chance) then when he removes the other empty one, and you swap you ALWAY get the box with the cash.

If you always stick, it is a 1 in 3 chance of picking the right one. Those odds are right!

TheYak87 Posted on 12/6 0:08
re: mathmatical question

you're making a choice between the one box you chose or the two remaining (one of which is opened to confuse the mathematically inept but it makes absolutely NO difference).
Your one box has 1/3 chance the other two have 2/3. Mathematically proven fact.

Scrote Posted on 12/6 2:15
re: mathmatical question

forgive me for diving in again but the point i was making all those months ago was that switching just increases your chance of winning - it doesn't guarantee it

and it only works as a mathematical proof over a series of identical tests - you can't prove anything from a single result

with ONE GO ONLY there are far more factors involved in real life than just basic mathematical skills

change it slightly to a more workable example - you roll a six sided dice (sides defined to shut up the ADnDers )

you get a four - what are the chances of getting a four next roll?? 1 in 6 i hear you all say

ok - you roll it and get a four again

in fact you roll it ten times and get a four every time

mathematically - what is the chance of getting a four on roll eleven?? 1 in 6 again i hear you cry

in real life what are the odds of a) getting a four again, b) someone betting money on not getting a four again, c) someone punching the owner of the weighted dice

how many rolls would you maths gurus have allowed before you did c in real life?

or at least asked to check the balance of the dice...

and now you are going to say that there are added parameters (i.e. the loading of the dice) which aren't in the original proposition - and to that i say "monty hall's body language" or "monty hall's nervous tic" or "monty hall's rather obvious semi on" etc.

real life is not mathematically sterile

having ONE GO ONLY at the 3 boxes renders the maths meaningless other than as a guide to what is 'better'

samizdat Posted on 12/6 2:36
re: mathmatical question

There is a pile of money to be made from those who believe that the odds are 50/50 after the first box is opened and shown to be empty. Of course I would prefer to raise my chances to a near certainty by betting against those who would stick with their choice if we were to start with a lot more boxes and open all but 2, showing them to be empty, but that is perhaps just being greedy. To those who are still unconvinced by Yoda the Coder's explanation I suggest you stay out of the casinos.

TheYak87 Posted on 12/6 6:02
re: mathmatical question

and my point is proved. Scrote believes that maths can't prove anything if the problem is a one off choice, his mathematical ineptitude fools him.
Your twice as likely to win if you switch, that can be mathematically proven. end of.

YodaTheCoder Posted on 12/6 17:58
re: mathmatical question

No, stick!

Mattyk50 Posted on 12/6 18:03
re: mathmatical question


Scrote Posted on 12/6 19:16
re: mathmatical question

yak - are you suggesting that maths can be used to predict a statistical certainty from one event?

do you fancy a game of guess the random dice number any time soon?

"Your [sic] twice as likely to win if you switch, that can be mathematically proven. end of."

my mathematical ineptitude is no match for your lack of basic english skills

but - your quote proves my point - you are twice as likely to win but not certain to - therefore switching only increases your chances in a mathematically sterile environment (e.g. a computer program running with a 100% random number generator (what was that at the back...))

if monty hall asks you if you want to switch and the producer stood in the background who is your dad's best mate is shaking his head what should you do?

--- Post edited by Scrote on 12/6 19:17 ---

YodaTheCoder Posted on 12/6 22:57
re: mathmatical question


YodaTheCoder Posted on 13/6 13:37
re: mathmatical question

No, switch! Definitely (not definately or definetly).

bororeddaz Posted on 13/6 13:39
re: mathmatical question

I fooking cringe everytime I see my spelling error.

YodaTheCoder Posted on 13/6 13:42
re: mathmatical question

There's more to cringe at in this thread than your typo...

"Stick. Switching makes no difference."

"Only a fool switches. Your odds are better if you stick."

"50-50 - It doesn't matter what you do."

TheYak87 Posted on 13/6 13:42
re: mathmatical question

So you admit that you're (thanks :P) twice as likely to win if you switch but you don't want to? We should set this up 'cos i'd make a fortune. Oh and you can invite your mate's dad, a psychic and a life coach along if you want. I'm sure it'll help. (frantically checks for punctuation and grammar)

Scrote Posted on 13/6 19:30
re: mathmatical question

Yak - i'm not saying i wouldn't (or other people shouldn't) switch

but the problem with this type of question is that it relies purely on maths to give an answer - my one and only point is that in real life the maths only matters if theres nothing else interfering with the individuals choice

i'm not disputing the fact that switching is the best option

Snickerdoodle Posted on 27/6 17:25
re: mathmatical question

No, switch! Definitely.

boro1607 Posted on 27/6 18:02
re: mathmatical question

You would be able to tell which box the money is in because it would have been heavier. Therefore if you pick a light box up to start with then and he opens an empty one, then I would swap because he would have the heavier.

Snickerdoodle Posted on 9/8 14:46
re: mathmatical question

Erm, stick?

showman21 Posted on 9/8 14:51
re: mathmatical question

You have to pick a box, if you choose the right box the money is yours.

I`d pick the box on the right!!

Snickerdoodle Posted on 9/8 14:53
re: mathmatical question


bororeddaz Posted on 9/8 14:54
re: mathmatical question

Ah, I guess block17 is up and running?

Snickerdoodle Posted on 9/8 14:55
re: mathmatical question

It is.


But for how long?...


showman21 Posted on 9/8 14:56
re: mathmatical question

why that daz?

Am i right

McCann_87 Posted on 9/8 14:56
re: mathmatical question

is the answer pokey bum vvank?