settle a bet

MrPillowTalk

Ok its been here before but there is some argument on the symantics.

you are on a game show where you have to choose between boxes 1, 2 and 3 in one of the boxes is a million quid in the other two is nothing. choose a box.

having chosen a box, the show host removes one of the other two boxes as it does not contain the prize, now he offers you the option of either choosing to stay with the box you originally chose or change to the remaining box.

What do you do and why?

SilverFox261

You take the swap.

When you were offered the original choice you had a 33.3% chance of getting it right, and 66.6% of getting it wrong. The game show host must remove one empty box. If you picked right the first time aka: picked the money, then both boxes that are left are empty and therefore it does not matter which box he picks. This will happen 33.3% of the time.

But, if you picked wrong the first time, which will happen 66.6% of the time, then the game show host will only have one possible box to throw away, leaving the box containing the money.

Therefore you're chance of winning goes from 33.3% to 66.6% if you take the swap.

Theres a good explaination in this vid:

http://www.youtube.com/watch?v=mhlc7peGlGg

hotspur

And it makes it easier to see if instead of 3 boxes you imagine it's 1 million boxes. Box you pick is 1 in a million shot, box remaining...not so much.

bohsman

Imagine there are 10 boxes, host removes 8 and offers the swap, easy decision, same reasons but odds are obv even more in your favour.

pistolpeter

SilverFox261 wrote: »

You take the swap.

When you were offered the original choice you had a 33.3% chance of getting it right, and 66.6% of getting it wrong. The game show host must remove one empty box. If you picked right the first time aka: picked the money, then both boxes that are left are empty and therefore it does not matter which box he picks. This will happen 33.3% of the time.

But, if you picked wrong the first time, which will happen 66.6% of the time, then the game show host will only have one possible box to throw away, leaving the box containing the money.

Therefore you're chance of winning goes from 33.3% to 66.6% if you take the swap.

Theres a good explaination in this vid:

http://www.youtube.com/watch?v=mhlc7peGlGg

well said

ianmc38

The Monty Hall Problem

RichieLawlor

Watch '21' the film, Kevin Spacey and the Nerdy guy explain it. It's called variable change

BobSloane

In before someone says it's 50/50, retardo debate and inevitable lock

strewelpeter

The Monty Hall problem came up as an analogy for explaining some commercial option in a company I worked for once. No matter how many ways or how many times I explained it (You'd think that Hotspurs 'imagine you start with a million boxes' should be idiot proof) the accountant in the company just refused to get it. He eventually accepted that he was wrong and from then on he'd come up to me every couple of weeks and have me explain it to him again.

Company went El Busto. Obv.

YULETIRED

Does anyone have a link to that one hotspur mentioned a few weeks ago, the one in the University where the correct decision oddswise was to guess the answer , I'd like to send it on email to some folk.


I remember explaining the MH problem to a copper friend of mine and he got it straight away, I'm shocked an accountant didn't get it, Could it be the fact my mates is a betfair degenerate

bops

obv bluff, take the box that they took away imo

Jesus Wept

Or throw Monty a box to the face and hotwire the car.

Anyhow, I'd prefer the 'joke prize' - the elephant goat.

Icarus152

A goat.

Davey Devil

I didn't realise MPT owned Louie's Bistro. I was in there for lunch a while back and it was top quality. Keep up the good work Eoin.

hotspur

YULETIRED wrote: »

Does anyone have a link to that one hotspur mentioned a few weeks ago, the one in the University where the correct decision oddswise was to guess the answer , I'd like to send it on email to some folk.

http://www.youtube.com/watch?v=icGaDA0hLMk 12mins in.

I'm sure it's in print if you google, but there are a lot of hat puzzle variants out there.

Whyno

Change your mind = change your luck imo.... whats the third option
Ah sh1t i`m confused...Whats the bankers offer :pac:

HeeHawsCantona

x

carfax

Mr PT & Gholi; is it possible for you two to have a few beers together without an argument over nothing?

MrPillowTalk

well played lads booked me the 1200 win.

lol at those that refuse to believe in variable change.

MrPillowTalk

Davey Devil wrote: »

I didn't realise MPT owned Louie's Bistro. I was in there for lunch a while back and it was top quality. Keep up the good work Eoin.

thanks we do our best.

pok3rplaya

€1200!!?? Damn I need to start hanging around with Gholi

Gholimoli

i cant pay up so ive decided to leave ireland for good.

the_cat_is_back

Interesting. So on the show 'Deal or No Deal' (Not that i watch that muck ) the contestant should always take the swap when offered it at the end (when there are two boxes left)?

the_cat_is_back

the_cat_is_back wrote: »

Interesting. So on the show 'Deal or No Deal' (Not that i watch that muck ) the contestant should always take the swap when offered it at the end (when there are two boxes left)?

Monty Hall is a zero sum game, while Deal or No Deal isn't really. You are going to win something, no matter what. The size of the reward is all that is in dispute. While you should always swap in Deal or No Deal the concept isn't wholly analogous for 3 reasons:

1. In the Monty Hall problem, no matter how many doors you expand it to, there is only 1 car and x number of goats. If we accept that the goats have no inherent value to the player then there is only one beneficial outcome. In Deal or No Deal there are a number of beneficial outcomes for the player.

2. When you initially select your box you have a 1 in 22 (I think that's the right number of boxes, not sure) chance of selecting the quarter million. However, when you are offered the swap at the end you may have only a 1p and a 10,000 box remaining. This alters the maths somewhat in a way that the Monty Hall problem does not have to deal with.

3. You don't HAVE to open your box. In the Monty Hall problem you must select a door and good luck to you; but the addition of the Banker's Offer means that you do not have to open the box you began with, or any box at all for that matter. While the Banker almost never offers a +EV for the player it is another factor to consider and one which further distinguishes the Monty Hall problem from any application to Deal or No Deal

Small Change

Deleted User wrote: »

While you should always swap in Deal or No Deal the concept isn't wholly analogous for 3 reasons:

I always felt that the option to swap was 0EV in Deal or No Deal.

At the start of the game, each box has a 1/22 chance of having each of the amounts.
Unlike the MH problem, nothing occurs in the course of the game that changes the odds for one box relative to the other remaining boxes.

For example, in a game where the two remaining boxes contain 250K and 1p respectively, each box should have a 50% chance of having the 1p.

the_cat_is_back

Deleted User wrote: »

Monty Hall is a zero sum game, while Deal or No Deal isn't really. You are going to win something, no matter what. The size of the reward is all that is in dispute. While you should always swap in Deal or No Deal the concept isn't wholly analogous for 3 reasons:

1. In the Monty Hall problem, no matter how many doors you expand it to, there is only 1 car and x number of goats. If we accept that the goats have no inherent value to the player then there is only one beneficial outcome. In Deal or No Deal there are a number of beneficial outcomes for the player.

2. When you initially select your box you have a 1 in 22 (I think that's the right number of boxes, not sure) chance of selecting the quarter million. However, when you are offered the swap at the end you may have only a 1p and a 10,000 box remaining. This alters the maths somewhat in a way that the Monty Hall problem does not have to deal with.

3. You don't HAVE to open your box. In the Monty Hall problem you must select a door and good luck to you; but the addition of the Banker's Offer means that you do not have to open the box you began with, or any box at all for that matter. While the Banker almost never offers a +EV for the player it is another factor to consider and one which further distinguishes the Monty Hall problem from any application to Deal or No Deal

Small Change wrote: »

I always felt that the option to swap was 0EV in Deal or No Deal.

At the start of the game, each box has a 1/22 chance of having each of the amounts.
Unlike the MH problem, nothing occurs in the course of the game that changes the odds for one box relative to the other remaining boxes.

For example, in a game where the two remaining boxes contain 250K and 1p respectively, each box should have a 50% chance of having the 1p.

Well let's say our only goal in Deal or No Deal was to get the 250k. We have 22 boxes, i.e a 1/22 chance of having the 250k. If we get down to the last two boxes and the two remaining amounts are 1p and 250k. We turn down the banker's deal as 250k is our only goal.

Originally we had a 1/22 (roughly 4%) chance of picking the 250k and 21/22 (roughly 96%) of not getting it. We are offered the swap. By sticking with the box nothing has changed. However, by swapping it we know that if we picked incorrectly the first time around (96% probable) the box left over is the 250k.

ArmaniJeanss

the_cat_is_back wrote: »

Well let's say our only goal in Deal or No Deal was to get the 250k. We have 22 boxes, i.e a 1/22 chance of having the 250k. If we get down to the last two boxes and the two remaining amounts are 1p and 250k. We turn down the banker's deal as 250k is our only goal.

Originally we had a 1/22 (roughly 4%) chance of picking the 250k and 21/22 (roughly 96%) of not getting it. We are offered the swap. By sticking with the box nothing has changed. However, by swapping it we know that if we picked incorrectly the first time around (96% probable) the box left over is the 250k.

No, above is different situation. Deal or No Deal is 50/50 under your example.

However If Noel Edmonds stopped the game with 3 boxes left, and told you which one of the 2 you hadn't originally picked contained 1p and asked you do you want to swap, now a swap becomes 66%.

Its the extra information which changes the probabilities.

doke

Mrs. Doke suggests jiggling the box you originally chose before you decide.

Then again, a pre-emptive jiggle is her approach to most of the big decisions.

Goodluck2me

the_cat_is_back wrote: »

Well let's say our only goal in Deal or No Deal was to get the 250k. We have 22 boxes, i.e a 1/22 chance of having the 250k. If we get down to the last two boxes and the two remaining amounts are 1p and 250k. We turn down the banker's deal as 250k is our only goal.

Originally we had a 1/22 (roughly 4%) chance of picking the 250k and 21/22 (roughly 96%) of not getting it. We are offered the swap. By sticking with the box nothing has changed. However, by swapping it we know that if we picked incorrectly the first time around (96% probable) the box left over is the 250k.

this is completely different to the MH problem as there is no change in information.

the_cat_is_back

ArmaniJeanss wrote: »

No, above is different situation. Deal or No Deal is 50/50 under your example.

However If Noel Edmonds stopped the game with 3 boxes left, and told you which one of the 2 you hadn't originally picked contained 1p and asked you do you want to swap, now a swap becomes 66%.

Its the extra information which changes the probabilities.

Goodluck2me wrote: »

this is completely different to the MH problem as there is no change in information.

I'm probably just looking at it wrongly. I thought that if in the original example where you had three options and one incorrect is eliminated you could apply the same logic to having 22 boxes with 20 eliminated. As in both cases you're left with two options, one wrong and one right. Brain working a bit slow today but I'll get it eventually

ArmaniJeanss

the_cat_is_back wrote: »

I'm probably just looking at it wrongly. I thought that if in the original example where you had three options and one incorrect is eliminated you could apply the same logic to having 22 boxes with 20 eliminated. As in both cases you're left with two options, one wrong and one right. Brain working a bit slow today but I'll get it eventually

In the Monty Hall problem you are not left with 2 boxes, you are left with 3 boxes one of which is known. In your DOND example you are left with 2 boxes neither of which are known.
In the MH problem you are obviously never going to pick the known 'bad' box, but its presence influences the overall maths. The situation at the end of DOND is never a comparable situation.

Mellor

Deleted User wrote: »

1. In the Monty Hall problem, no matter how many doors you expand it to, there is only 1 car and x number of goats. If we accept that the goats have no inherent value to the player then there is only one beneficial outcome. In Deal or No Deal there are a number of beneficial outcomes for the player.

2. When you initially select your box you have a 1 in 22 (I think that's the right number of boxes, not sure) chance of selecting the quarter million. However, when you are offered the swap at the end you may have only a 1p and a 10,000 box remaining. This alters the maths somewhat in a way that the Monty Hall problem does not have to deal with.

3. You don't HAVE to open your box. In the Monty Hall problem you must select a door and good luck to you; but the addition of the Banker's Offer means that you do not have to open the box you began with, or any box at all for that matter. While the Banker almost never offers a +EV for the player it is another factor to consider and one which further distinguishes the Monty Hall problem from any application to Deal or No Deal

While all of these are true, they aren't the reason that the DOND situation is different to the MH. It's actually a lot simpler.

the_cat_is_back wrote: »

I'm probably just looking at it wrongly. I thought that if in the original example where you had three options and one incorrect is eliminated you could apply the same logic to having 22 boxes with 20 eliminated. As in both cases you're left with two options, one wrong and one right.

The main reason for taking the swap in the monty hall problem (and the one I always use to explain it) it that when you swap you always switch your prize. This is because the host chooses which boxes to drop based on additional knowledge he has. This is the key point, other wise the odds don't change.

In DOND, you pick a random box, and you drop random boxes. Which results in two random boxes at the end, it makes no difference if you take the swap (except for the mind games aspect of a swap being offered or not)

However, if you picked a box, and Noel Edmunds (or the banker) eliminated all but one remaining boxes, and then said the remaining prizes are $250k and $1, and offered you the swap, not it is best to swap. As the choices of which boxes to be removed was made by Noel, who was aware where the top prize was. This would be the same as the MH problem.

boxclever

I understand this theory but I dont really agree with it. The maths are obviously correct but it doesnt improve your chances, as when one box is taken away (im not refering to the deal or no deal senario) the odds are the same for both boxes. You have a greater chance of having the correct box now either way, whether you switch or not.

A switch does not actually improve your chance of having the prize. If the person was offered the choice of either box he could still pick the origional box and the same odds would be true

Lurker1977

boxclever wrote: »

I understand this theory but I dont really agree with it. The maths are obviously correct but it doesnt improve your chances, as when one box is taken away (im not refering to the deal or no deal senario) the odds are the same for both boxes. You have a greater chance of having the correct box now either way, whether you switch or not.

A switch does not actually improve your chance of having the prize. If the person was offered the choice of either box he could still pick the origional box and the same odds would be true

*sigh*. It was only a matter of time I suppose.

Small Change

boxclever wrote: »

I understand this theory but I dont really agree with it. The maths are obviously correct but it doesnt improve your chances, as when one box is taken away (im not refering to the deal or no deal senario) the odds are the same for both boxes. You have a greater chance of having the correct box now either way, whether you switch or not.

A switch does not actually improve your chance of having the prize. If the person was offered the choice of either box he could still pick the origional box and the same odds would be true

Imagine the following scenario;
-You have a million boxes, one contains €1 million and the other 999,999 are empty.
-You pick one at random
-The host removes 999,998 empty boxes leaving your box and one other

Do you think your original (one in a million) choice is now 50% likely to be correct?

boxclever

no, but, it is as likely to have it as the other remaining boxes

boxclever

Sorry I misread. Yes I believe it now has a 50% chance of containing the prize

boxclever

How does any past choices play a part in the odds for this? If the game started with just these two final boxes the odds would be 50-50.
At the end of the 1000000 boxes with two remaining (forget about which box you have chosen) the odds are still 50 - 50 for each.
If you were then given a choice to swap you are not improving your chances whatsoever but doing the maths you could aruge your box is still against the 999,998 boxes but in reality it is not

Small Change

boxclever wrote: »

doing the maths you could aruge your box is still against the 999,998 boxes but in reality it is not

Ah, I see where I am going wrong now


...but these go up to 11

boxclever

Small Change wrote: »

Ah, I see where I am going wrong now


...but these go up to 11

I dont understand your post. If you are sarcasically pointing out the fact that I say maths can be argued when that is what you were doing then you misread.

My point is that the math argument is not effective in this instance

Small Change

Last post on this, as I suspect I may be feeding a troll....

Two Scenarios:
Scenario 1:
-There are a million boxes, one contains €1 million and the other 999,999 are empty.
-You pick one at random
- At this point you have one box and there are 999,999 boxes left
- Also at this point there are at least 999,998 empty boxes among the remaining 999,999
-The host gives you the option give up the single box you have in exchange for the 999,999 other boxes
- If you choose to switch, you now have 999,999 boxes, of which 999,998 are empty
- As you are about to start opening boxes, the host helpfully speeds up the process by removing 999,998 empty boxes
What are the odds that your remaining box from the 999,999 contains €1 million?

Scenario 2:
-There are a million boxes, one contains €1 million and the other 999,999 are empty.
-You pick one at random
- At this point you have one box and there are 999,999 boxes left
- Also at this point there are at least 999,998 empty boxes among the remaining 999,999
- The host helpfully speeds up the process by removing 999,998 empty boxes
-The host gives you the option give up the single box you have in exchange for the one remaining other box
- What are the odds that the remaining box from the 999,999 contains €1 million?

The process for getting down to two boxes in the above scenarios is exactly the same (you choose one, the host removes 999,998 empty boxes from the remaining 999,999).
The only difference is in the questions the host asks you which have no impact on the physical contents of the boxes.

boxclever

That was not the argument as I understood it. The origional theory states when the other options were removed you knew they were empty so the two left obviously had the prize and the other nothing

ArmaniJeanss

Sim it, there are only 9 possibilities.
e.g., you pick A, the good prize in is B etc.

You'll see that swapping wins 6/9.

mormank

i was thinking about this the other day and for this theory to hold you need to know that the host will do this everytime! if the host merely picks and chooses when he will do this at a whim then i dont think it actualy does hold true...and dont get me wrong, i understand the theory perfectly, i just think it does need to be a constant scenario

gosplan

While we're on the subject of probable outcome style teasers:

The Truel
(It's a famous game theory problem - one of the first I think.)

Tis basically the same as a pistol duel but with three people.

You're the worst shot and only have a 33.3% success rate when it comes to hitting the target. Opponent no1 is next with a 66% success rate and Opponent no2 is a dead shot and never misses. (We assume that all successful shots are fatal)

Your opponents however are very sporting and agree to carry out the truel in turns. As the worst shot, you go first followed by opponent no2 and finally the dead shot guy. Whoever is left alive after the first round fires again and so on till there's only one left.

The question is what do you do with your first shot and why?

reilly110

could someone please expain this AGAIN ???

Small Change

You miss....and you are guaranteed to get at least one shot against the remaining player (if no 2 misses, the deadshot will kill him and you get the next shot, if he hits you still get the next shot)

Killing either of the others will result in the remaining one getting a free shot at you

monoP54

I want to shoot myself.

Brayruit

You are shown two envelopes with money in them.

You are told to pick one of them.

You open the envelope and see that there is an amount X.

You are then offered another envelope that you can swap for the X and you are told that there is either double the amount in the second envelope (2X) or half the amount (0.5X).

So you decide that the expected amount in the other envelope is 1.25X so you swap because the EV of swapping is 0.25X.

But lets say you were told this whole set up in advance, just not having any idea what the values X and 2X (or X and 0.5X) are.

Then you would conclude that you should always swap envelopes - right?

But then why didn't you pick the other envelope in the first place?

Mellor

boxclever wrote: »

A switch does not actually improve your chance of having the prize. If the person was offered the choice of either box he could still pick the origional box and the same odds would be true

You must of missed this last time.
This is the best way I have found to explain it.

3 boxes, 1 has a car, the others a goat each.

You pick a box, you have a 33% chance of getting the car, and 66% chance of getting a goat. Can we agree on this much?

The host, who knows what's in each box, removes a box that contains a goat. He chooses this box, it wasn't at random.
He now offers you a switch.

If you have the car already, switching will obviously give you a goat.
If you have a goat, then switching will give you the car, as the other goat is gone.
So, switch you switch the box, you always switch the prize
Seeing as we were 66% to originally pick a goat. switching gives you the car 66% of the time when we switch.

You must get it now,


Otherwise, hi DBC

Brayruit

Mellor wrote: »

Otherwise, hi DBC

This, obv.