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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.

boxclever

Mellor wrote: »

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

I understand what you are saying but I disagree. When a box is removed which contains a goat the odds that you have picked the car origonally improve from 33% to 50%.

That was the point I was trying to make. I agree about all the percentages but the box you are asked to switch to also had a 33% chance of containg the car starting out and has also improved to having a 50% chance of having the car

boxclever

boxclever wrote: »

I understand what you are saying but I disagree. When a box is removed which contains a goat the odds that you have picked the car origonally improve from 33% to 50%.

That was the point I was trying to make. I agree about all the percentages but the box you are asked to switch to also had a 33% chance of containg the car starting out and has also improved to having a 50% chance of having the car

This is the mistake most people make and you are wholly incorrect.

Stop thinking about there being 2 boxes left. There are still 3 boxes.

You get 2 chances to make a decision. The first time you will get to see 1 out of 3 boxes (33.3% chance of picking the car). Once the host shows you a goat you will now get to see 2 out of the 3 boxes (66.6% chance of picking the car).

The problem is used to demonstrate that information changes the probability but not always how you think it does and also that the chances of picking the box with the car remain static unless you make a new decision.

Let me put it this way. If the host showed you a goat but DIDN'T offer the switch, you still had a 1 in 3 chance of being right when you picked your box. That hasn't changed. So why would it change if you declined the swap? The effect is the same. However when you take the swap you are making a new decision based on 2 out of 3 pieces of information. That is where your probability of being correct comes from.

Also, it's not a matter of agreeing with any of this or not. This isn't an opinion piece, it's a maths problem and the solution Mellor gives is the correct one.

BobSloane

https://www.youtube.com/watch?v=GmU_q5xrnto

gondorff

Lol at 'understanding but not agreeing':pac:

boxclever

Stop thinking about there being 2 boxes left. There are still 3 boxes.

It is irrelevant the amount off boxes as once you know what is in the box you know with 100% confidence the two remaining boxes comprise of a car and a goat.

You also know that there is an equal probability that the car is in each box. The only way you have a 66% chance of picking the correct box (with the car) is if you knew the identity of one of the boxes from the begining.

When the host asks you if you want to swap your odds of having the correct box are not 33%. When you are offered the swap there is no argument to say the other box holds more of percentage change of containing the car and thats were this argument is redundant.

Of course you have more chance now than when there were 3 boxes where the contents of all 3 were unknown. But whether you take the swap or not you have this greater chance

Lurker1977

BobSloane wrote: »

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

.

5starpool