Modified Monty Hall problem

[Okamiden]

The original problem, for those unfamiliar:

Monty Hall problem - Wikipedia

en.m.wikipedia.org

The variant:

You have 100 closed doors in front of you. Only one door has a grand prize in it. $10,000 and a limited edition version of whatever your favorite game is. The other doors have N-Gages.

The game is in two phases. You pick one door at random. It stays closed. No peeking.

Then, the host opens 98 doors at random, but never opening yours. He doesn't know what's behind the doors he opens. If he opens the prize door, oh no! You lose. Unless you want the N-Gage.

But on this particular game you're in, by sheer luck, he manages to open 98 doors without a single one being the prize door. He picked at random, so that's pretty lucky.

There are now two doors left. You get an ultimate choice. You can now decide to switch doors with the other door left, or keep yours. What do you decide, and why?

Yeah, I was bored.

Spoiler

I expect people to solve this quickly so maybe consider using spoilers

 

[jdmc13]

I switch.

Spoiler

1% chance I picked the correct door the first time which means there is a 99% chance it is in the only other remaining door.

 

[B-Dubs]

I switch doors.

 

[fr0st]

Pretty sure you always switch since the chance of you getting the prize is extremely high

 

[Chopchop]

Isn't that

Spoiler

just 50/50 then? In the original problem, the host knows which door has the prize. If the host doesn't know, then it's just pure luck, right

?

 

[Joni]

I take the 98 N-Gages.

 

[Poppy]

pretty sure you should always switch

 

[Septimus Prime]

It's still the same solution as the regular version of the problem.

 

[RecRoulette]

It's just a 50/50 at that point since he doesn't know?

Initial thought was "always switch" but in this case it's always ngage.

 

[Platy]

i stay on my own door because the retro market is bordering on saturation and overpricing and considering how low is my country's money compared to the dollar, if I sell the n-gage on the ebay i will get a pretty big moeny

 

[Okamiden]

It's not like the original problem. Show your work, people!

 

[WedgeX]

If I had the time to rebuild my Monte hall problem worksheet in excel, I'd tell you.

 

[Stalker]

You switch. The probability you got the correct door out of 1 hundred is extremely low when all the doors have been eliminated (and the game is still going) the higher probability is the door you didn't pick.

 

[gutshot]

I don't understand why there needs to be 100 doors at the start if the end result is just a coin flip?

 

[thefro]

It's still the same solution as the regular version of the problem.

Click to shrink...

Yep, and the original problem is silly because it assumes the host always does the same thing or is acting in good faith, which doesn't fit the actual "Let's Make a Deal" show.

It's basically the original version of one of those shitty Facebook memes where they have a problem with confusing instructions.

 

[Deleted member 4367]

Him not knowing which doors to open makes it 50/50.

 

[Mr.Awesome]

The Monty hall problem is driven off the fact that they are giving you an incorrect door.

I dont see that in the op so it's just 50/50.

 

[Border]

I will probably go to my grave never really understanding the logic behind this problem.

The idea is that my first choice only had a 1% probability of being correct. But if I switch after all the reveals, that choice would have a 50% probability of being correct. But why doesn't my original choice also have the same 50% probability at that point? Isn't "the decision not to switch" a choice that is independent of my original selection?

 

[Okamiden]

Yep, and the original problem is silly because it assumes the host always does the same thing or is acting in good faith, which doesn't fit the actual "Let's Make a Deal" show.

It's basically the original version of one of those shitty Facebook memes where they have a problem with confusing instructions.

Click to shrink...

Well in this case, the host is picking completely at random. So there's no bad faith.

 

[B.K.]

That's just Deal or No Deal.

 

[Okamiden]

The Monty hall problem is driven off the fact that they are giving you an incorrect door.

I dont see that in the op so it's just 50/50.

Click to shrink...

Giving you an incorrect door? I don't understand what you mean.

 

[Stalker]

Giving you an incorrect door? I don't understand what you mean.

Click to shrink...

In the monty hall problem you a given a door by the host that doesn't have the car.

3 doors

you pick one and the host gives you one as proof it doesn't have a car. The host knows so he didn't pick the car one. You had a lucky chance out of 3 doors but now your being told it's not 1 of the doors your odds increase if you switch

 

[Burly]

In this situation I don't think it matters. The original problem always had the addendum that the host would never open the grand prize door, it would always be "other" doors. If both picks are truly random, then I don't think it matters.

 

[Cat Party]

This has nothing to do with the Monty Hall problem. It's 50/50 if you should switch.

 

[Stinkles]

Trick question -- you sidetalk to open all the doors simultaneously.

True Story I once had an n-gage and every available game.

 

[Threadkular]

Probability is my math nemesis

 

[Stalker]

True Story I once had an n-gage and every available game.

Click to shrink...

I'm Sorry

 

[Fugu]

The odds here are 50/50 so it doesn't matter if you switch or not. The host, by pure fluke, opened nothing but ngage doors.

What you did was took the "easy Monty Hall problem explanation" and changed a critical element to make the original reasoning behind switching no longer true. In that problem there's a billion minus one doors with goats behind them and one with a car, so the host offers to open all doors but two for you, with the key difference being that the host isn't allowed to open the prize door. The reason switching is advantageous is because it leverages the host's knowledge against him. To not switch is to take the one in a billion odds that you nailed the prize right at the outset.

 

[Okamiden]

This has nothing to do with the Monty Hall problem. It's 50/50 if you should switch.

Click to shrink...

Spoiler

That's the point. It's a variant to play on people's expectations.

 

[Mona]

It's still the same solution as the regular version of the problem.

Click to shrink...

Yea

 

[Border]

Yeah, the original problem is kinda dependent on the fact that they are intentionally revealing all the "Losing" doors and only leaving you with your original choice + 1 alternative.

In this variant, it's merely luck or happenstance that the winning door was not revealed. It's unlikely that you would be able to pick the Winner out of 100 choices, but it's also highly unlikely that they would be able to reveal 98 doors and never hit the winner.

 

[ABIC]

I will probably go to my grave never really understanding the logic behind this problem.

The idea is that my first choice only had a 1% probability of being correct. But if I switch after all the reveals, that choice would have a 50% probability of being correct. But why doesn't my original choice also have the same 50% probability at that point? Isn't "the decision not to switch" a choice that is independent of my original selection?

Click to shrink...

Yeah, I think you're mostly there.

Your first choice has a 1% probability of being correct also means there's a 99% probability the correct door is some other door.

One other way to think about it is to run this simulation hundreds of times -- imagine if Bob and Paul had to play this game 500 times, for example.

You see Bob and Paul pick a door, and then 98 other doors disappear leaving one.
Bob never switches.
Paul always switches.
They do this again and again and again and again for 500x.

It becomes more logical to see that Paul will win more often than Bob. In fact, if you ran this simulation through code, that's exactly what would happen.

 

[HommePomme]

Yeah, the original problem is kinda dependent on the fact that they are intentionally revealing all the "Losing" doors and only leaving you with your original choice + 1 alternative.

In this variant, it's merely luck or happenstance that the winning door was not revealed. It's unlikely that you would be able to pick the Winner out of 100 choices, but it's also highly unlikely that they would be able to reveal 98 doors and never hit the winner.

Click to shrink...

Yeah I think it doesn't matter that it''s "lucky" that they opened 98 doors and never hit the winner, you need to switch to take advantage of that luck

 

[Okamiden]

Trick question -- you sidetalk to open all the doors simultaneously.

True Story I once had an n-gage and every available game.

Click to shrink...

Wow. How many games were there even, on the N-gage?

 

[ItIsOkBro]

think of it from the perspective of the host. there are two doors left and he's just yolo opening doors. let's say he could open your door or the other door. he wouldn't have different odds for opening your door. it's 50/50 for him and 50/50 for you if to switch.

 

[Parthenios]

I still believe the correct answer is to switch.

When you picked the first time, your odds of choosing correctly were 1/100, and those odds never change regardless of how many doors are revealed. If you switch, your odds of having picked correctly change to 1/2.

In the classinc Monty Hall, switch actually is correct 2/3 of the time, not 1/2, so maybe that specific part is changed based on the host knowing or not knowing, but the correct choice is switch regardless.

 

[Phatmanny]

The fact that you said he chose at random means they both have an equal chance of containing the prize 1/100

it's because in the original problem he deliberately opens a losing door is what modifies the odds.

 

[Phatmanny]

No new information is added by opening doors at random. While if he chose 98 "losing" doors deliberately then it modifies the odds

 

[Fugu]

Yeah, I think you're mostly there.

Your first choice has a 1% probability of being correct also means there's a 99% probability the correct door is some other door.

One other way to think about it is to run this simulation hundreds of times -- imagine if Bob and Paul had to play this game 500 times, for example.

You see Bob and Paul pick a door, and then 98 other doors disappear leaving one.
Bob never switches.
Paul always switches.
They do this again and again and again and again for 500x.

It becomes more logical to see that Paul will win more often than Bob. In fact, if you ran this simulation through code, that's exactly what would happen.

Click to shrink...

This logic is applicable to the real Monty Hall problem, but not the question the OP is posing.

 

[LakeEarth]

I will probably go to my grave never really understanding the logic behind this problem.

The idea is that my first choice only had a 1% probability of being correct. But if I switch after all the reveals, that choice would have a 50% probability of being correct. But why doesn't my original choice also have the same 50% probability at that point? Isn't "the decision not to switch" a choice that is independent of my original selection?

Click to shrink...

The OP example is messed up because the host randomly opened the doors, and just so happened to not open the one with the prize. In the real version of the Monty Hall problem, the host knows what door has the prize and purposely never opens it when opening all other doors.

So with the real Monty Hall problem, it's not 50/50 because it isn't REALLY a choice about switching from your door to the other unopened door. It's a choice between the door you picked, and the other 99 doors. Him opening the 98 doors with no prize before allowing you to switch is masking this fact.

If you still have a problem imagining this, expand it. There are a million doors now. You pick one of them. Monty Hall eliminates 999,998 of the doors he knows don't have the prize in them. Is switching to that lone door he didn't open still a 50/50 shot to win now?

 

[ss_lemonade]

Isn't this just the exact same problem? You still have a significantly higher chance of getting it right if you switch

 

[ABIC]

This logic is applicable to the real Monty Hall problem, but not the question the OP is posing.

Click to shrink...

Yeah, you're right, but I was mainly replying to a poster not the OP.

The OP question is more interesting with the host being able to open the prize door.

 

[Okamiden]

Isn't this just the exact same problem? You still have a significantly higher chance of getting it right if you switch

Click to shrink...

It's different because the host picks at random and just so happened to not open the prize door.

 

[Septimus Prime]

This logic is applicable to the real Monty Hall problem, but not the question the OP is posing.

Click to shrink...

How does the math differ? Do you open 1 door or 99 doors, randomly? Host intent doesn't change that he already opened 98 losing doors for you.

 

[Ellyshia]

I went through this problem years ago cause of deal or no deal. Ultimately, because it's random chance since the host also doesn't know, the probability is 50/50. The easiest way to see this is to draw out the tree of possibilities with 3 prizes.

1 - You pick the winner(A) host picks a loser(B) = Don't switch
2 - You pick the winner(A) host picks a loser(C) = Don't switch
3 - You pick a loser (B), host picks a loser(C) = Switch
4 - You pick a loser (B), host picks a winner(A) = Doesn't happen in this reality
5 - You pick a loser (C), host picks a loser(B) = Switch
6 - You pick a loser (C), host picks a winner(A) = Doesn't happen in this reality

So basically there are 6 possible outcomes, but the premise already removes 2 of those outcomes. Of the 4 remaining outcomes, 2 of them you want to switch, 2 of them you don't. So 50% chance. It's a coin flip.

 

[Nappuccino]

The fact that you said he chose at random means they both have an equal chance of containing the prize 1/100

it's because in the original problem he deliberately opens a losing door is what modifies the odds.

Click to shrink...

Aren't you just describing why people fall for the Monty Hall problem? Intentionally revealed or not, you've still seen them only open losing doors.

 

[Stinkles]

Wow. How many games were there even, on the N-gage?

Click to shrink...

I don't know how many it ended up with but I had like 20+

I didn't really play any...

 

[Sybil]

The real problem is the fact this game show managed bought and kept 99 N-Gages. Nothing good can happen from hoarding all of those. We have to stop them, 1 N-Gage at a time

 

[Lumination]

It doesn't matter because the chance is the same. The original problem leverages the fact that the host knowingly avoids the prize. So you are essentially making a 1% guess (with 100 doors) vs a 99% guess. Since the host doesn't know and is just very unlucky, the chance of either door containing the prize is both the original 1/100, or 50/50 when it's down to the last two.

 

[Mezentine]

This actually makes the Monty Hall Problem much easier to understand, oddly enough. The math of why it works isn't quite as counterintuitive, and as someone else says this makes it clearer that what's happening when you switch is you're taking advantage of the other person's "luck" (in a neutral sense) by not revealing the prize with their picks.