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Post by amidf on Dec 21, 2017 18:01:16 GMT -5
This is a popular math / logic problem based loosely on an American gameshow called "Let's Make a Deal" and its host Monty Hall. The setup is this (and it is very crucial to note the details):
- The contestant is given a choice of 3 doors. Behind 2 of the doors are goats. Behind one of the doors is a grand prize.
- The placement of the prize is determined randomly, but the host knows the current position.
- There is no sensory means (sight, sound, smell) for the contestant to determine which door has a goat or grand prize behind it.
- The goal of the contestant is to select the door with the grand prize, no matter how much they might also like a goat. If your home country really values goats, imagine the grand prize is 10 goats.
- After the contestant selects a door, the host always opens one of the other two doors to reveal a goat.
- The contestant is always given a choice to either stick with the door originally picked or to switch to the remaining door the host did not open.
- Once the contestant has made the choice to switch or not, their door is opened and they receive the prize behind it.
The questions for our War Robots Forums audience (feel free to Google the solution; you may find some wrong answers though):
- What are the odds the door originally picked by the contestant has the grand prize behind it?
- What are the odds the other door left unopened by the host has the grand prize behind it?
- What is the best strategy to win the game - always stick with the original choice, always switch, or it makes no difference?
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polykillr
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Post by polykillr on Dec 21, 2017 18:03:38 GMT -5
Forgive me if I'm being stupid or dumb, but if the host opens one door, you have a 50% chance of being right or wrong if you stick or change door, surely?
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Post by amidf on Dec 21, 2017 18:07:01 GMT -5
Forgive me if I'm being stupid or dumb, but if the host opens one door, you have a 50% chance of being right or wrong if you stick or change door, surely? If you're right, you might answer "It doesn't matter" to the poll question. Let's let a few more people vote before we reveal and then argue about the answer. |
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Post by Deleted on Dec 21, 2017 18:08:07 GMT -5
Interesting. Should be a 33% chance, and shouldn't matter if you switch or not. But you have me doubting that logic, since with a dummy prize removed it feels like a 50/50 shot. When does the arguing start though? amidf smells like a goat. There, it's on! |
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Post by amidf on Dec 21, 2017 18:12:44 GMT -5
Interesting. Should be a 33% chance, and shouldn't matter if you switch or not. But you have me doubting that logic, since with a dummy prize removed it feels like a 50/50 shot. When does the arguing start though? amidf smells like a goat. There, it's on! If there's a 33% chance if you switch and 33% if you don't, where did the other 33% go? What prize do you win that 33% of the time? I hope it's a Haechi! |
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polykillr
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Post by polykillr on Dec 21, 2017 18:16:36 GMT -5
Also it must be rigged anyway, because otherwise you could have host picking the door with grand prize and opening that ??
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polykillr
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Post by polykillr on Dec 21, 2017 18:19:40 GMT -5
No, I change my mind. Surely, your first pick will have a 1 in 3 chance of being the correct door, but if you change door you will have a 2 in 3 chance? So there's no way to guarantee getting it, but you're more likely if you change?
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Post by Pulse Hadron on Dec 21, 2017 18:28:47 GMT -5
When you initially pick a door there’s a 1/3 chance it’s the grand prize and a 2/3 chance it’s not the prize. When the host reveals one of the other doors it has to be a goat. At this point there’s still a 2/3 chance the door you picked is NOT the prize, but because the host revealed a specifically non-prize door that 2/3 chance is now all on the one other door. It’s better to switch.
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Post by ♧SGT FURY 24/7♧ on Dec 21, 2017 18:32:20 GMT -5
What's wrong with goats?
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Post by Deleted on Dec 21, 2017 18:34:10 GMT -5
When you pick, its 33%. Then they toss out a goat, so there's 2 doors left. So you have a 50/50 shot if you pick at that point. All this thinking make Hulk head feel bad. Want smash now!
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Post by amidf on Dec 21, 2017 19:17:31 GMT -5
polykillr and Pulse Hadron have it right. The best strategy is to switch, which will win 2/3rds of the time. Sticking with the original door wins 1/3rd of the time. It doesn't go up to 50% just because the host showed us a goat. There is always at least one goat in the doors not chosen. When the host eliminates a door, the odds for switching do change! Now the 2/3rd chance you picked the wrong door to begin with are concentrated in that final door, making the switch much more likely to help you. There are other ways to explain it. Let's see if we need them!
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Post by ezekielcrow on Dec 21, 2017 22:05:06 GMT -5
Hmm. The odds may change from 3 doors to 2 but wont it just increase the odds that the door you picked could be the right one as well? It seems as though your original pick would have 50% chance since one of the three doors would have been eliminated anyway.
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polykillr
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Post by polykillr on Dec 22, 2017 2:25:28 GMT -5
Hmm. The odds may change from 3 doors to 2 but wont it just increase the odds that the door you picked could be the right one as well? It seems as though your original pick would have 50% chance since one of the three doors would have been eliminated anyway. Yeah, that's what I thought at first, but then I looked at it from a more mathematical angle. The first door you pick, you clearly have a 1/3 chance of winning, right? 1 door, 3 options? Then one door is opened, but your chance of winning is still exactly the same, 1/3 because you don't change. If you change, the other door has a 2/3 chance of winning because one door is gone. You're more likely to win if you change everytime, than staying. |
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Post by Deleted on Dec 22, 2017 2:41:22 GMT -5
3 doors. you choose 1, at 33%.
One door is opened.
To choose one of the doors is now 50%
However. If you stick with your first pick, it's still only 33% to be right - as that's what it was when you first chose it - that you've seen a goat behind another door, doesn't change that chance.
If you change doors, your percentage to be right goes up.
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polykillr
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Post by polykillr on Dec 22, 2017 3:01:03 GMT -5
3 doors. you choose 1, at 33%. One door is opened. To choose one of the doors is now 50% However. If you stick with your first pick, it's still only 33% to be right - as that's what it was when you first chose it - that you've seen a goat behind another door, doesn't change that chance. If you change doors, your percentage to be right goes up. I think it's 67% rather than 50% because you still have 3 doors, you just know what's behind one of them. |
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Post by TheWWRNerdGuy on Dec 22, 2017 3:41:09 GMT -5
I have been taught this problem, but there are logical flaws in the answer.
It's no longer a 1 in 3 chance because there are only two doors where it is possible to get the grand prize.
Let's say that you already know where the goat the host is going to reveal is. Does that mean you have a 66.6666666 etc. % chance of being wrong?
What you didn't choose has a 33% chance of being right, and a 67% chance of it being wrong. Is there still a 67% chance of it being wrong after one wrong answer is revealed?
Here's another logic problem based on twisting logic:
3 friends were leaving a budget hotel. They paid £30 to the hotel person (I'll call him 'Bill').
When the money was returned to the hotel manager he told Bill that it was meant to be £25. So,
Bill decided to give them each £1 and take the last £2 for himself, because he wanted the money to be evenly spread between them.
The friends paid £27 in total, and if you add the £2 that Bill took, then it is £29. Where did the extra £1 go?
In amidf 's answer, the logic was twisted, and the logic in my question was also twisted. It doesn't make sense if you look at it carefully.
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Post by Deleted on Dec 22, 2017 6:52:18 GMT -5
Tax  |
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Post by ezekielcrow on Dec 22, 2017 7:19:20 GMT -5
Isnt this how governments craft their budget?
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Post by amidf on Dec 22, 2017 9:28:36 GMT -5
TheWWRNerdGuyThe friends paid £27 total = £25 + the £2 they let Bill keep. Their change from £30 was £3. No logical problem. Your puzzle has a trick, as there is no reason to add the £2 Bill kept to what the men paid. The Monty Hall problem doesn't have a "trick" per se. if you actually run a simulation you will find that switching is twice as good a strategy as staying with the first door picked. The "trick" is wrapping your head around why. Here's another way people learn to overcome their mental block to believing the odds are other than 50-50: Imagine the problem with 100 doors. You pick one. Monty opens 98 others every time to reveal goats. Should you switch now, or does the original door still have equal odds to the other that survived 98 openings of its compatriots? If you run the game from the perspective of the host you will very quickly see how it works. Back to 3 doors. You are the host. The contestant picks a door with a goat 2/3rd of the time. You have to open the other door with the goat. If the contestant switches, he will win. The contestant picks the grand prize door 1/3rd of the time. If he switches, he will lose. Therefore, the contestant who always switches will win 2/3rds of the time. Now play it out in your head for a contestant who never switches.
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Post by Pulse Hadron on Dec 22, 2017 11:06:02 GMT -5
When I made sense of this problem I found using amidf’s technique of exaggerating to 100 doors helped, and also considering the odds of only ‘the door you picked’ vs ‘all other doors’.
So with 100 doors there’s a 1/100 chance you picked the prize and a 99/100 chance the prize is amongst all the other doors. When the host reveals doors the odds are still the same, a 1/100 chance you picked the prize and a 99/100 chance the prize is amongst all other doors, however there is only 1 other door now.
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[AurL] Valiant
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Post by [AurL] Valiant on Dec 22, 2017 12:41:21 GMT -5
Yes if we follow the problem with the intent to determine what the chance of picking correctly on the first selection the math seems to defie logic (as you said devils in the details) but that is because we are answering a very specific question and then using it as the basis for the rest. However there are some real world logical flaws in this problem even if we chose to follow the math all you have to do is change the wording.
- The contestant is given a choice of 3 doors. Behind 2 of the doors are goats. Behind one of the doors is a grand prize.
- The placement of the prize is determined randomly, but the host knows the current position.
- This is a two round challenge, the contestant is asked to pick a door to save from elimination.
- The host then always opens one of the remaining doors he knows to have a goat.
- The contestant is then asked to pick between the two remaining doors and what ever is behind the door is his prize.
- What is the contestants chance of picking the grand prize?
So what did I change? Simple I reset the choice after the first door was eliminated and made the question what is the contestants chance of picking the grand prize, in this set up the contestant always had a 50/50 chance of getting the grand prize.
The first pick is arbitrary as we know the host will eliminate one wrong answer from our final selection no matter what so if we picked a goat first the host will eliminate the other goat leaving our final choice between a goat and the grand prize 50/50 or pure chance. If we pick the grand prize the host will still eliminate a goat leaving us the same 50/50 odds when we pick between the two remaining doors.
Think of it this way let's say there are 3 doors, 2 wrong and one right the host simply eliminates a wrong door then you pick, what are your chances of picking the correct door? 50%
The logic of the problem doesn't change as you were always interested in picking the right door and what your chance to pick that door is but the math of the problem does.
Let's play with the hundred door scenario and let's say you pick 5 out of the 100 doors the host then always eliminates 94 doors that were wrong leaving you 6 remaining doors (your 5 + 1). Next the host asks you to pick 3 of the six remaining doors if the prize is in one of the 3 you win if not you loose. So what is your chance of winning? Well as the host knows the correct answer and isn't allowed to eliminate it he is improving your odds each time from 6 out of 100 to 3 out of 6 or 50%. Now if the host didn't know where the grand prize was and it could be eliminated then you were 6 out of 100 or 6%
This does not take into account the human element, any leading or misdirection by the host or anyone's point of view other then yours - as the contestant.
So the answer to your questions:
1) Roughly 1/3 in that original choice
2) There is a 50% chance that either one of the two remaining doors has the grand prize behind it now
3) It makes no difference
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Post by amidf on Dec 22, 2017 12:43:40 GMT -5
If we can wrap out minds around the original problem, we can explore other surprising outcomes. For example, imagine there are originally 6 doors. The contestant picks one. The contestant is now asked to divide the remaining 5 doors into a group of 3 and a group of 2 and then Monty will open all but one door from each group (it is important this is spelled out ahead of time so that Monty isn't changing his behavior based on his knowledge of the grand prize location). Should the contestant switch, and if so, should he go with the door remaining from the set of 3 or the set of 2?
The answer is that the door remaining from the set of 3 has a 50% chance of having the prize, the door remaining from the set of 2 has a 1/3rd chance, and the originally selected door has a 1/6th chance. The explanation is the same, but if you haven't yet grasped how the probabilities collapse as new information is revealed, this is going to be very counterintuitive.
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Post by amidf on Dec 22, 2017 12:59:16 GMT -5
Yes if we follow the problem with the intent to determine what the chance of picking correctly on the first selection the math seems to defie logic (as you said devils in the details) but that is because we are answering a very specific question and then using it as the basis for the rest. However there are some real world logical flaws in this problem even if we chose to follow the math all you have to do is change the wording. - The contestant is given a choice of 3 doors. Behind 2 of the doors are goats. Behind one of the doors is a grand prize. - The placement of the prize is determined randomly, but the host knows the current position. - This is a two round challenge, the contestant is asked to pick a door to save from elimination. - The host then always opens one of the remaining doors he knows to have a goat. - The contestant is then asked to pick between the two remaining doors and what ever is behind the door is his prize. - What is the contestants chance of picking the grand prize? So what did I change? Simple I reset the choice after the first door was eliminated and made the question what is the contestants chance of picking the grand prize, in this set up the contestant always had a 50/50 chance of getting the grand prize. The first pick is arbitrary as we know the host will eliminate one wrong answer from our final selection no matter what so if we picked a goat first the host will eliminate the other goat leaving our final choice between a goat and the grand prize 50/50 or pure chance. If we pick the grand prize the host will still eliminate a goat leaving us the same 50/50 odds when we pick between the two remaining doors. Think of it this way let's say there are 3 doors, 2 wrong and one right the host simply eliminates a wrong door then you pick, what are your chances of picking the correct door? 50% The logic of the problem doesn't change as you were always interested in picking the right door and what your chance to pick that door is but the math of the problem does. Well, you're wrong about at least the original scenario you describe, but that assumes I understood it. As I found it confusing, I don't want to bring in your other scenarios yet. What have you changed from the original problem in the scenario you described? As far as I can tell, nothing of importance. The contestant still saves a door from elimination and then is asked to choose between that first door and one other. If the doors have not been shuffled, i.e. the contestant still knows which one he originally picked, then switching will win 2/3rd of the time. If the doors have been shuffled, then he has lost the information that gives him an advantage and his chance is 50%. Which did you mean? BTW, the reason I expected lots of arguing in this thread is because this problem has confounded many bright minds even among mathematicians and probability theorists. It is a wonderful problem! But run some simulations and you will find out what the correct answer is even if you cannot grasp the theory. You can play this game with cups and a ball with your kids at home. Play as the host and ask them to always switch or always stay with the first choice. You can track the results (run at least 50 to reduce the effect of chance) but after a few games, you will quickly recognize the different mechanics in play between the always switch and always stay methods. |
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Post by [AurL] Valiant on Dec 22, 2017 13:53:21 GMT -5
I just asked a different question I don't care what your chance of picking the correct door originally was all I asked was for you to pick when you have two doors remaining and what your chance of winning at that point was.
This math problem appears to be a matter of perspective if I chose to calculate your chance of picking the correct door originally it is 1 in 3 if I use this to determine if you should stay or switch your math makes since and it would appear that you should switch but if I chose to calculate your chance of wining based on your selection between the final 2 doors it is 1 in 2 or 50% both can be mathematically verified at which point it doesn't seem to matter what you do.
Theoretically the contestant never has an advantage no matter if you shuffle the doors or not, assuming the contestant understands the rules of the game, if you give them the opportunity to switch and simply say now pick between these two doors the probability of picking correctly is now roughly 50% - the original choice is arbitrary and has no effect on your chance to win (no benefit to picking one door over the other). If you apply that variation to a mathematical simulation I would expect we would see the results of pure chance, though the math as you have stated it indicates I should see a clear advantage in winning by taking the original door into account.
As I said before all this is assuming we can eliminate the human variable.
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[AurL] Valiant
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Post by [AurL] Valiant on Dec 22, 2017 13:56:23 GMT -5
Why are my post quarantined until reviewed by staff?
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Post by amidf on Dec 22, 2017 15:02:48 GMT -5
Why are my post quarantined until reviewed by staff? This is to stop spam from new users. I am one of the people enabled to allow posts out of quarantine, but since I use the mobile version I don't even know they are under quarantine. I switched to desktop, and from what I can tell, your posts are out now. |
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Post by amidf on Dec 22, 2017 15:12:28 GMT -5
[AurL] Valiant Still not sure I am following you. Perhaps you mean if the contestant picks randomly to switch or not to switch, they will have 50-50 odds? That is correct. If the contestant ignores the info, then there are two doors - one with a 2/3rd chance and one with a 1/3rd chance. If he picks randomly he will get it right one half the time. That is consistent with and uses the same math as the other choices I described: always switch or always stick with original. Original door odds: 1/3 Switch door odds: 2/3 Contestant always sticks with original: 1/3 wins Contestant always switches: 2/3 wins Contestant switches half the time: 1/3 x 1/2 + 2/3 x 1/2 = 1/6 + 2/6 = 1/2 wins
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Post by TheWWRNerdGuy on Dec 22, 2017 19:09:54 GMT -5
There are 100 doors, only one containing a balanced version of War Robots, you pick one, and the host opens the other 99, which are all goats, leaving only the door you picked. Is there still a 99% chance that your original choice contains a goat?
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Post by amidf on Dec 22, 2017 19:54:59 GMT -5
There are 100 doors, only one containing a balanced version of War Robots, you pick one, and the host opens the other 99, which are all goats, leaving only the door you picked. Is there still a 99% chance that your original choice contains a goat? I can't be convinced that Pix would ever again develop a balanced version of War Robots. Ergo, they're all goats. |
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Post by Deleted on Dec 22, 2017 20:38:53 GMT -5
When I was presented to this, many years ago, I couldn’t get my head around it for more than an hour. Insisted that it didn’t matter. Felt really stupid when I realized that I was wrong, as did the (allegedly many) math scientists who fell into the same logical trap.
It is really simple though: the door the host opens isn’t eliminating a third of the originally available options. He’s eliminating the negative option from the two non-selected choices, not a random option from the non-selected choices.
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