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You are presented with 3 boxes. One of the boxes contains $300. The other 2 are empty boxes. You will win the contents of the box you choose.
Now... After you have chosen a box, one of the un-chosen boxes is opened and revealed to be empty.
You are now presented with the option of keeping your original choice or switching to the remaining box.
1) What is the EV of not switching?
2) What is the EV of switching?
3) What do you do?
always switch. when you pick you're 1/3. when one is revealed you are now 1/2.
most people think it doesn't matter to switch, but it's much easier to see when you replace 3 doors with 300.
partial credit; not completely accurate.
WHY
Let the other critters play, too.
EDIT: Your answer was my first answer, but the solution is subtle and interesting.
There's a 2/3 chance of not picking the 300 first try, so by always switching you have a 2/3 chance of getting dat 300.
What do I win?
Haven't we done this before? Always switch.
OH I SEE
The simplest and most intuitive answer is that to win by not switching, you have to pick the 300 box first, which is a 1/3 chance. To win by switching, you have to not pick the winning box first time, which is a 2/3 chance.
1. $100
2. $200
3. Switch.
Luco wins the internets.
[1] [2] [3]
The probability is 1/3 to choose the box with the money the first time, and 2/3 that the money is in a non-chosen box.
( [1/3] ) ( [1/3] [1/3] )
The revealed box is always empty, so the 2/3's chance of containing the money has collapsed into one box.
( [1/3] ) ( [2/3] [0] )
1) The EV of not switching is $100
2) The EV of switching is $200
3) Always switch.
Quote:
Originally Posted by MadMojoMonkey
Switch, If you stick with your first choice, its your ego .
You had a 3/1 shot the first time. now its 2/1 .
If you stick with it, you are calling station and doomed.
If you switch , youll beat the stakes and move up.
I know I've heard this before, and I remember, "you should always switch." But no matter what, I cannot understand why.
The logic seems like picking tails, because last the last coin flip resulted in heads.
You have a 1/3 chance to pick the right box. Once you make your initial pick, one of the, or the, empty box is revealed and removed. You now get to chose between the two remaining boxes.
What information has been revealed about either of the remaining boxes by removing the empty one from play? The initial pick was left in because it was the prize, or because it was empty. The alternative box, the one not revealed to be empty was left in because it is either the prize or it is empty.
wait. wat. why?Quote:
The revealed box is always empty, so the 2/3's chance of containing the money has collapsed into one box.
Ah, ok, I get it. I'm not going to ninja edit, because that's lame.
You actually do gain information, because your box is never going to be revealed to be the prize or be empty, only one of the other two. So essentially you are being offered the potential contents of both boxes you did not choose instead of the potential contents of the box you chose.
you have to think of it in two separate parts.Quote:
Originally Posted by boost
1) 1 out of 3 boxes is $300. 2 out of 3 boxes is $0.
2) 1 out of 2 boxes is $300. 1 out of 2 boxes is $0.
if you have your choice between the two, obviously you would pick #2.
The first "coin toss" was three side, whereas the next one is two sided.
Also, if it helps, try replace 3 boxes with 300 boxes. Then go through the problem again:
- Choose 1 box out of 300
- Of the remaining 299 boxes, 298 boxes are revealed as $0
- Do you stick with your box (at 1/300) or pick the new box (at 1/2)?
It's called the monty hall paradox or something stupid like that.
Math is still wrong here. It's still a 3-side cointoss, but now two of them net you $300. You basically get to choose between your first choice, or 'both of the others'.Quote:
Originally Posted by givememyleg
And in the example of 300 boxes, you go from 1/300 to 299/300 chance.
we've had this thread, not even that long ago :(
jesus it was two years ago. consolation prize for bad time memory though: people said nice things about me! this is probably why it was still so large in my memory.
http://www.flopturnriver.com/pokerfo...on-185673.html
It is not 50/50 on the 2nd choice; it's 33/67. This is a common mis-assessment of the effect of the revealed information.Quote:
Originally Posted by celtic123
In this case, the revealed box is ALWAYS empty. There is 0% chance the revealed box contains the money.
This is correct. You do NOT go to 50/50 on the 2nd choice.Quote:
Originally Posted by jackvance
In your second part, there's still no reason to pick Box B over Box A. Yes it's a better situation to be in than Situation One, but you're in this new better situation 100% of the time and your choice comes now, your choice doesn't come first and put you in the new better situation.Quote:
Originally Posted by givememyleg
The way the OP is worded, it's slightly ambiguous. The classic Monty Hall problem is based on the host knowing the contents of all boxes and always choosing an empty box from the set of two unchosen boxes. The way the OP is worded, it isn't clear if the box that is opened was opened at random between the two, or if the "host" knew which boxes were empty before hand and will always deliberately choose an empty box.
IMO, the easiest way to visualize the EV of the classic Monty Hall problem is this:
You choose one box from the three. The chance of this box being correct is 1/3. The combined chance it is in one of the two other boxes is 2/3. When the all-knowing host opens one of the non-chosen boxes now the probabilities are:
Your chosen box: still 1/3
The empty box the host opens: 0
The remaining box: 2/3
If the "host" isn't all-knowing opens one of the non-chosen boxes at random and it happens to be empty, then the probabilities become 1/2 and 1/2. This iteration would also involve the game being voided if the "host" opens the box with the prize.
What is ambiguous about that?Quote:
Originally Posted by MadMojoMonkey
Why does the puzzle need a "host"?
Didn't read any responses, but it depends if the box is chosen at random or not. If the box is chosen in a way to always present you with an empty box, then you switch, etc.
Yeah, the trick here is understanding why the EV of the discarded box gets incorporated into the unpicked box. The reason is that the picked box has been removed from the possibility of being eliminated. By picking whichever box, we have eliminated it from being revealed. So the two remaining boxes are separated by our initial pick from the initial pick itself. These boxes combined have a probability of holding the prize of 2/3. When the empty one is revealed, the two unpicked boxes still together have a 2/3 chance of holding the prize.
Even if it's chosen at random, your odds are still the same as the original choice, so you can never be worse off than you were by switching.Quote:
Originally Posted by spoonitnow
If the one unchosen box is picked at random, it can still be revealed to be empty.Quote:
Originally Posted by MadMojoMonkey
Yeah but the answer to your initial question is different. If it is opened at random and is empty then your EV is the same whether you switch or not, so it doesn't matter.Quote:
Originally Posted by MadMojoMonkey
3 cases:
1 you chose the $300, opened empty box
2 you chose empty box, opened empty box
3 you chose empty box, opened $300 box
Since #3 didn't happen, it's between #1 and #2 so 50/50 and your EV is $150.
Yep, I call bullshit.
Depends on whether or not an empty box or goat behind a door or whatever MUST be revealed (as in the classic Monty Hall problem), or it just so happens to be. This is a crucial detail that is often missed by people who know the solution to the MH problem but who don't truly understand it.
I rarely, if ever, get pms.
I'm not sure I follow. Of course he could reveal the prize, but then of course you'd change your pick, so long as you were allowed to still pick the prize.Quote:
Originally Posted by Lukie
He can only reveal the prize if he picks at random, in which case there is no benefit in switching. If he picks at random than you aren't making the choice of one initially and then effectively two boxes later. You'd instead be switching your randomly selected box for another randomly selected box.
Wait, I can do better than that.
Firstly think of the three scenarios available.
1. You picked box 1, the winning box. Host opens empty box 2.
2. You picked an empty box 1, host opens an empty box 2.
3. You pick an empty box 1, host opens the winning box 3.
Let's imagine he chooses randomly.
You are at the point of deciding whether to switch.
That means that of the three available scenarios, you are at scenario 1 or 2. Originally there were 3 equally likely scenarios, but we now know it isn't scenario 3, so it leaves 2 equally likely scenarios of either number one or two.
If the host always intentionally picks the empty box, then scenarios 2 and 3 merge to give the same outcome.
Which box has the hole in the bottom to put your penis in?
I feel like the OP has left out important details :/
Right. That's the whole point.Quote:
Originally Posted by boost
I don't watch any game shows anymore but we have all seen enough over the years. There are some really weird ones out there. Spin this, win this, choices, non-choices, call a friend, poll the audience, etc. Like at the end of wheel of fortune you pick a choice and win that prize. There's no game theory problem or whatever involved.
It is disingenuous to present the problem while withholding a key detail. The point of the Monty Hall problem is to point out a counter-intuitive but mathematically correct principle, not to test or assume that someone else would design a game show in the same way.
"Oh, look, we revealed the prize! Would you like to switch? *crowd laughs* haha. Ok, now run over there and spin that wheel!!!........"
Also I'm not really a fan of the answer, "well you should always switch, since it can't hurt you", especially when the whole point of presenting a Monty Hall problem is usually to try to 'catch' someone who has never heard of the problem (and thus isn't familiar with what should and shouldn't be included). So if you present a question like in the OP, if the box is revealed empty but done so randomly, you have a 50/50 shot regardless of whether you switch. Many people argue this and are correct. You can't just say "well you should always switch because it will never hurt". The entire point is arguing over switching gives you a 1/2 or 2/3 chance of winning the prize. Another way to look at it is, you can switch, but we will add $20 to your prize if you stay with the box you first selected. Then whether or not the box was revealed accidentally or not has a material effect on your decision.
This is some Derren Brown shit in here.
It's basically just a mindfuck.
Recreate the problem on your desk with 3 equal sheets of paper, with 1 lightly marked with a pencil face down, and the solution and explanation becomes pretty obvious.
This puzzle is clear and open. As GMML points out.
Try this.
We have 300 boxes, all empty, except 1 with $$£££.
You chose 1 box and the host opens 298 , revealing them to be empty.
We are left with 2 boxes , the one you chose at the beginning and the other one that the host hasn't opened yet.
Do you think you picked the money box from the off ?
Or do you switch ?
Jeeze.
lol, if host chose at random it's still pointless switching.Quote:
Originally Posted by celtic123
Grunch
Switching is essentially choosing every other box but the one you're abandoning. Many will be empty and emptied before the final choice, but fuck 'em.
You ain't foolin' no one 'rand her'
http://i.imgur.com/jYEUDYZ.jpg
Yes, however in this case we have some other considerations. 298 boxes opened at random and none of them find the prize which is in the 299th box, the chance of that happening should be abysmal, our intuition tells us. So you're probably already sitting on the prize? Let's calculateQuote:
Originally Posted by rong
Odds of randomly opening only 298 empty boxes=
298/299 *297/298 *296/297 * ... *3/4 *2/3 *1/2= 1/299
This is a higher chance than your needed 1/300 to bink the prize, so it's still just 50/50.
WTF GUYS I /THREADED THIS 37 POSTS AGO
Due to ambiguous wording in the OP, you have only correctly answered one interpretation of the question.Quote:
Originally Posted by Luco
If I ever become a supervillian, this is the game I'd play with police and superheroes. Except the boxes would contain babies or something, and failing to find the baby lights it on fire. Because evil and what not.
This thread is still getting bumped? Are people just saying over-and-over again, "I've already heard this problem, and the answer is to switch"?
check out the history of people talking about this problem Monty Hall problem - Wikipedia, the free encyclopedia
ol marilyn vos savant started some shit
some clicking landed me to this one, lets call it part 2 of the thread
A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
Somebody save the babies!
A lawyer says he has two new babies to show you, but he doesn't know whether they're male, female, or a pair. You tell him that you want only a male, and he telephones the fellow who's setting them on fire. "Is at least one a male?" he asks him. "Yes!" he informs you with a smile. What is the probability that the other one is a male?
1/3Quote:
Originally Posted by gabe
Epic! <3 gabe
Oh shit, here we go
Lol fire. I get how we come to the answer but I still find it confusing.
Mindfuck. Thought process in spoiler tags:
Spoiler:
So intuitively my brain is like "1/2" but probably this Problem wouldn't be a Problem if that were true so I kept thinking (although it could be a level given the thread it's posted in). Couldn't get away from the 1/2 answer until I started writing things down, at which point 1/3 being the correct answer immediately became apparent. However, my brain still really wants the answer to be 1/2, so I kept thinking on it in case it was somehow some kind of tortoise and arrow trick.
When I asked myself what about coinflips, you flip two coins and one is heads, what're the chances the other is also heads, my brain is perfectly happy to accept 1/3 and sees that as logical and intuitive.
So now I'm sitting here wondering if the reason I can accept the cointoss but still cannot properly accept the puppyfire is that when the former was presented to me my gut (snap-reaction) said 1/2, whereas when the latter was presented to me I already knew the correct answer and so my gut (snap-reaction) said 1/3.
Really I'm just posting this whole thing because it alarms me if my brain is willing to betray me like this - hang onto something that it knows is wrong just because it was its first feeling on the matter.
edited to add: a) thanks for posting! and b) it's also possible the cointoss thing is easier for me (and others?) to see through because it's so easy to contrast "I flip two coins and one is heads, ..." with "I flip a coin and it's heads. I flip it again, ..."
Spoiler:If you flip two coins and one is heads, surely the other it's 1/2? So confuseddddddd
Is this a Jyms question? otherwise I don't understand why we are cooking the babies
In response to Pascal:
Spoiler:Also this is the same as for the puppies so if you don't want that spoiled, don't read on.
If you flip two coins and the FIRST is heads, the second being heads is 50/50. But if you flip two coins and EITHER one is heads the possibilities are:
First coin: H, Second coin: H
First coin: H, Second coin: T
First coin: T, Second coin: H
First coin: T, Second coin: T
except without TT since neither there is heads, leaving only:
HH
HT
TH
so only in one of the three equally likely cases is the other also heads.
Spoiler:The 4 possibilities are:
First coin: H, Second coin: H
First coin: H, Second coin: T
First coin: T, Second coin: H
First coin: T, Second coin: T
But you know that the first coin is heads, so you can remove both the predictions which involve the first coin being heads. Which therefore leaves:
First coin: H, Second coin: H
First coin: H, Second coin: T
Therefore 1 in 2 chance?
Spoiler:The "trick" or whatever to the whole thing is that it's not that the first coin is heads but that either one of the coin is heads. We don't stop flipping when the first coin is tails, because only the second coin being heads still meets the requirement of "at least one of the coins/puppies is heads/male".Quote:
Originally Posted by me
More succinctly: We don't know that the first coin is heads, we only know that one of either coins is heads, and therein lies the mindtrickery.
Spoiler:Ahh, that makes sense I think. I'm looking at the probability AFTER we've flipped the coin, whereas we should be looking at the probability from BEFORE we've flipped the coin?
Spoiler:The trick to the thought process IMO is that both babies were born at the same time, or both coins were flipped at the same time. So when you ask the nice chap setting the puppies on fire, you don't know whether you got the first one or the second one, so all 3 scenarios are still viable?
This would fail a common logic test though, because you know you can eliminate FF when you hear that at least one of them is Male.
Grunch, and I hadn't heard this before but I'm damn-near 100% certain this is correct:Quote:
Originally Posted by gabe
EDIT: Apparently we're doing spoiler tags now.
Spoiler:
33% if they checked both dogs and concluded that at least one had a penis; 50% if they only looked at one and that one certainly had a penis, so we have 0 information on if the other one does.
All of the possible permuations are as follows:
1. Dog 1 has a penis and Dog 2 has a penis; (25%)
2. Dog 1 has a penis but Dog 2 does not have a penis; (25%)
3. Dog 1 does not have a penis but Dog 2 does; and (25%)
4. Neither has a penis. (25%)
In the scenario that the bather checks both dogs, at least one will have a penis 75% of the time (the aggregate of the first 3 permutations). Only one out of those three permutations (33% of the time within that subset), though, do both of the dogs have a penis.
In the scenario where the first dog they check has a penis, and then it's up in the air whether the second one does or not, we're talking exclusively about permutations 1 and 2. In one out of THOSE two permutations, Dog 2 has a penis, which makes sense because we've been given zero information whatsoever about the dog, so its nature's probability of penis-ownership is unchanged.
This one feels like less of a paradox. We pretty clearly get less information out of someone saying they checked multiple dogs and at least one of them was a male than we get out of checking one dog and find that it is male.
EDIT after reading responses:
Spoiler:This may have been less of a mindfuck for me than it was for others to think it through methodically and find a different reaction. I think my various learning disabilities makes it so that I'm never surprised that thinking it through methodically gets me a different answer than the first one I blurt out, so I'm just like, "Surprise, surprise, my initial reaction was wrong like it so frequently is."
The Monty Hall problem, though, did fuck my mind for a long time just because it's SO counter-intuitive. It's much harder for me to wrap my brain around the concept of two doors' probabilities' combining than it is for me to wrap my head around the fact that we get weaker information from knowing that at least 1 out of 2 of the 50% probabilities are realized than we do out of knowing that 1 out of 1 of them are, so the chances of the 25% probability being realized are more marginally improved.
Spoiler:I also think this question is worded poorly. I was so hung up on what the hell you meant by "the other one" that by the time I realized what it was supposed to mean, I'd kinda realized that there were a couple of different scenarios and realized why they have different probabilities associated with them.
Wording it as, "What are the chances that both of the dogs are male, now that you know that at least one of them is?" "The other one" makes it so ambiguous because it seems to heavily imply that we're talking about one dog being male and one other one being in question, which kind of isn't really what's going on.
More males are born than females, at least among humans.
Yeah but it's like 53% male or something like that. Well within the boundaries of a coinflip like AK vs QQ.Quote:
Originally Posted by Lukie
Pretty sure the female babies are more flammable.
M, MQuote:
Originally Posted by gabe
F, F
M, F
F, M
If one is male
M, M
M, F
F, M
1/3
F babies taste better though