Aces High Bulletin Board
General Forums => The O' Club => Topic started by: Anaxogoras on May 13, 2009, 10:03:32 AM
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There are three doors. Behind one of those doors is a million bucks. Each of the other two doors hides a donkey. Here's how the game works: You pick a door at random, and then I open a different door and show you a donkey. You now have the option of keeping your original pick, or switching. What should you do?
A) Keep my original choice
B) Switch
C) It doesn't matter, the two options are equivalent
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Switch
Odds are much better.
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How angry are the bucks and how much ammo do I have?
Seriously, though.. Fresh off a statistical analysis test?
Your first choice implies a 33.33% chance of success.. with your second choice being 50% chance of success. Theoretically, you should rechose, since your odds are 'better'.
However, since your final choice is 50/50, then it's hit or miss.
I suggest eeny-meeny algorithm started by coin toss.
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I would guess C. The remaining 2 options should both yield a 50% chance of success.
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I know the correct answer is to switch, yet I have never been able to really understand why. I've always been good at math, (85-90th percentile) but this simple probabilty question always threw me for a loop.
Edit: So my answer is B.
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I'm in the same boat as Druski. Normally I have no problem with math but statistics are another story.
I understand the chances, your first pick you have a 1/3 chance and your second pick you have a 1/2 chance. Where I get lost is since one donkey is taken away, why does your first pick not automatically go to a 1/2 chance?
In other words once there are only 2 doors left, why do you have to re-pick to have your chances go up? If there are only 2 doors left and you have picked one, your chances are now 1/2 weather you re-pick or not.
I say option C. Even though the chances have changed, they changed when one door was eliminated, not when you re-pick.
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So you are saying there is a 66.66666666666666666667% chance of getting those rabbits or deer.
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I pick the door you opened. I always wanted a donkey! :D
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I pick the door you opened. I always wanted a donkey! :D
That's the spirit! I declare a winner.
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no benefit to rechoosing as the only difference that has happened is that you may have chosen one of the doors that didnt have a donkey behind it on your first attempt, thus eliminating the one opened with the donkey in it. switching now would not increase your odds of success nor would it diminish your odds of success. You currently have 50/50 odds with your choice of the original 3 doors, with only 2 remaining.
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no benefit to rechoosing as the only difference that has happened is that you may have chosen one of the doors that didnt have a donkey behind it on your first attempt, thus eliminating the one opened with the donkey in it. switching now would not increase your odds of success nor would it diminish your odds of success. You currently have 50/50 odds with your choice of the original 3 doors, with only 2 remaining.
See now this makes logical sense to me, but I'm certain I've heard people say switching increases odds. Can someone please give the definate answer here?
Edit: I'll answer my own question. I guess this is the best explanantion I've found to date. I still don't like it.
At first you have 33.33% chance of choosing the right door and there is 66.67% chance of the prize being somewhere else. You know that Anax is going to open an empty door so when he does, this should not change a thing about your belief of your door being the right one. You still have 33.33% chance of having selected the right door. Thus there must be 66.67% chance of the prize being somewhere else (behind the last door).
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Since you don't seem to have a big enough headache..
Suppose Anoxagoras RANDOMLY chose the door and it had a donkey. Now, what are the odds that you have bucks or a donkey behind your door?
If you keep the door you chose, then you be betting that the first 2 doors Anoxagoras chose would both be donkeys. Probably less likely than one of each.
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Thanks for the replies. I've always thought this was a fun problem. I can remember arguing about it for almost an hour with a buddy of mine until we actually tried a version of the game with playing cards. Once it becomes empirically obvious that switching is better (evident after only 20 trials or so), you become more open to the mathematical answer, counter-intuitive though it seems.
When you first choose a door, your odds are necessarily 1/3. When I open a door and show you a donkey, the odds that your initial guess was correct are still 1/3, but the other (unseen) door must have a million bucks 1/2 of the time. Therefore you switch.
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When you first choose a door, your odds are necessarily 1/3. When I open a door and show you a donkey, the odds that your initial guess was correct are still 1/3,
This is the part I don't get.
If you show me one donkey, then the odds of my initial guess being correct automatically go up to 1/2.
There are 2 doors left, and I have 1 picked, how are my odds not 1/2? Weather I pick again or stay with my original, the numbers are the same. 1/2.
Another part that adds to the confusion is why is simply 'switching' better? Why is the opposite of what you have currently in the two remaining options is somehow better?
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If you show me one donkey, then the odds of my initial guess being correct automatically go up to 1/2.
Nope! :D Your odds are not 1/2 because you made your guess with 3 possible choices. What I show you has nothing to do with the probability of your first guess being correct.
Try the game with playing cards. Do 20 trials where you keep the same choice, and then do 20 more where you switch.
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Nope! :D Your odds are not 1/2 because you made your guess with 3 possible choices. What I show you has nothing to do with the probability of your guess being correct.
My odds are 1/2 because there are 2 doors, and I have picked one.
One door has the money, the other does not. After you eliminate the first choice, I now have a 1/2 shot of winning weather or not I switch.
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I picked the first door. Where's my donkey? :noid
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I just did the experiment, but I am by myself so I had to modify it.
I placed 3 pieces of cardboard face down, only one has a money sign drawn on it.
I shuffled the cards, then picked one at random.
Next I eliminate one card, and note if it was the money card or not.
Then I turned over the card I picked and noted if I won or not.
After all this I repeated the process, except I turned over the other remaining card that I did not pick.
My results are pretty much even. About 1/3 of the time I win, 1/3 of the time I loose, 1/3 of the time I accidentally eliminate the money card, ruining that specific test. If I eliminate all of the tests that were ruined, 1/2 the time I win.
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All I ask is for someone to explain mathematically why: I have one door picked out of 2 remaining doors. One has the prize, one does not. No matter if I picked the door before the others were eliminated, after, or was simply randomly assigned that door at any time, my chances are still now 1/2.
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Glass half full versus half empty if you ask me.
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You need to do the experiment with a friend, and try it from both sides.
Here's another way to think about it. Instead of 3 doors, there are 100 doors. You pick one, and I open 98 doors and show you 98 donkeys. Do you want to switch or keep your original choice? What are the odds that out of 100 doors, your choice is correct when I've eliminated 98 possibilities and left one extra remaining? About 99:1. ;)
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3 Doors, 2 donkeys.
I chose one.
Anax opens a door - donley.
2 doors left, one has a donkey.
The chance my door has the million bucks is NOW at 50% - not 33%. Even without having changed my choice. The fact alone that Anax had opened a door has increased my chances.
The original problem is more complex, as the hosts knowledge and behaviour plays an important role. See "Monty Hall problem" ;)
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You need to do the experiment with a friend, and try it from both sides.
Here's another way to think about it. Instead of 3 doors, there are 100 doors. You pick one, and I open 98 doors and show you 98 donkeys. Do you want to switch or keep your original choice? What are the odds that out of 100 doors, your choice is correct when I've eliminated 98 possibilities and left one extra remaining? About 99:1. ;)
Yahhh... ummm.. errrr.. no.
Ultimately, you have 1 last choice between 2 doors. You could have 100 billion doors/donkeys before hand, but the final choice is a fresh one between 2 doors.
You still have a 50/50 chance of choosing the right door.
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There are three doors. Behind one of those doors is a million bucks. Each of the other two doors hides a donkey. Here's how the game works: You pick a door at random, and then I open a different door and show you a donkey. You now have the option of keeping your original pick, or switching. What should you do?
A) Keep my original choice
B) Switch
C) It doesn't matter, the two options are equivalent
B. When you switch who have a 50/50 chance.
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Yahhh... ummm.. errrr.. no.
Ultimately, you have 1 last choice between 2 doors. You could have 100 billion doors/donkeys before hand, but the final choice is a fresh one between 2 doors.
You still have a 50/50 chance of choosing the right door.
Exactly. And even if you stay with your original door, after all the other doors are eliminated you still have a 1/2 chance during the final draw.
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Watch "Deal or No Deal".
As the suitcases get eliminated, and the million dollar winning case is still unaccounted for, the offers for the case you have already picked go up.
Thats because as there are less cases left, there becomes a higher chance that you may have the big money. If there are only two cases left, and its either $1 or $1 million, the 'bank' will likely offer you somewhere around $500,000, reflecting the current odds you have the million. They do NOT offer you $20,000, which would be reflecting the 1/50 chance you would have at the very start (before 48 cases are eliminated).
That means in the final draw of 2, even after a thousand cases or doors are eliminated, and assuming the winner is still unaccounted for, you have a 1/2 chance of winning WITH OUT re-picking.
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And just to put it one other way in case this is not confusing enough.
My uncle had 8 boys. He said he must have some luck to have 8. I explained that even though he had already had 7 boys in the past, when his 8th was conceived he still had exactly a 50/50 chance of it being a boy or girl.
That means despite preconditioning circumstances, when you are figuring the odds you can only consider the current possibilities. If there are two doors left, and one is the winner, despite overcoming any previous odds, your current odds are 1/2 no matter what door you pick, nor when or how you picked it.
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I say there's a 25% chance. :devil
1.) You stay after a donkey is shown and you win the money.
2.) You stay after a donkey is shown and you lose.
3.) You switch after a donkey is shown and you win the money.
4.) You switch after a donkey is shown and you lose. Making you look like a real...uh...donkey :)
4 options / 1 choice = 25% ! Am I right lol?
Seriously though. Cherry picker happens to be right.
It's all based off the likelihood that you chose wrong the first time because the 66% chance you missed your first guess is > the 33% chance you nailed it.
The probability of 1 of those 3 doors having the money is 100%. 1/3 + 1/3 + 1/3 . So when you make your first guess you have 1/3 that you're right and 2/3 you're wrong. When you are shown a donkey it doesn't magically become a 50/50 shot because that's not the probability you started with(you're working with thirds!). What you do have though is two doors, your original with a 1/3 chance of having the money and the other with a 2/3 chance of having the money.It really does work if you do the experiment. And if you're like hammer and would chose the door he showed you with the donkey, well there's no helping you.
-Zap
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When you are shown a donkey it doesn't magically become a 50/50 shot because that's not the probability you started with(you're working with thirds!).
No, it doesn't "magically" change, it changes because one possibility was completely eliminated, changing the odds of the current situation.
Again, like I say why does the deal or no deal "bank" make higher and higher offers for your case as other cases are eliminated and the million doesn't show? If its like you guys say, then the offer should be $20k from the start and never change as cases are eliminated (1 choice divided by 50 possibilities multiplied by 1 million dollar max possibility = $20k). After 47 cases are eliminated, and the million is still on the board, they offer roughly $333k for your original choice. Then if one more is eliminated, the offer goes up to $500k for your original choice. This is because every time they eliminated a case that was not a winner, your odds of winning actually go up.
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If you are watching a game of single elimination musical chairs with 50 people in Vegas, you have a slim chance of picking the winner, the odds would pay 50 for 1. If you pick a person, and he makes it to the final round, and you go back to bet more money on that person, the odds would pay 2 for 1. Thats because the odds of that person winning are now 50%.
This doesn't change the fact that you still won 50-1 on your first bet, but in that final match bet, your chances of winning were exactly 50%. Therefor; even if you did NOT bet in the final match, and only let your initial bet ride, your chances winning the final match (assuming you made it that far) would still be exactly 50%.
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The situation is different because you can't change cases after every round of eliminations. Wonder why?
-Zap
EDIT: I guess what I'm trying to say is that in the donkey example you have 1 chance in 3 of winning anything. But because you can still make 500,000 in deal or no deal they approach it from a different mathematic angle. Deal or no deal and the donkey game arn't so closely related. Here is a website where you can give the original question a try.
http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html (http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html)
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(http://i138.photobucket.com/albums/q257/candlexlight/3401099.png)
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Argh, I've been shown the witty picture :( .
-Zap
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The two things that make this confusing are:
1. The "host" will always show you a donkey. This makes the odds of the other two doors, one of which is yours 2 in 3 of having the $
2. Rush said it best, "If you choose not to decide, you still have made a choice."
Meaning, if you switch you choose. If you don't switch. you also choose. 50-50.
I have a headache
wrongway
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you have a 1 out of three chances if you pick one door....... YOU only have qa 50% chance if you make a new choice...
think of it backwards if you stay with you first choice your chance is still 1 out of three even if your shown a donkey.....
hope this helps
Ill give you something else to think about that is counter intuative......... Your refigerator does NOT cool your food..... Can any one answer with what it does do?
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http://www.youtube.com/watch?v=mhlc7peGlGg&feature=related
:aok
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three words.
Drugs are bad.
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you have a 1 out of three chances if you pick one door....... YOU only have qa 50% chance if you make a new choice...
think of it backwards if you stay with you first choice your chance is still 1 out of three even if your shown a donkey.....
hope this helps
Ill give you something else to think about that is counter intuative......... Your refigerator does NOT cool your food..... Can any one answer with what it does do?
removes warm air
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http://www.marilynvossavant.com/articles/gameshow.html
shows how switching is better odds.
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I reckon there's 3 donkeys and no money
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The airplane won't fly!
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The airplane won't fly!
Ha! Exactly. Though I admit I didn't understand why it would at first...took me a bit to get that one. This one I'm not sure I'll ever get. :uhoh
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think of it backwards if you stay with you first choice your chance is still 1 out of three even if your shown a donkey.....
hope this helps
Think of it sideways. Two doors. One with the money, one with the goat.
You pick a door and the host does nothing but asks if you want to switch.
How does this make your odds better to switch? It's the same thing, isn't it? The goat is behind one of the two remaining doors. The same as when you originally picked.
With the three doors, they still open one of the goat doors no matter what. So, in effect, the first pick is 50/50 too after they open the first goat door.
I get the 2 in 3 chance to pick a goat in the first round but like I tried saying with the "even if you don't make a choice" thing is, now there are two doors, 50/50 either way.
Keep the original, 50/50.
Choose the other, 50/50.
That's why I can't wrap my head around the "choose again" concept.
wrongway
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There are three doors. Behind one of those doors is a million bucks. Each of the other two doors hides a donkey. Here's how the game works: You pick a door at random, and then I open a different door and show you a donkey. You now have the option of keeping your original pick, or switching. What should you do?
A) Keep my original choice
B) Switch
C) It doesn't matter, the two options are equivalent
Sounds like someone just got done watching 21.
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What is confusing everyone is that they think their odds go up to 50/50 after the third choice is eliminated.
Think about it like this: You randomly pick one of the 3 choices, so your odds of getting it right are 1 in 3. If you correctly picked, (which should occur about 1/3rd of the time) and you switch (which you should) you lose. However, the odds are that 2 out of 3 times you pick the wrong one so that switching should give you a winning percentage of 66%. (TWICE your original odds of 1 in 3)
Clear as mud..
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What is confusing everyone is that they think their odds go up to 50/50 after the third choice is eliminated.
Think about it like this: You randomly pick one of the 3 choices, so your odds of getting it right are 1 in 3. If you correctly picked, (which should occur about 1/3rd of the time) and you switch (which you should) you lose. However, the odds are that 2 out of 3 times you pick the wrong one so that switching should give you a winning percentage of 66%. (TWICE your original odds of 1 in 3)
Clear as mud..
Exactly!!
wrongway
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I figured out the whole misunderstanding yesterday.
It's all relative to what percentages you're describing.
I agree that the chance that your first choice is correct is 1 in 3 or 33.3333.. %
I also agree that by that principle, the odds would dictate that you should switch your selection, since you were most likely wrong on your first choice.
However, when you say 'what are your chances of being correct' after the first donkey is exposed, the result is 50/50.
It's really how you word it.
Sorta like 2+2=5, for large values of 2.
90 is 10% less than 100... but 100 is 11% more than 90.
Woohoo!! Happy Friday!!
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After the donkey is shown your chance is still 1out of 3 not 50/50
Your first choice locks you in at 1/3 just because you see a donkey does not change the choice you made... Its the same choice.. The same choice is the same odds of 1/3......
think of it this way if you were shown the money instead of a donkey..... you would only be right 1/3 of the time........ To get the money 50% of the time you would still have to switch doors. If you stay with your first choice your only going to get the money 1/3 of the time...
<S>
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My conundrum (still) is that the third door is irrelevant as is your first choice.
2 doors. 1 donkey. 1 $1million. Pick one door.
Are my odds still 1 in 3?
Say I picked door #1 with three doors. 66.6% donkey. One donkey is gone now. Door #1 = 50% donkey, just like the other door that is left.
If you have two doors to begin with and choose one, and were offered to switch, would you?
I get the "three doors, switch" thing but I, and I'm sure other naysayers, see the second choice as a choice of two doors starting from scratch.
(http://www.smileyvault.com/albums/basic/smileyvault-atomic.gif)
wrongway