Non-Poker Probability Paradox |
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Posted Mon Jan 23, 2006 10:24 pm GMT by xDiamond_CutteRx
Ok, consider these two situations and tell me if you can figure them out, because I can't... it's rather paradoxical.
1. You're on a game show where you get to keep whatever is behind one of three doors for you to choose from. Two doors have goats behind them, and one door has a car behind it. When you go to choose door #1, the host stops you, shows you there is a goat behind door #3, and asks if you want to switch your choice. Should you? It seems like your new choice should be 50-50, but theoretically, you should change, because you were probably wrong the first time. But why should this matter?
2. You're a prisoner of war, and there are 2 other prisoners, Smith and Jones, in your cell. The guards tell you that 2 of you will be shot by the end of the week. You should therefore have a 33% chance of survival. But, you ask one of the guards which of your cell-mates will die, and he tells you that Smith will die (you know for sure that at least ONE of them will be shot). Are your chances of survival now 50%? Either way, you end up with a paradox.
Supposedly, the answer is mathematical, but it defies conventional logic.
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Posted Mon Jan 23, 2006 10:32 pm GMT by supafrey
The first one is easy -
Original decision = 1/3 chance of it being right.
When he basically gets rid of a choice, it's now 1/2. =)
Posted Mon Jan 23, 2006 10:42 pm GMT by xDiamond_CutteRx
| supafrey wrote: | The first one is easy -
Original decision = 1/3 chance of it being right.
When he basically gets rid of a choice, it's now 1/2. =) |
Yes, but that's not the problem. It shouldn't matter whether you change or not because it's 50-50, but the mathematician who published this said it does matter, and that you have better EV by changing to door #2.
Posted Tue Jan 24, 2006 1:11 am GMT by Sean_in_NJ
It's the Monty Hall problem.
It's been discussed here in the past.
Posted Tue Jan 24, 2006 1:46 am GMT by supafrey
And exactly for the reasons I just said. What don't you get diamond?
Posted Tue Jan 24, 2006 7:35 am GMT by suitedaces84
Supa,
You're wrong. At first I didn't get it either. Use the link Sean posted and you'll understand why.
Posted Tue Jan 24, 2006 8:02 am GMT by tame_deuces
Nice link, I've read up on this thing in the past, but it was nice to see it 'in action'.
Posted Tue Jan 24, 2006 9:50 am GMT by supafrey
oh. so it's even better odds than I first surmised.
With my supa-luck though, my advice was still right.
Posted Tue Jan 24, 2006 1:11 pm GMT by Geno
I do accept the theory but that links proves nothing as it might be rigged for all we know 
Posted Tue Jan 24, 2006 2:38 pm GMT by TallBrad
| xDiamond_CutteRx wrote: | Ok, consider these two situations and tell me if you can figure them out, because I can't... it's rather paradoxical.
1. You're on a game show where you get to keep whatever is behind one of three doors for you to choose from. Two doors have goats behind them, and one door has a car behind it. When you go to choose door #1, the host stops you, shows you there is a goat behind door #3, and asks if you want to switch your choice. Should you? It seems like your new choice should be 50-50, but theoretically, you should change, because you were probably wrong the first time. But why should this matter?
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Somebody was watching the television show Numb3rs last weekend. 
Posted Tue Jan 24, 2006 4:26 pm GMT by Geno
I finally sat down and worked out why this paradox is true. It took all of 2 minutes but I got there. Yay the new me, boo the old me :D
Posted Wed Jan 25, 2006 2:34 am GMT by Cyberhwk
The Monty Hall Paradox is some cool stuff!!! :D
Also think about this...
If there is infinity unique points between 1 and 0 (i.e. 1.1, 1.01, 1.001, 1.0001, etc.), then it stands that there should also be infinity points between 1 & 2. But wouldn't there be 2 X Infinity points between 1 & 2? Can't be because infinity would have to be finite, which it is not by definition.
...and don't even get started about .999999999 equaling 1. 
Posted Wed Jan 25, 2006 12:12 pm GMT by galderon
| xDiamond_CutteRx wrote: | | 2. You're a prisoner of war, and there are 2 other prisoners, Smith and Jones, in your cell. The guards tell you that 2 of you will be shot by the end of the week. You should therefore have a 33% chance of survival. But, you ask one of the guards which of your cell-mates will die, and he tells you that Smith will die (you know for sure that at least ONE of them will be shot). Are your chances of survival now 50%? Either way, you end up with a paradox. |
I've never heard the Monty Hall problem posed as the three prisoners before, so I figured I'd work it out.
Your chances of survival are still 33%. There are three choices:
Smith and you
Jones and you
Smith and Jones
Assume the guards roll a six-sided die to choose:
1-2: Smith dies, and YOU DIE.
3-4: Jones dies, and YOU DIE.
5-6: Smith and Jones die, and YOU LIVE!
Since you don't care about Smith or Jones, you can simplify it to:
1-4: YOU DIE.
5-6: YOU LIVE.
They roll the die, and then tell you that Smith is gonna get it. You know they rolled a 1, 2, 5 or 6. Now, they're ready to shoot the second guy. They DON'T reroll. They look at the die they rolled before, and it's still 1-4 you die, 5-6 you live. The confusing part is that the probability is revealed to you in stages, even though there is just one probability.
Posted Wed Jan 25, 2006 12:22 pm GMT by galderon
| Cyberhwk wrote: | | If there is infinity unique points between 1 and 0 (i.e. 1.1, 1.01, 1.001, 1.0001, etc.), then it stands that there should also be infinity points between 1 & 2. But wouldn't there be 2 X Infinity points between 1 & 2? |
I think you mean 0 and 2 here?
| Cyberhwk wrote: | | Can't be because infinity would have to be finite, which it is not by definition. |
I have no idea what this statement means.
| Cyberhwk wrote: | | ...and don't even get started about .999999999 equaling 1. |
I think you mean .9 repeating here. For two numbers to be different, you need to be able to find a number in-between them. There is no number between .9 repeating and 1, therefore, they are the same number.
Posted Wed Jan 25, 2006 12:55 pm GMT by KingOHearts
This isnt probablity per se, but its a pretty well known little math teaser. Drove me nuts several years ago the first time I heard it.
More fun can be found here -> http://www.jimloy.com/puzz/puzz.htm
Three people are eating at a restaurant. The waiter gives them the bill, which totals up to $30. The three people decide to share the expense equally ($10 each), rather than figure out how much each really owes. The waiter gives the bill and the $30 to the manager, who sees that they have been overcharged. The real amount should be $25. He gives the waiter five $1 bills to return to the customers, with the restaurant's apologies. But, the waiter is a dishonest man. He puts $2 in his pocket, and returns $3 to the customers. Now, each of the three customers has paid $9, for a total of $27. Add the $2 that the waiter has stolen, and you get $29. But, the original bill was $30. What happened to the missing dollar?
Posted Wed Jan 25, 2006 2:21 pm GMT by TallBrad
| KingOHearts wrote: | This isnt probablity per se, but its a pretty well known little math teaser. Drove me nuts several years ago the first time I heard it.
More fun can be found here -> http://www.jimloy.com/puzz/puzz.htm
Three people are eating at a restaurant. The waiter gives them the bill, which totals up to $30. The three people decide to share the expense equally ($10 each), rather than figure out how much each really owes. The waiter gives the bill and the $30 to the manager, who sees that they have been overcharged. The real amount should be $25. He gives the waiter five $1 bills to return to the customers, with the restaurant's apologies. But, the waiter is a dishonest man. He puts $2 in his pocket, and returns $3 to the customers. Now, each of the three customers has paid $9, for a total of $27. Add the $2 that the waiter has stolen, and you get $29. But, the original bill was $30. What happened to the missing dollar? |
There is no missing dollar. The customers have $3, the waiter has $2 and the restuarant has $25. That totals $30.
Posted Wed Jan 25, 2006 2:55 pm GMT by Dave B
Or, you could say:
each person was -10, rest +30 = 0
after:
each person now -9, rest +25, waiter +2 = 0
The trick is when people try to add the -9s to the plus 2 and ignore the +25
Posted Thu Jan 26, 2006 2:53 am GMT by Cyberhwk
| galderon wrote: | | Cyberhwk wrote: | | If there is infinity unique points between 1 and 0 (i.e. 1.1, 1.01, 1.001, 1.0001, etc.), then it stands that there should also be infinity points between 1 & 2. But wouldn't there be 2 X Infinity points between 1 & 2? |
I think you mean 0 and 2 here?
| Cyberhwk wrote: | | Can't be because infinity would have to be finite, which it is not by definition. |
I have no idea what this statement means.
| Cyberhwk wrote: | | ...and don't even get started about .999999999 equaling 1. |
I think you mean .9 repeating here. For two numbers to be different, you need to be able to find a number in-between them. There is no number between .9 repeating and 1, therefore, they are the same number. |
Yes.
What it means is...if there is infinity points between 0 and 1, then (you're correct that's a typo), there should be 2 X infinity points between 0 and 2. But , how can one have more than infinity? (Yeah, this one's kind of lame. Solution = infinity is a abstract concept, not a number).
Ahhhhhh...I know there's a proof but that's a good simple way to explain it, thanks.
Posted Thu Jan 26, 2006 11:11 am GMT by galderon
| Cyberhwk wrote: | | What it means is...if there is infinity points between 0 and 1, then (you're correct that's a typo), there should be 2 X infinity points between 0 and 2. But , how can one have more than infinity? (Yeah, this one's kind of lame. Solution = infinity is a abstract concept, not a number). |
Now you're getting into advanced infinity math! Well, for one thing, there are different kinds of infinity, but the two you describe aren't different. Infinity is kind of a number, but it doesn't behave the same way with regards to math. This is where your false paradox comes from (because Infinity * 2 = Infinity).
I could go on, but I won't, since we're really far from poker at this point.
Posted Sun Jan 29, 2006 7:40 pm GMT by snoogins47
| xDiamond_CutteRx wrote: | | Supposedly, the answer is mathematical, but it defies conventional logic. |
Or, it doesn't defy any form of logic, it's just not intuitive to most people.
btw, supa, your first few posts were missing the point. The point isn't that your odds increase, that's pretty obvious: with one choice narrowed down, a completely random choice will fare better than when there were all three doors. The point is that switching your door selection is far and away a better strategy than sticking with your original choice.
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