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Caliban Special user 727 Posts |
Quote:
On 2010-07-02 17:48, Philemon Vanderbeck wrote: Actually, the odds in the Monty Hall Problem DON'T change. There was a 1 in 3 chance that your box contained the money (or whatever) at the time you chose it - so there is still the same 1 in 3 chance that it contains the money after another empty box has been eliminated. The point of it is that people think the odds should have changed to 1 in 2 because there now are only two boxes - but actually the odds remain exactly the same. I appreciate that to anyone who is not already familiar with the Monty Hall Problem, this all makes no sense whatsoever! |
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Philemon Vanderbeck Inner circle Seattle, WA 4694 Posts |
However, the odds that the box you didn't choose goes up to 2 out of 3.
Professor Philemon Vanderbeck
That Creepy Magician "I use my sixth sense to create the illusion of possessing the other five." |
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Garrette Special user 926 Posts |
Quote: I may be mistaken on the technicalities, but I only began to grasp the Monty Hall problem when I recognized that this is not the case. The odds of the prize being in the SET of two doors you did not choose remains at 2 out of 3; since you've been shown part of that set--and that part does not have the prize--the remaining door can be seen as having 2 out of 3 odds but only because it still represents the whole set you did not pick, not because it I a single door with increased odds.
On 2010-07-02 18:37, Philemon Vanderbeck wrote: All of which is dependent on Monty knowing that the door he showed you would be empty; it changes if he did not know and simply showed you a door which happened to be empty. I could be wrong, though. |
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Philemon Vanderbeck Inner circle Seattle, WA 4694 Posts |
Yes, the Monty Hall problem is based on the principle that the revealer of the options knows where the prize is hidden.
So, to apply this to a Russian Roulette type routine, the performer "knows" which chamber contains the bullet.
Professor Philemon Vanderbeck
That Creepy Magician "I use my sixth sense to create the illusion of possessing the other five." |
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Garrette Special user 926 Posts |
Hmmmm..... I'm not sure.
If the performer knows where the bullet is then it is akin to Monty offering himself the prize for picking the right door. That's not the case, so I think for it to be similar to Monty Hall, then Monty must be involved. Let's discuss two scenarios: 1. The performer does not know where the bullet is and pulls the trigger on five consecutive chambers. In this case, the odds are as discussed, i.e., 5:1 then 4:1 then 3:1 then 2:1 then 1:1. 2. The performer does not know where the bullet is but Monty Hall is observing and does know. The perfomer picks five chambers to pull the trigger on, but Monty chooses in what order he does it. Further, Monty will always choose an empty chamber for the trigger pull if an empty chamber is still available. In this case, the odds remain 5:1, 5:1, 5:1, 5:1 and 5:1. That's my layman's take anyway. Perhaps Caliban or another will jump in to point out if I have it right or wrong. |
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Caliban Special user 727 Posts |
Quote:
On 2010-07-03 08:22, Garrette wrote: If I understand your idea correctly, surely the odds of actually being shot would be zero on the frst 4 shots (because at least 4 of the 5 chambers selected must be empty and he will always choose an empty chamber if one is available) - but the odds of the last shot being the real bullet are 5:1. I think ... |
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Garrette Special user 926 Posts |
Good point. I didn't think it through.
I suppose to make it work I (let's assume I'm the performer) would spin the cylinder with the understanding that the trigger would be pulled on the first five chambers before it is pulled on the sixth, but the order of the first five is up in the air. I get ready to pull it on the first chamber but you (Monty) say "Wait. Let me show you that chamber two is empty." In that instance, the odds on my chamber are still 5:1. They have not decreased to 4:1. If you also show me that chamber 3 is empty, the odds on the first chamber are still 5:1 and so on. Howzzat? Edit: Should I have said "increased to 4:1" instead of "decreased to 4:1?" |
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acesover Special user I believe I have 821 Posts |
The odds are only 5 to one if you look at the final chamber first. If you were to wager on a 5 or 6 to 1 bet where each person puts up $6 and the person starts to cllck on the first chamber and if it is not there he gives the other person a dollar if not in the second chamber he gives the person another dollar etc. but if the bullet comes up the other person gives him $6. this is where this goes awry and is not an 5 or 6 to one deal because as soon as the bullet chamber comes up the changing of money is over. Which means that on average the bullet will be in the last chamber 1 in 6 times and the other 5 times it will be in the 1 to 5 chamber which will come up a minium of 5 times and a max of 25 times in theory. So it is only a 5 to 1 if you try and pick the bullet chamber first and end it there.
So if this is done Russian Roulette style there are going to be a lot more than 5 to one losers lying around. I think.....
If I were to agree with you. Then we would both be wrong. As of Apr 5, 2015 10:26 pm I have 880 posts. Used to have over 1,000
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Garrette Special user 926 Posts |
No. 5 in 6 times the bullet will be in one of the first five chambers, regardless. 1 in 6 it will not.
You can test this manually with numbered envelopes and a six-sided die. Roll the die to indicate which envelope holds the virtual bullet. Roll it again to choose on which chamber you will first pull the trigger. 5 in 6 times, the bullet will be in one of the first five envelopes. 1 in 6 times it will not. To simulate your scenario you have to add a second person who will roll the first die for you so that you don't know where the bullet is. Roll your own die to select an envelope to start with, pull the virtual trigger and ask your partner if the bullet fired. If not, continue. You will find the same results as before: 5 times in 6 you will fire the bullet in one of your five chambers. |
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acesover Special user I believe I have 821 Posts |
I think we have said the same thing but in different ways. It is hard to explain.
What I do know is that I have a headache.
If I were to agree with you. Then we would both be wrong. As of Apr 5, 2015 10:26 pm I have 880 posts. Used to have over 1,000
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Caliban Special user 727 Posts |
Quote:
On 2010-07-03 20:48, acesover wrote: Hang on ... let me get this straight. So you have two players - and as it's 5:1 - let's say they each put up $5. In your scenario, player A is essentially making a $5 bet on the bullet being in the last chamber, while player B is placing a $1 on each of the other chambers. If chamber 1 is empty then player B gives a dollar to player A and so on ... but if the bullet comes up in chamber 3, then player A gives his $5 to B. Is that is? That would mean, although it's a 5:1 bet, there are two reasons why the money doesn't work out under those rules. 1) Player B is not betting $5 at 5:1 - he's making 5 separate 5:1 bets of $1 each. As soon as you understand that, B's odds make perfect sense. 2) If the bullet is in the final chamber, then player A will only have won $5 from player B - the five $1 bills. BUT he bet $5 on that chamber at odds of 5:1. Really, of course, he should win $25. If you pay him the true winnings for the odds then the money all works out. Posted: Jul 4, 2010 9:30am Or to put it another way ... let's simplify things by only having one player and classing the other person as the banker. I think Aceover's point is this: if the player bets a dollar on each chamber and he loses his dollar every time an empty chamber is fired but wins $5 if it's the bullet - then 5 times out of 6 he will win the $5 before he's gambled $5. So he's guaranteed to make a profit. The problem with this analysis is that the 5:1 odds are for the entire sequence of firing 5 empty shots and the bullet being in the last chamber. So, for the odds to be 5:1, the player must only bet on one chamber that he commits to before the first shot is fired. An alternative is that he could make separate 5:1 bets on several different chambers - but then he would only win the $5 if the bullet was fired from one of the chambers he had agreed to put a dollar on before the first shot was fired. That means he would only win $5 on the bullet being in chamber 4 if he had agreed in advance to put a dollar on that chamber. That also means, of course, that he would lose a dollar on chamber 4 if the bullet was fired before that chamber was reached. The reason the player comes out ahead in Aceover's example is that he is paid out at 5:1 on bets that he doesn't place until part of the sequence has already happened - and the odds are different by that point. For instance, if the bullet were in the 5th chamber, he would bet a dollar on that chamber and win $5 (5:1). But really he should only win $1 because he didn't commit to placing the bet on that chamber until AFTER the first 4 chambers had already been fired. By that point in the sequence, the odds are 1:1. |
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