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(...) For any still not getting this one (it's tricky!) here's another way to analyze the problem. List all outcomes: Prize Choice Door Revealed Stay or Switch? 1 1 2 stay! 1 2 3 switch! 1 3 2 switch! 2 1 3 switch! 2 2 1 stay! 2 3 1 switch! 3 1 2 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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Dave, you rock. I am finally convinced. :) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) I think your analysis is incorrect, precisely because you did *not* list all outcomes. Choose Prize Door Revealed Possible choices 1 1 2 Stay (win). Switch (lose). 1 1 3 Stay (win). Switch (lose). 1 2 3 Stay (lose). Switch (win). 1 3 2 Stay (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Bruce Schlickbernd writes: <snip> (...) So, like others suggest about knowing that one of the marbles is red, the chance that the other one is red... the one that we know becomes basically irrelevant to the chance--there (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Liking it or not is irrelevant. The probabilities are what they are. And the answer is still 2/3 if you switch, given the premises. No matter whether you like it or not. Go to the site I gave you and read what is written there. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Actually, it supports the 50/50. Host will always offers regardless of whether you chose correctly or not (basic premise - choose the "host does not know"): simulator runs 50/50. If it's a screw you scenario, then yes, you switch, but I (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) You're also being incompletely rigorous. :) If you want to do an "all possible outcomes" probability check, you have to add in all the 'door revealed=prize' options which exist for the sake of the probabilities involved, but are never invoked (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Got it... Goes into a fairly comprehensible walk-thru, but the point that makes it obvious (as stated) is: " Imagine that there were a million doors. Also, after you have chosen your door; Monty opens all but one of the remaining doors, (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Actually, I explained (albeit briefly) why this isn't correct. You see, you now have 2 listings for "Choose 1, Prize 1", "Coose 2, Prize 2", and "Choose 3, Prize 3", when each deserves *equal* probability to each other possibility, like (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Which is why those are left out: there never enter the scenario. Only those that would actually happen are "invoked". And the simulator supports my conclusions. Only if the host "cheats" does the 2/3 option become statistically correct. :-) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) The host does cheat: he always reveals a losing door. That's part of the basic premise. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) This really bugged me about all the links Larry sent. They all claimed that there was some sort of "loophole" if you assume that the host has the option to either: - reveal the door you picked, showing a goat (in which case, yes, it's 50/50) - (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) That doesn't make sense at all-- If I choose door 1 and the prize is behind door 1, the host can choose to show me door 2 *or* door 3 - 2 possibilities If I choose door one and the prize is behind door 2, the host can *only* show me door 3 - (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Whew. You are way off in your probablity and statistics analysis: if you want to double everything, then you would have to list 112 and 113 twice also. They are *not* the same selection. (...) and 1 1 2 (...) and 1 1 3 (...) (etc. snip) (...) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) <mucho snippage> (...) Bruce - I see the confusion. You aren't solving the problem as presented: "Now, here's a trickier one: Ever watched the old game show "Let's Make a Deal"? Sometimes, they'd offer people things similar to this scenario. (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) That is not a cheat. -->Bruce<-- (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Doh. This is just coming back to haunt me. I should never have included the "door revealed" column. It's entirely irrelevant, and is only serving to confuse. A better list: Prize Choice Switch or Stay? 1 1 stay! 1 2 switch! 1 3 switch! 2 1 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Thank you for repeating the message I already read and applied to the situation. I did illuminate that it is not a valid Let's Make a Deal scenario, but it's still the one I'm talking about. -->Bruce<-- (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) See this one for a clearer picture: (URL)No, because he won't use the same algorithm since he then becomes (...) That's not the problem presented, though. According to the problem, it's KNOWN that he ALWAYS reveals a zonk prize door that you (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) If I were to think about it like a sequence of events, logically leading up to the end choice, it would go something like this-- The event starts-- Contestant picks a door -- 1, 2, or 3 If this is all there is to the contest, we have 33 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) though I admit it has been a long time since I have taken my probablity and statistics class. (...) Gosh, everyone is seizing on this: I didn't say it was and it has nothing to do with my answers!!! :-) (...) Sigh. Yes, I know that. (...) Yes, (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) think that way, though I'll note I passed Calculus). I come up with 2/3 if you switch. A 1/3 chance that you have the right choice initially. The other two doors are an aggragate 2/3 chance. Even though one is eliminated, it is a wrong one, so (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) This one here? (URL) many trials? It is a *probability* after all, no guarantees about a short run of trials. I ran 30 "stay" and 30 "switch" and got roughly 40 and 60 as the two percentage results. Moving to the serious number cruncher, which (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) I think I may have found the fallacy... (indulge me if you please) Here's a bag with 3 marbles in it--2 are red (goats, whatever), one is white (brand new car) Pick a random one out of the bag and hold it in your hand--don't look at it! What's (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Nope, you're assuming that the marble pulled out of the bag is r2. It might be r1. Better to leave off the numbers, as the red marbles are not distinct. hand bag r r w r r w w r r remove a red from the bag and it reduces to: hand bag r w r w w (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Basically, each column decreases in liklihood as more choices are added. Start with one colum: Prob. Prize 1/3 1 1/3 2 1/3 3 Now, you get to choose a door. Since the prize door doesn't affect the probability: Prob. Prize Choice 1/9 1 1 1/9 1 2 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Clear as mud, for then if the red is irrelevant, then there cannot be (...) there can only be (...) for if we remove either red ball from the bag, that very same red ball in this specific instance cannot be in the person's hand. We cannot take (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Why not? There are two different red balls, so each of those red balls is equally likely as the white ball to be in my hand. (...) No, because you only tried to remove r2. We are only removing *a* red ball from the bag; because there are 2 red (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Bleargh! typo. r2 w (r1 is the only red ball, it is removed) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) If you keep going on and on and on... well I probably still will, unless someone explains why I should not. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) This analysis assumes there is a percievable difference between the red balls, which as originally stated, there isn't. It's similar to the difference between combinations and permutations. The two permutations you've listed with a red ball in (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Saw that ;) but understood... but if we're talking one instance--one event--and that's what we're, indeed, talking about--you are on a game show, one event--you have one chance of winning one car behind one door.... so you have 3 doors I won't (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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(...) I think it's a great idea! Dave K (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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(...) Saw that ;) but understood... but if we're talking one instance--one event--and that's what we're, indeed, talking about--you are on a game show, one event--you have one chance of winning one car behind one door.... so you have 3 doors I won't (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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(...) Ok, here's where I'll take it from. The car is behind one of the doors, so the probabilities must add up to 1. 3 doors, 1/3 of a chance for each door. You pick door 1. There is a 1 in 3 chance the car is behind that door. There is a 2 in 3 (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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(...) Didn't mean to offend. Sorry 'bout that. I read the site thru again and your synopsis of the explanation above and I concur that it should be 2/3, but I like to work things thru... :) If I were to walk in off the street right after one of the (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
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