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k, here's something that came across my desk... you have a bag with 4 marbles in it. 2 of the marbles are red 1 marble is white 1 marble is blue you reach in and grab 2 random marbles you open up your hand a bit to only see the colour of one of (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) <long snip> you're right, in both senses. Your logic for the first solution is valid, and your identification of his error in the second solution (which would lead to the same answer) is also correct. It is, indeed, 33% Adrian -- (URL) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) I'm not a stats person, but I agree with your friend. r(2) r(1) is the same as r(1) r(2), because order doesn't matter. If it does, then you end up with more combos (w r(2), etc)... -- that's what I was going to say. But, I decided to run a (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) Here's his rationale, hopefully explained more fully (but I still think it's not quite kosher...) The time of the event--you grabbed 2 marbles there are 6 possible combinations in your hand r(1) r(2) r(1) b r(1) w r(2) b r(2) w b w those are (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) My knowledge of stats gives me less than 1 in one zillion chance of answering any random stats question correctly, but I likewise agree: The essential question is "If you reach into a bag with a red, a white, and a blue marble, then what are (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) I think he's wrong because he's trying to reason about partial knowledge, and that always causes problems. Look at it as a draw without replacement instead. The order you reveal things doesn't influence outcomes, they were what they were. you (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) Hmm... I just thought of something. Are there two separate questions at work here? Question 1: Having drawn two marbles from a bag of two red, one white, and one blue marbles, what are the odds that the two you have drawn are both red? (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) Yep, you're right, your friend is wrong. The problem is that you *wouldn't* count "r(2) r(1)" if order didn't matter. But then again, the thing he's forgetting is that if you don't count "r(2) r(1)", then you also CAN'T count *both* "r(1) b" (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) The original question: 4 marbles in a bag, 2 are red, one is white, one is blue You reach into the bag and grab 2 of them. You see that one of the marbles in your hand is red. What is the chance (either fraction or percent) that the other (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) Wow! You wrote a program! Thanks Dan! Again I'm impressed with the knowledge and skills of those that partake in LUGNET! Adding to my case, Dave K (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Larry Pieniazek writes: <snip> (...) Thanks Larry, I'll add it to the rationale. I failed gr. 12 advanced math once--was too busy being the editor of the yearbook that particular year (and playing cards in the cafeteria) It (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) Your friend has assumed that there is a difference between the reds, and you know which one you have seen in your hand. As there is no information in the original problem to substantiate this assumption, you have to assume it's not the case, (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Ross Crawford writes: <snip> (...) I like driving home from work--gives me time to think about stuff... And after reading all the great replies, I'm going to propose the following solution: If you have a bag with 4 marbles, (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, David Eaton writes: <snip> (...) I'd say 1 in 2, or 50 percent, but maybe I'm missing something. Good question though. Dave K (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) I don't quite get why this works-- if you're doing rand($#bag), won't that always pick r, r, or w for the 1st value? Hence, theoretically, you could pick "wb", which isn't in your sample set... on the other hand, you're increasing the chance (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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In lugnet.off-topic.geek, Dave Schuler writes: <snip> (...) I'd say that me drawing a red one is 1 in 3, and you drawing a red one is one in 3 so the chances are for us to draw a red at the same time is 1 in 9. The chances to draw a blue one at the (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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(...) doh, you're right :) Also, I was calculating the chance of getting 2 red marbles, in general, not checking the chance given the first one was red... :) That's what you get when you do an experiment, and have an expected solution in your (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) I think the two answers depend on a more accurate statement of the problem. If the problem is stated that you pick two marbles, and given that the first one you pick is red, what is the chance the second one is red, then order matters, and the (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Actually, there's a 2/3 chance that it's behind door Z! A very sneaky problem, in fact. It was shown to us in high school as the subject of a very minor controversy. After being published as "2/3 chance" in a math publication, some college (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) I already proved that I should stay out of this... but. When you had 3 doors to pick from, it was 1/3 that whatever door you choose is the correct one. However, after opening one of the doors you didn't pick (assuming the host is not opening (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Nope. The analysis that I have seen that argues for a different outcome than 1/2 has to do with the fact that you're being (or so it is argued) conveyed more information than it seems. That is, they KNOW which door is right, and by showing you (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) Inasmuch as you get to pick a second time, the odds are 50/50. If you did not get to pick a second time, the odds would be 1/3, not 1/2. Of course, this is leaving out the poker aspect: if your initial pick was incorrect, would they even offer (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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(...) See the link I posted and follow some of the links it has. You are in good company, lots of mathemeticians came to the same conclusion. (and roasted Marilyn Vos Savant about it at the time it first got a lot of publicity) That was before they (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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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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In lugnet.off-topic.geek, Larry Pieniazek writes: Another site I happened to stumble across (which is so chock full of math gadgets and java applets that I added it to the header...): (URL) this writeup of the problem, along with a simulator for you (...) (22 years ago, 2-Apr-03, to lugnet.off-topic.geek)
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(...) "Performing these experiments not only gives you some clues, it also slows you down from the common frenzy of everyday life, so you can focus on just one thing for a period of time." ROSCO (22 years ago, 2-Apr-03, to lugnet.off-topic.geek)
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