Log-in members only
RE: OT On the subject of logic & riddles page 9This is a discussion thread · 105 replies ramsey: [nq:1]Maybe somebody can explain it in a new way so I "get it".[/nq]Simplify the problem. You get to pick a door. Then Monty will ALWAYS offer you the opportunity to EITHER open that door OR open ALL the other doors. Suppose there are 1million doors. Your chance of picking the right door is 1 in a million. If you switch there is a 99/100 chance of getting the prize. So the switch must be right. If there are 3 doors there is a 1/3 chance of picking the right one. There is a 2/3 chance it is behind one of the 2 doors you didn't pick. The fact that Monty opens some doors and they are empty simply has no relevance providing that the wording of the problem is such that he knows where the prize is, and he will never open the door with the prize behind. ramsey
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
ramsey: R8480 (Email Removed) writes[nq:1]I explained it... what is not clear? Let's say A is the prize and B and C are gags. Situation ... or C (one of the gags). If you switch, you get the other gag and lose the prize you picked.[/nq] Although there is nothing in the wording of the problem that says he will pick B or C at random. If he always opens B if it doesn't contain the prize then when he does open B there is no advantage in switching (and you should keep A if there is a cost associated with switching). Of course in this situation when he opens C the prize is guaranteed to be in B. ramsey
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
KSias: The way I see this problem is if you look at the first choice - you pick door A. If you switch you effectively get doors B and C pre pick. Whichever one Monte shows you you would pick the other one and get both for 2/3 odds. In the example of 100 doors and you pick door 42 - you have a 1% chance of winning ( think everyone can agree to that). If he shows you 98 gag gifts and gives you the option of winning you do not have a 50% of getting the grand prize if you switch - you have a 99% chance. In that case you only miss 1% of the time - the time you pick correctly the first time. Every other option you would get the prize if you switch.If the doors he picked were random (i.e. he could select the prize) then it at times he would select the prize but since he probably never did - he is always showing you a gag gift. In the case of 100 to me it clear you would always switch (unless of course you think this is a rigged game but assuming it is straight) you would have a 99% chance of winning if you switch. The same is true for the 3 door case. Ken
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
R8480: Hi Buzz...Isn't that the wonderment of most things??? That two sides can believe so completely in such opposite conclusions and be amazed at the other's inability to understand. Fortunately (or unfortunately) this one has a mathematical proof and can be demonstrated. I need to run out, so I will address more of your post a little later. But two things I want to address. 1. I emphasized 'always' because that is a classic problem with thestatement of this puzzle. You are astute that you undersood this element from the original problem, but it was not worded as clearly as it might have been, so I predicated my answer on the facts of the classic puzzle as they should be stated just to avoid the issue (which WAS in fact misunderstood by some). I had no ulterior motives in doing so. 2. My moniker (R8480) is the result of trying to pick a screen name at apoker site way back when. All of the names I tried were taken and I got frustrated and typed something randomly and that is what came out. So that is what I have stuck with. I wish I were that young. Here is my deal, which should be very profitable for you if you believe your position so strongly). We will play the puzzle out. You be Monte, I will be the contestant. You set three cards to represent doors, one red (the prize) two black (the gags). I will pick one card without looking. You then show me a black card. And I will ALWAYS switch to the third card on the board. Everytime my final choice is the red card, you give me $1000. Everytime it is black, I give you $1010. If it is a 50/50 proposition, then after a couple hundred iterations, you will be up $1000. But in reality I will be up around $66,000. I will make it better for you to avoid silly randomness. If we are within 5% of 50/50, I will give you your 1000 dollarses back. But we will not be. R8480
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
DPCondit: Buzz, you have demonstrated many times that you are intelligent, and also generally have developed a good working knowledge of probabilities. This I will not dispute.The riddle, in the first post, was inadequately expressed. It should specifically state that Monty Hall will: A) Always reveal what is behind a door that you did not pick. B) Always reveal a goat. C) Always offer you a choice of switching, or remaining with your original choice. For completeness sake, John Herbst added another twist to the question which I will not expound upon here, but let me add another condition: D) When Monty Hall is faced with revealing one of two remaining doors, and behind both doors are goats, the door he reveals will be picked either at random, or using a method of which you have no knowledge of. Your proof proves that one choice will be correct, and that one choice will be incorrect, it does not prove the frequency of correctness of each choice. The truest proof of all, which you have apparently not done yet, is to take two pennies and a quarter, and arrange them in whatever order you like, then choose one, pretending you don't know what it is. Then reveal a penny that you did not pick, and record what happens when you switch, and when you don't switch. Do this once where you choose each coin one time. This will provide the proof that you will always do better by switching, at least when the parameters of the riddle are expressed adequately. I know you have not done this yet, because you would have changed your position on this. Good luck, Don
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
DPCondit: [nq:1]This will provide the proof that you will always do better by switching,at least when the parameters of the riddle are expressed adequately.[/nq]Actually, I should have said you will be correct twice as often by switching, not "always do better by switching". Sorry for the mistake. I know
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
Stephan Lemonjello Jr.: [nq:1]Isn't that the wonderment of most things??? That two sides can believe so completely in such opposite conclusions and be amazed at the other's inability to understand.[/nq]I think the people who believe that switching doesn't improve their odds should post their score on the math section of the SAT. I'd bet they would only average about 400 or so, because this is a pretty simple problem. One time I got into a big argument about an even simpler problem. On 9/11/2002, the number 911 was drawn in the NY state lottery (the 3-digit daily drawing). Since there were TWO of the 3-digit drawings in NY on that day, I casually stated that the odds of 911 being drawn on that day were about 1 in 500. Amazingly, a bunch of people on a message board disagreed and said the odds were 1 in 1000, even though there were two drawings. After arguing back and forth for a few days, most of them were still convinced that I was wrong. Only after a letter from a math professor was posted did they finally change their tune.
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
HeatSearch: Bill, you are missing the point of my post. It seems you are trying to combine the two situations into one. The first being when we must pick one of the three doors. The second we must pick one of the two remaining doors. The two problems are seperate from each other and one has no bearing on the other. The bottom line is you can choose either of the two doors and will win half of the time regardless of which you pick.Regards, David
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
Joe Long: [nq:1]R8480 - I did read the whole thread before my second reply to Don. I had already read your first response before my first reply to Don. I responded to Don because I know Don and like him.[/nq]Then you must have missed the two posts where I said that I programmed a computer simulation of this problem, and in the simulation not switching won 1/3 of the time and switching won 2/3 of the time. Don't take my word for it, do one yourself. If you don't know how to program a simulation, just do your poker-chip simulation with a friend. It shouldn't take all that many trails before the 1/3 - 2/3 pattern begins to emerge. Joe Long jlong (at) rnbw (dot) com
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
R8480: Wanna bet David??R8480
This thread originates from within 'usenet', and as such the content and users are to have been moderated by our community.
Show more
| Have a question? People are waiting to help. Interesting stuff |
All RSS Feeds