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Josh the Superfluous Inner circle The man of 1881 Posts |
A 66% chance against his 50%. If the guy holding the other closed box's odds don't change, doesn't that tell you something about your logic?
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 |
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TomasB Inner circle Sweden 1144 Posts |
He believes I have 33% chance to win (the times John doesn't get the X) if I always stick with my card, so the game should really be in Craig's favour...if he is correct.
His calculated winnings would be 2/3*1.5-1/3*1=2/3 dollars per game played where John doesn't get the X. /Tomas |
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Josh the Superfluous Inner circle The man of 1881 Posts |
Is it over simplistic to say: 3 boxes each have a 1 in 3 chance. 1 box is elimintated and the remainder have a 1 in 2 chance. if box C were elliminated the chances of you holding box A or box B would be even.
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 |
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Craig Krisulevicz Special user Philadelphia, PA 647 Posts |
Tomas,
I wouldn't know how to write one of those algorithms. I'm terrible with computers. Josh, We're comparing apples to oranges on that actually, since they ar two totally different and independent situations. Guys, we could argue this all day. In fact it's been argued for a while. Some PhD agree with it, others don't. If you guys were in front of me, and I had a pad and pencil, I could convince you. Here's a little simulator that takes the guesswork out. It tracks all four possible scenarios. It's just like this $1, $1, $100 problem. Notice how the ratio of blue to (blue +red) converges to 2/3 http://www.cut-the-knot.org/Curriculum/P......ll.shtml And here is a continously running situation http://www.remote.org/frederik/projects/ziege/empirie.html
Who is John Galt?
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Josh the Superfluous Inner circle The man of 1881 Posts |
Quote:
On 2006-03-02 11:04, magicjohn2278 wrote: This is the situation. Are you saying it'd be to my advantage to switch with Mike. But if I were Mike, it would be advantageous to switch with Josh? Craig I doubt you'd convince me, but who knows. This stuff is fun but it makes my brain spin. Nice talking with all of you. Happy friday!
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 |
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TomasB Inner circle Sweden 1144 Posts |
Craig, forget about any algorithms and random generators. I'd be ok if you just randomly pressed 1, 2 and 3 100 times. That'd be good enough for me if in the end there is about the same amount 1s, 2s and 3s. As long as you don't happen to press one digit twice as much as any of the others I'm ok.
Let's say that 1 means that I get the X, 2 means that Josh get the X and 3 means that Mark gets the X. Any objections? Is this a fair model of the game? /Tomas |
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Craig Krisulevicz Special user Philadelphia, PA 647 Posts |
Quote:
On 2006-03-03 17:18, Josh the Superfluous wrote: If you had three players You (Y) Mike(M) and John (J) And were Y, you should switch with M, but if you were M, you should switch withY Basically, you should switch with the other guy who did not open his box if had the option to switch.. Happy Friday? I hear that! Time to go out folks, see you all over the weekend
Who is John Galt?
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Craig Krisulevicz Special user Philadelphia, PA 647 Posts |
Quote:
On 2006-03-03 17:36, TomasB wrote: If we played what you just describe, everyone would have a 33% chance each time. Where's my ability to switch? We'll work on this more over the weekend. I'm out for the night. Cheers
Who is John Galt?
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Josh the Superfluous Inner circle The man of 1881 Posts |
I'VE GOT IT!!!!!!!! The puzzle is all about the ability to switch. This option only presents it self when John opens up his box and loses. Otherwise he'd get $100 and switching wouldn't be considered. So the puzzle starts with Johns chances of winning at 0%.
(I didn't read Craig's above post before typing this. Posted in haste as dinner and my wife waited)
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 |
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Josh the Superfluous Inner circle The man of 1881 Posts |
Not to mention 2 boxes and 1 - $1 bill and 1 -$100 bill. ......50/50 eh?
(By the way, if your playing with 2 card cheats and one of them folds, switch with the other.)
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 |
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TomasB Inner circle Sweden 1144 Posts |
Quote:
On 2006-03-03 18:07, Craig Krisulevicz wrote: Craig, I told you that your string of numbers represents ME, John and Mark. I will be given the opportunity to change with Mark each time the game is played (which is when John doesn't get the X). But I PROMISE you that I will not change but stick with my original box. According to you I'd lose 33% of the time. So you can reread the game I offered you: Each time the game is played (when John doesn't find an X in his box) I will give you one dollar when I don't get the X and you should give me 1.5 dollars each time I do get the X. If I according to you lose 33% of the time you will on an average collect about 67 cents from me in each game played (which is only the games where John _doesn't_ find an X in his box). Let me know if you have any questions regarding this game. One way for you to create the string is to write 33 "1" in a row, then insert 33 "2" among them in any order you like and finally put in 34 "3" among those. I don't even care if you make it look random or not. I just want their individual frequencies to be about 33%. /Tomas |
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MR2Guy Regular user Nashville 179 Posts |
I love this logic problem.
It does matter if the first card eliminated is picked by random or not. 3 people touch 3 cards. Someone, somehow by random, is eliminated from the bet by deciding that his/her card is turned over first. That leaves 2. At this point, odds are 1/3 for all cards being the prize card. If you ask the remaining 2 if they would swap their card for the other 2 BEFORE you reveal the random card, and they agree, both have a 2/3 chance of winning. If you turn over the 3rd card, and it's the prize card, they BOTH win. If you flip the random 3rd card and it's not the prize card, and ask them if they want to swap, their odds are 50/50. You can't disregard the times you randomly pick a card and it's the prize. There is inherent information transmitted IF someone knows where the prize is. This is why only one person can play. Regardless of which card they pick, the prize knower is ALWAYS able to show a non prize card, since they have the knowledge of where the prize card is. If 2 play, it only leaves a single, random card to flip, so the knower does not have control over showing a losing card every time up front. Turning over a card doesn't matter. Example: I give you a choice of 1-3, I can always show you a non prize card, and now ask if you want to swap. Actually, I'm asking you if you want to swap your one card for two, it doesn't matter that I turned a card over, you get both, a 2-3 proposition. Same applies if I don't know where the prize card is, pick a card, and your choice is take the other two, or keep your one. Example: I know where the 7 of Hearts is in a deck. I spread them and ask you to pick one (52-1) I turn over the remaining 50 and ask you to pick either your original or the remaining face down card. Actually, your choice is the original choice, 1-52, or the REST of the cards, turning over 50 cards doesn't change anything, since I know where the 7H is, you get either 51 cards, or 1. Now lets say I take out 10 cards, A-10 of Spades, shuffle them (honest!), and deal 5 cards to each of us. What are the odds you have the AS? We take turns. I turn over the top card of my stack. It's not the AS. Now you get to choose to either keep your 5 cards, or swap your 5 with my 4. Should you? Would your odds improve if I didn't know where the AS is? Jason
Question every rule.
There are no absolutes. |
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TomasB Inner circle Sweden 1144 Posts |
Ooops, I can't edit my previous post, but the times I wrote "lose 33%" it should of course be "lose 66%" since that is what Craig believes I'd do if I don't switch after John hasn't found the X.
/Tomas |
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magicjohn2278 Special user Isle of Man UK 544 Posts |
Quote:
On 2006-03-03 15:32, Nir Dahan wrote: You are right! I have now had to reconsider my original solution to the problem! MR2Guy is of the opinion that John opening his box at random and "finding a $1" is different to someone who knows John has only $1 and telling him to open it. Basically, everyone has a 1 in 3 chance of getting the £100. If you know which player doesn't have the $100 and you switch with the other, then the only time you lose is the 1 in 3 times that you hold the $100 to start off with. However, in real life, if one of the other players, say John, opens his box at random, then there is a 1 in 3 chance he will reveal the $100, so switching with Mike is pointless as you can't win. If John DOES show that he doesn't have the $100, Mikes chances of holding the $100 are exactly the same as yours, so why switch? ..so can anyone explain the solution to me!!!? |
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Josh the Superfluous Inner circle The man of 1881 Posts |
I misread something above. So please excuse the yelling. But I'll leave it here for fun. enjoy:
Everyone does NOT have a 1 in 3 chance!!!! The problem states "John opens his box and finds a $1.". The fact that John opens his box and finds a $1 is now a given. At the point of the option to switch, John has a 100% chance of removing 1 box and one $1 bill from the equation, and a 0% chance of having the $100 bill. In a card dealing simulation with 2s and a Queen, you'd have to leave one of the 2s on the spot marked for John. Since any time you dealt him the queen, the hand would have to be discarded for not meeting the set up (and I'll say it once more in hopes of someone addressing it) ***John opens his box and finds a $1.***
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 |
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magicjohn2278 Special user Isle of Man UK 544 Posts |
…and the answer is ………………………………………………………………………………………………………………………
It makes no difference whether you switch or not! Why? Well before I try to explain, perhaps I should add, that I live in the UK, and up until a day or so ago, I had never heard of Monty Hall and his three doors! Monty is the key to the whole conundrum, and doesn’t exist in the original question. In fact, in retrospect, I posed the question wrong! But the answer above IS the correct answer to the question I posed! The question should have been - “You, John and Mike are each given and identical box. One contains a $100 bill, the other two contain a $1. Monty knows who is holding the $100, and asks John or Mike to opens his box to show $1. Before you open your box, and keep the contents, you are given the opportunity to exchange boxes with the other. What do you do? Does it make any difference? Now the answer is Yes, you should switch boxes, as it changes your odds of winning from 1/3 to 2/3. Why?? Because Monty is helping you. Why does switching boxes not help the other person equally? Because Monty is helping YOU! How? Well here it gets a little more complicated! We will assume that you ARE always going to switch. 1 in 3 times, by chance, you are going to give away the $100 which you already hold. And lose. (A natural loser.) 1 in 3 times, by chance, you are going to switch with the person holding the $100, and win. (A natural winner.) And 1 in 3 times Monty is going to help. (An assisted winner.) Now Monty can’t influence the outcome of either a natural winner, or a natural loser, these occour naturally, 1 in 3 times each by chance, so how does he help? Imagine Monty looks in then opens John’s box, it contains $1. That is the end of Monty’s involvement. We are in the situation where we have either a natural loser (You have the $100.) or a natural winner (Mike has the $100.) It is pointless him opening Mikes box to see if he has the $100, whether he has or hasn’t, Monty can’t change the outcome of your natural win or loss. A third of the time Monty is going to look in John’s box, and see $100. Now, he is going to open Mike’s box and show you that it contains $1. This is the equivalent of him telling you which box the $100 is in. So Monty has just changed the “abandoned game because John revealed the $100 at the beginning” to an assisted win for you when you switch with John. Obviously, the assisted win situation never helps the person you switch with. With Monty playing, your win rate is 2 in 3, John never wins, and Mike wins 1 in 3. So what is the difference between the original scenario, and the situation where Monty has looked in Mike’s box, seen the $100, and then opened John’s? Isn’t this an “assisted win” for you? Well yes, but it only ever puts John at a disadvantage, Mikes chances of winning are not affected, are exactly the same as yours so there is no point switching. In the Monty game, Mike is disadvantaged every time Monty helps YOU by opening Mike’s box when he finds the $100 in John’s. From this position, Mike has lost, and you have won. As Monty doesn’t exist in the original scenario, and it is never Mike’s box which is opened first then Mike is not disadvantaged, so his chances of winning are identical to yours and there is no point switching! It might interest UK readers to learn that I set this problem as a result of an “off the cuff” remark by Derren Brown in his television series a few months ago. He was doing a “just chance” routine with three ring boxes and two girls, after opening one of the girl’s boxes, he asked the other is she wanted to switch with his, when she declined he commented, that statistically she should have done, as it was twice as likely the either his or the other girl’s box contained the ring, and as she now knew which box was empty, she should switch. This got me thinking…. And still has, as it turns out, he was wrong! |
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jgravelle Loyal user Milwaukee (Head shown not actual size) 270 Posts |
Quote:
On 2006-03-05 14:05, magicjohn2278 wrote: Hey man... I might live in the U.S., but in the interest of international outreach, I've gone out of my way to familiarize myself with the dignitaries of your culture: Benny Hill, Boy George, Dame Edna... I mean, the least you could do is familiarize yourself with our illustrious celebrities like Mr. Hall. I'd also advise you to learn about contemporary American icons like Martha Stewart, Mel Gibson and Arnold Schwarznegger who've made this country what it is today. Now if you'll excuse me, I'm helping my son do a book report on Canada and I have to go find a map of the southern hemisphere... Regards, -jjg |
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Josh the Superfluous Inner circle The man of 1881 Posts |
Wow! Josh was right from the very start and with each subsequent posting. I for one am in awe of his brilliance.
(whoops! I forgot to change my username.)
What do you want in a site? "Honesty, integrity and decency." -Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 |
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LobowolfXXX Inner circle La Famiglia 1196 Posts |
Quote:
On 2006-03-04 22:41, Josh the Superfluous wrote: Yes, John's 33 1/3% chance of finding the hundred has evaporated, and is equally added to the chances of the other two. They each have their original 33 1/3% AND their 50% share of John's 33 1/3% (16 2/3%). Thus, their new chances are 33 1/3% + 16 2/3% = 50%.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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idris New user St. Louis, MO 38 Posts |
The problem I've always had with the Monty Hall problem is that it is assumed he will always open a door for a losing prize. It's been 30 years since I've seen the program, but I don't believe that he always opened one of the doors. Therefore, the Monty Hall problem as posed has never been correct.
Jerry
Jerry
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