Page 13 of 25 FirstFirst ... 312131423 ... LastLast
Results 421 to 455 of 864

Thread: Do you know the correct answer?

  1. #421
    Join Date
    Nov 2017
    Location
    Gulf Islands B.C.
    Posts
    3,882

    Default Re: Do you know the correct answer?

    Quote Originally Posted by peb View Post
    The table in post 6 is NOT incomplete. The first two columns give all 9 possibilities. 3 spots where the car can be, 3 guesses. 3 times 3 equals 9. There are 9 rows, each row shows 8ne possible game scenario.

    It us a complete mathematical proof.
    Yes it is incomplete. It groups two possibilities as single. And it does this three times.

    To not count them because the result is the same is like saying it doesn’t matter if you remove one or four empty chambers from the Russian roulette revolver.
    Last edited by Decourcy; 06-12-2022 at 06:14 PM.

  2. #422
    Join Date
    Apr 2019
    Location
    Massachusetts
    Posts
    1,472

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Decourcy View Post
    Let’s play a different game. Russian roulette. One bullet in 6 chambers. 1 in 6 chance yes? One squeeze and a “click”. Don’t rotate the cylinder. Remaining 5 are now 1in 5 yes? Squeeze again “click”. Remaining are now one in 4. Same same.

    The point being that of course probabilities change as options reduce



    The loaded chamber was always the loaded chamber but the probability that it will be the next chamber increases until, if it’s the last chamber, it’s 6/6 or 100%

    But also, if you intend to pull the trigger 6 times without spinning the cylinder, the the odd are 6 in 6 (100%) that you will shoot yourself by the time you are you reach the 6th pull if not sooner. Your example looks at the probability for a single pull while counting how many many pulls have already beet taken.

    Going back to Monty. The contestant makes one choice and has a 33% chance Monty had two doors - 66% chance. The contestant does not open the door until Monty has opened both of his doors. That means there is a 66% chance that one of his doors is the car. If he does not trade either door with the contestant, then he still has 66% chance. His chances haven't changed, because he has not changed either of the two doors in his possession. But, his odds don't change after he has opened the first door assuming it is a goat because we are saying it is a 66% chance that one the doors has a car. It could be either one and ther ren=mains a 33% chance that neither door has the car: i.e. the contestant is the winner.

  3. #423
    Join Date
    Oct 2008
    Location
    Walney, near Cumbria UK
    Posts
    58,096

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Tom Montgomery View Post
    .
    I did miss your edit.

    Are you saying that you DO accept his first, formal, explanation of the probabilities?
    Nope, it has the same result as the Informal one, just less clarity in amongst the waffle. If you switch, you have one chance of being correct 1/3. You can only switch once, not twice. Remember, one choice, with its 1/3 chance, has been taken out of the equation.
    It really is quite difficult to build an ugly wooden boat.

    The power of the web: Anyone can post anything on the web
    The weakness of the web: Anyone can post anything on the web.

  4. #424
    Join Date
    Apr 2019
    Location
    Massachusetts
    Posts
    1,472

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    Nope, it has the same result as the Informal one, just less clarity in amongst the waffle. If you switch, you have one chance of being correct 1/3. You can only switch once, not twice. Remember, one choice, with its 1/3 chance, has been taken out of the equation.
    I just don't see this being settled here. The answer is as clear to me as yours is to you. I think I'm going to call it a day. Have a good night all!

  5. #425
    Join Date
    Sep 1999
    Location
    Norwalk, Ohio
    Posts
    34,274

    Default Re: Do you know the correct answer?

    .
    I just played the game 20 more times using the "always switch" strategy. The contestant scored 14 wins and 6 losses. A winning percentage of 70%.

    I have now played the game 80 times with the contestant scoring 53 wins and 27 losses for a winning percentage of 66.25%.
    "I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly

  6. #426
    Join Date
    Oct 2003
    Location
    Valley of the Sun
    Posts
    117,179

    Default Re: Do you know the correct answer?

    you didnt have to play it out
    you could use maths
    Quote Originally Posted by Tom Montgomery View Post
    .
    I just played the game 20 more times using the "always switch" strategy. The contestant scored 14 wins and 6 losses. A winning percentage of 70%.

    I have now played the game 80 times with the contestant scoring 53 wins and 27 losses for a winning percentage of 66.25%.
    Simpler is better, except when complicated looks really cool.

  7. #427
    Join Date
    Oct 2003
    Location
    Valley of the Sun
    Posts
    117,179

    Default Re: Do you know the correct answer?

    ​...
    Simpler is better, except when complicated looks really cool.

  8. #428
    Join Date
    Feb 2004
    Location
    Texas
    Posts
    14,331

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Decourcy View Post
    Yes it is incomplete. It groups two possibilities as single. And it does this three times.

    To not count them because the result is the same is like saying it doesn’t matter if you remove one or four empty chambers from the Russian roulette revolver.
    I have ignored all Russian Roulette posts, sorry, that's my quirk. When I was in high school the boyfriend of a girl in my class killed himself by actually playing Russian roulette when drunk one night. I can't hardly stand to read the term in a small town, that caused a huge amount of trama.

  9. #429
    Join Date
    Sep 1999
    Location
    Norwalk, Ohio
    Posts
    34,274

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Paul Pless
    you didnt have to play it out
    you could use maths
    Yeah. But I rather enjoyed playing.

    I just completed 100 plays of the game employing the "always switch" strategy.

    The first twenty plays resulted in 10 contestant wins and 10 losses for a winning percentage of 50%.
    The second twenty plays resulted in 15 contestant wins and 5 losses for a winning percentage of 75%.
    The third twenty plays resulted in 14 contestant wins and 6 losses for a winning percentage of 70%.
    The fourth twenty plays resulted in 14 contestant wins and 6 losses for a winning percentage of 70%.
    The fifth twenty plays resulted in 15 contestant wins and 5 losses for a winning percentage of 75%.

    The final total was 100 plays resulting in 68 contestant wins and 32 losses for a winning percentage of 68%.
    "I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly

  10. #430
    Join Date
    Nov 2008
    Location
    New Hampshire
    Posts
    38,930

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Tom Montgomery View Post
    .
    So I just played the game 20 more times following your requirement. I turned over the card to my left (the contestant card) every time. The I discarded a Joker from the two cards to my right. And then always making the switch.
    I may have misunderstood your earlier post. If you find the ace in the two cards held by Monty, then I will agree that's a win for switch.

    I just played it 45 times and it came out closer to 50:50, but in the process I realized that it's actually difficult to shuffle 3 cards well. I started seeing patterns in the outcome that I'm not sure are random.
    "Where you live in the world should not determine whether you live in the world." - Bono

    "Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers

    "Those are my principles, and if you don't like them... well, I have others." - Groucho Marx

  11. #431
    Join Date
    Feb 2007
    Location
    Port Stephens
    Posts
    25,329

    Default Re: Do you know the correct answer?

    Quote Originally Posted by bamamick View Post
    I like goats.

    Mickey Lake
    Me too and I already have a car.
    Rick

    Lean and nosey like a ferret

  12. #432
    Join Date
    Mar 2009
    Location
    2 states: NJ and confusion
    Posts
    43,259

    Default Re: Do you know the correct answer?

    Suppose you had only two choices to begin with. You'd have a 50/50 chance. Now a third choice is added before you choose. Now you've a 1 in 3 chance, no?

    Once the 3rd choice is eliminated, it's no longer a factor as to odds. The host may have some sort of established pattern. If the prize is behind the even numbered curtain, he eliminates one of the odd numbers, and if that's known by the contestant, then the contestant is more likely to choose the even numbered curtain. That would show up in long term results.
    "alternative facts (lies)" are a cancer eating through a democracy, and will kill it. 1st amendment is not absolute.

  13. #433
    Join Date
    Nov 2017
    Location
    Gulf Islands B.C.
    Posts
    3,882

    Default Re: Do you know the correct answer?

    Quote Originally Posted by John Smith View Post
    Suppose you had only two choices to begin with. You'd have a 50/50 chance. Now a third choice is added before you choose. Now you've a 1 in 3 chance, no?

    Once the 3rd choice is eliminated, it's no longer a factor as to odds. The host may have some sort of established pattern. If the prize is behind the even numbered curtain, he eliminates one of the odd numbers, and if that's known by the contestant, then the contestant is more likely to choose the even numbered curtain. That would show up in long term results.
    The host does have a pattern. He always eliminates one of the goats. The contestant knows that

  14. #434
    Join Date
    Aug 2002
    Posts
    2,632

    Default Re: Do you know the correct answer?

    I’m seeing the ‘problem’ as a paradox of time, with both arguments being correct.

    Looked at as a singular event/game sample, the odds at the second choice – to switch, or not – are 50-50. However, with increased frequency of the events/samples, over time, say, playing 100 games, it would seem to be logical to choose the switch, because statistically the first choice would have a 66% probability of being wrong. Therefore, the ‘correct answer’ is to switch if playing multiple games, but in a singular instance it’s 50-50.

    But, there is randomness, intuitiveness, woo-woo, in the first selection – how do you put numbers to that?

  15. #435
    Join Date
    Dec 2003
    Location
    Portland, Oregon
    Posts
    81,892

    Default Re: Do you know the correct answer?

    Quote Originally Posted by J P View Post
    I’m seeing the ‘problem’ as a paradox of time, with both arguments being correct.

    Looked at as a singular event/game sample, the odds at the second choice – to switch, or not – are 50-50. However, with increased frequency of the events/samples, over time, say, playing 100 games, it would seem to be logical to choose the switch, because statistically the first choice would have a 66% probability of being wrong. Therefore, the ‘correct answer’ is to switch if playing multiple games, but in a singular instance it’s 50-50.

    But, there is randomness, intuitiveness, woo-woo, in the first selection – how do you put numbers to that?
    Prolly not.

    When I shared the puzzle with the mob at dinner, Son1 said - quickly and with no hesitation - 'switch, every time'. He said he visualized the 2/3 chance being compressed into the 'switch' door in terms of quantum physics. He tried to explain it... but it was a leap too far, for me. I suspect he was wrong also. Or correct for the wrong reasons.
    David G
    Harbor Woodworks
    https://www.facebook.com/HarborWoodworks/

    "It was a Sunday morning and Goddard gave thanks that there were still places where one could worship in temples not made by human hands." -- L. F. Herreshoff (The Compleat Cruiser)

  16. #436
    Join Date
    Feb 2010
    Location
    northern Georgia, or Mississippi Delta USA
    Posts
    26,221

    Default Re: Do you know the correct answer?

    Wikipedia has a good page on the problem. https://en.wikipedia.org/wiki/Monty_Hall_problem

    When you first choose a door, you have a 2 in 3 chance of being wrong. Even after seeing in the second door, you still have the 2 in 3 chance of being wrong. Opening the door doesn't improve your chances if you stick with your first choice..

  17. #437
    Join Date
    Sep 1999
    Location
    Norwalk, Ohio
    Posts
    34,274

    Default Re: Do you know the correct answer?

    .
    Twelve pages over the course of 32 hours!

    When I started this thread yesterday I had no idea….
    "I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly

  18. #438
    Join Date
    Sep 1999
    Location
    Armada, MI, USA
    Posts
    528

    Default Re: Do you know the correct answer?

    Quote Originally Posted by peb View Post
    If you play my the rule to always switch
    You only lose if you picked the right door. That's a 33% chance of losing. It is very, very simple.
    I thought this was a very compelling argument made by peb way back at post #123. It seems to have gone unnoticed. I would sure like to see one of the 50/50 guys comment on it.

  19. #439
    Join Date
    Feb 2010
    Location
    northern Georgia, or Mississippi Delta USA
    Posts
    26,221

    Default Re: Do you know the correct answer?

    About 10 or 12 pages in the middle, i didn't bother to read.

  20. #440
    Join Date
    Feb 2010
    Location
    northern Georgia, or Mississippi Delta USA
    Posts
    26,221

    Default Re: Do you know the correct answer?

    But some seem to still not believe that.

  21. #441
    Join Date
    Dec 2009
    Location
    Frankfort, MI
    Posts
    10,729

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Tom Montgomery View Post
    .
    Twelve pages over the course of 32 hours!

    When I started this thread yesterday I had no idea….

    “I returned, and saw under the sun, that the race is not to the swift, not the battle to the strong, neither yet bread to the wise, nor yet riches to men of understanding, nor yet favour to men of skill;

    ^^^^^^^ Maybe not, but it’s still the way to bet…

    Jeff C
    Don’t expect much, and you won’t be disappointed…

  22. #442
    Join Date
    Feb 2004
    Location
    Texas
    Posts
    14,331

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Jimmy W View Post
    Wikipedia has a good page on the problem. https://en.wikipedia.org/wiki/Monty_Hall_problem

    When you first choose a door, you have a 2 in 3 chance of being wrong. Even after seeing in the second door, you still have the 2 in 3 chance of being wrong. Opening the door doesn't improve your chances if you stick with your first choice..
    It is the simplist explanation.

  23. #443
    Join Date
    Feb 2001
    Location
    Seattle, WA USA
    Posts
    16,179

    Default Re: Do you know the correct answer?

    This is amazing to me, that people don't understand this.


    Here's a simulator I wrote in the Go programming language: https://goplay.tools/snippet/l-0dNVMAhlq

    It's in the [a] Go Playground, so you can compile the code and run it. Currently set up to do runs of 1,000,000 plays of the game.

    The rules of the game are

    - There are 3 doors. One of them a valuable prize behind it (a car!). The other two has worthless prizes (goats!).
    - The contestant selects a door. It remains closed.
    - Monty Hall, the preceptor of the game show, opens one of the two remaining doors. It cannot be the door containing the valuable prize, as that would ruin/end the game. It's always a worthless prize.
    - The contestant is offered the chance to switch to the remaining door, and can either switch, or hold.

    The question is, which strategy is more effective? Switch to the last door?, or holding on to your initial choice?

    When the contestant first makes their selection, they have a 1/3 chance (1:2 odds) of selecting the door with the car behind it, and a 2/3 chance (2:1 odds) of selecting a door with a booby prize behind it.

    After Monty Hall has opened the door, and revealed the booby prize behind it, he has taken 1 booby prize out of contention, because his pick was not random.

    That makes the remaining door coalesce into a 1/2 chance (even odds) that it is the valuable prize, because it's either behind that door, or it's behind the door selected by the contestant. And the door selected by the contestant is still sitting there with a a 1/3 chance of it being the valuable prize.

    It's not so much a probability & stats problem as it is

    Here's output from a sample run:

    Code:
    Strategy: Switching
    +---------+-----------+---------+
    | OUTCOME | FREQUENCY | PERCENT |
    +---------+-----------+---------+
    | Win     |    666164 |  66.616 |
    | Lose    |    333836 |  33.384 |
    +---------+-----------+---------+
    |             1000000 | 100.000 |
    +---------+-----------+---------+
    
    Strategy: Holding
    +---------+-----------+---------+
    | OUTCOME | FREQUENCY | PERCENT |
    +---------+-----------+---------+
    | Win     |    332717 |  33.272 |
    | Lose    |    667283 |  66.728 |
    +---------+-----------+---------+
    |             1000000 | 100.000 |
    +---------+-----------+---------+
    This is the guts of the thing:
    Code:
    func main() {
    
      // Our sample size is 1,000,000 plays of the game
      const sample = 1_000_000
      
      // for each possible strategy (switching or holding)
      for _, strategy := range []string{SWITCHING, HOLDING} {
        // zero the counters
        win, lose, total := 0, 0, 0
        
        // play the game, over and over...
        for n := 0; n < sample; n++ {
          
          // play the game
          didWin := play(strategy)
          
          // record the outcome as either a win or a loss
          if didWin {
            win++
          } else {
            lose++
          }
          
          // record the total number of games played (wins+losses)
          total++
          
        }
    
        // report the outcome. This just formats it as a nice table.
    
        tbl := tablewriter.NewWriter(os.Stdout)
        tbl.SetAutoMergeCells(true)
        tbl.SetHeaderLine(true)
        tbl.SetHeader([]string{"Outcome", "Frequency", "Percent"})
        tbl.SetColumnAlignment([]int{tablewriter.ALIGN_LEFT, tablewriter.ALIGN_RIGHT, tablewriter.ALIGN_RIGHT})
        tbl.SetFooterAlignment(tablewriter.ALIGN_RIGHT)
        tbl.Append(row("Win", win, total))
        tbl.Append(row("Lose", lose, total))
        tbl.SetFooter(row("", total, total))
    
        // render the table with a nice header to tell us which strategy was in use.
        fmt.Println()
        fmt.Printf("Strategy: %s\n", strategy)
        tbl.Render()
       
      }
      
      return
    }
    
    func play(strategy string) (didWin bool) {
    
      // populate the doors. The car is placed behind door at random and
      // every other door has a goat behind it.
    
      doors := []string{GOAT, GOAT, GOAT}
      doors[randomDoor()] = CAR
    
      // The contestant selects a door at random from the 3 available doors.
    
      contestantSelection := randomDoor()
    
      // Monty Hall opens a door that is neither the winning door
      // or the door selected by the contestant.
      //
      // We don't really care what door he opened, as that is guaranteed
      // not to be of any interest (he's not going to open the winning door),
      // but we do care about the single door that is neither selected by the
      // contestant or opened by Monty Hall.
      
      _, remaining := montyHallOpensDoor(doors, contestantSelection)
    
      // if our strategy is switching, we switch to the remaining door,
      // otherwise we stand pat.
    
      if strategy == SWITCHING {
        contestantSelection = remaining
      }
    
      // The contestant wins if his selected door has the car behind it.
    
      didWin = doors[contestantSelection] == CAR
    
      // return the win/lose indicator to the caller
      return didWin
    }
    You would not enjoy Nietzsche, sir. He is fundamentally unsound. — P.G. Wodehouse (Carry On, Jeeves)

  24. #444
    Join Date
    Feb 2001
    Location
    Seattle, WA USA
    Posts
    16,179

    Default

    This diagram show why this is true.

    You would not enjoy Nietzsche, sir. He is fundamentally unsound. — P.G. Wodehouse (Carry On, Jeeves)

  25. #445
    Join Date
    Sep 2005
    Location
    new zealand
    Posts
    5,843

    Default Re: Do you know the correct answer?

    From a link to a Reddit page in the link that Tom first posted - it sort of flips the problem on it head, and made more sense of it to me than all the charts and tables .

    Quote:
    To me, the clearest explanation is that the only way to get it wrong by switching is to have picked the correct door in the first place. The odds of picking the correct door first are 1 in 3.


    Pete
    The Ignore feature, lowering blood pressure since 1862. Ahhhhhhh.

  26. #446
    Join Date
    Oct 2008
    Location
    Walney, near Cumbria UK
    Posts
    58,096

    Default Re: Do you know the correct answer?

    I have laid it out in the form of a white board, as in Toms video above.
    The error that the statistician with his sharpie made is that he did not model all the facts in his working. He ignored that we know that Monty's door ALWAYS has O probability of revealing the car. The probability that the car is behind the other two doors is equal, we simply don't know which door.
    This is how it should be worked, without the waffle.

    White board.jpg

    I hope that is clear enough.
    Last edited by Peerie Maa; 06-13-2022 at 04:21 AM.
    It really is quite difficult to build an ugly wooden boat.

    The power of the web: Anyone can post anything on the web
    The weakness of the web: Anyone can post anything on the web.

  27. #447
    Join Date
    Apr 2000
    Location
    Southampton Ont. Canada
    Posts
    7,293

    Default Re: Do you know the correct answer?

    I will change my answer to 'switch'.
    Contestant's chances to win are 1 in 3
    Monty's chances are 2 in 3.
    If you switch,you get Monty's chances ,minus the door that he opened.
    Not unlike Monty allowing you to pick 2 doors.
    R
    Sleep with one eye open.

  28. #448
    Join Date
    Oct 2008
    Location
    Walney, near Cumbria UK
    Posts
    58,096

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Ron Williamson View Post
    I will change my answer to 'switch'.
    Contestant's chances to win are 1 in 3
    Monty's chances are 2 in 3.
    If you switch,you get Monty's chances ,minus the door that he opened.
    R
    Nope that suggests that Monty's decision affects the location of the car, which is impossible.
    Take out Monty's 1:3 and set it to zero, and you are left with 1:3 and 1:3, no more, no less.
    It really is quite difficult to build an ugly wooden boat.

    The power of the web: Anyone can post anything on the web
    The weakness of the web: Anyone can post anything on the web.

  29. #449
    Join Date
    Mar 2011
    Location
    Christchurch NZ
    Posts
    512

    Default Re: Do you know the correct answer?

    Nick. How does Monty's removal of on of the doors that you did not pick retroactively change the chance of you choosing the right door first, from 1/3 to 1/2 ?

  30. #450
    Join Date
    Oct 2012
    Location
    Poznań, Poland
    Posts
    3,235

    Default Re: Do you know the correct answer?

    I'll admit not having read through all 14 pages, so sorry if I'm just repeating an already stated point (which I probably do)


    Quote Originally Posted by CWSmith View Post
    Okay, Nick, I'll see if this old dog can learn a new trick:

    Contestant Picks this Door Prize is Behind This Door Monty Shows This Door Stays Switches
    1 1 2 Win Lose
    1 1 3 Win Lose
    1 2 3 Lose Win
    1 3 2 Lose Win
    cut for brevity
    Win % 0.5 0.5

    Okay, Tom, I told you what I thought was wrong with your chart. Now I'd like you to tell me what is wrong with mine.
    What is wrong is the assumption that all options are equally plausible. Corrected version:

    Contestant Picks this Door Prize is Behind This Door Monty Shows This Door Scenario probability Stays Switches Weight
    1 1 2 1/6th Win Lose .5
    1 1 3 1/6th Win Lose .5
    1 2 3 1/3rd Lose Win 1
    1 3 2 1/3rd Lose Win 1
    cut for brevity other scenarios analytically redundant
    Weighted win ratio 2/3rds 1/3rd 3
    WszystekPoTrochu's signature available only for premium forum users.

  31. #451
    Join Date
    Oct 2008
    Location
    Walney, near Cumbria UK
    Posts
    58,096

    Default Re: Do you know the correct answer?

    There are lots of ways to fiddle the math to fit the desired outcome.
    But you need to justify why you chose those numbers. Why is the probability of any pick a variable. The probability of picking 1 from 3 is always the same.
    It really is quite difficult to build an ugly wooden boat.

    The power of the web: Anyone can post anything on the web
    The weakness of the web: Anyone can post anything on the web.

  32. #452
    Join Date
    Mar 2011
    Location
    Christchurch NZ
    Posts
    512

    Default Re: Do you know the correct answer?

    So if your original pick remains at 1/3, the probability of the prize being one of the other two choices must be 2/3 and you now know not to pick one of those.

    If, as you assert, that it was not an advantage to change now the probability of either staying or changing must be 1/2. Which means that the probability of you originally making the correct choice must have changed, which is obviously nonsense

  33. #453
    Join Date
    May 2017
    Location
    Cumbria, UK
    Posts
    328

    Default Re: Do you know the correct answer?

    I was in the 50/50 camp when I first saw the problem but now would switch.

    Putting aside all the trees and statistics, the nub of the issue is what do you know after Monty opens the door that you did not know before. Answer nothing, Monty will always have the option to open a door with a goat therefore nothing has changed.

    There are 2 sets, my first choice 33% the other 2 doors 66%. Monty selecting a goat tells us nothing about what is in the second set but there is still a 66% chance that the second set had the car.

    Were Monty to open an other door at random and picked a goat then it would be 50/50.

    Apologies to all posters who have already made the point, I do not have time to read all 470 replies.

  34. #454
    Join Date
    Oct 2012
    Location
    Poznań, Poland
    Posts
    3,235

    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    There are lots of ways to fiddle the math to fit the desired outcome.
    But you need to justify why you chose those numbers. Why is the probability of any pick a variable. The probability of picking 1 from 3 is always the same.
    Good point, I should've explicitly written that:
    Only when a contestant picks correct gate, which will happen in 1 of every 3 games, Monty has to choose a goat to be revealed. Simply put, the scenarios 1-1-2 and 1-1-3 have to "share" the same 1/3rd chance of happening. They are, naturally, equivalent, meaning 50% chance of happening *within* the initial 1/3rd thus being half as likely as the other two scenarios where Monty has no choice in which goat to reveal.
    If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.

    Quote Originally Posted by oldcodger View Post
    Were Monty to open an other door at random
    This is the key: he's not. There is 0% chance he'll show you the car. This is the introduced bias.
    WszystekPoTrochu's signature available only for premium forum users.

  35. #455
    Join Date
    Oct 2008
    Location
    Walney, near Cumbria UK
    Posts
    58,096

    Default Re: Do you know the correct answer?

    Quote Originally Posted by WszystekPoTrochu View Post
    Good point, I should've explicitly written that:
    Only when a contestant picks correct gate, which will happen in 1 of every 3 games, Monty has to choose a goat to be revealed. Simply put, the scenarios 1-1-2 and 1-1-3 have to "share" the same 1/3rd chance of happening. They are, naturally, equivalent, meaning 50% chance of happening *within* the initial 1/3rd thus being half as likely as the other two scenarios where Monty has no choice in which goat to reveal.
    If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.


    This is the key: he's not. There is 0% chance he'll show you the car. This is the introduced bias.
    But but but. . . Monty always has to open a door with a goat, whatever the contestant picks. So the probabilities are the same whichever the contestant picked, car or goat.

    If we're listing out every scenario in the table, the 1-1-2 and 1-1-3 have to be valued as input at half of what 1-2-3 and 1-3-2 are. Doing otherwise introduces bias, which coincidentally matches the false '50-50 on switching' solution.
    Which is what I meant, choosing numbers to agree with a desired outcome.
    It really is quite difficult to build an ugly wooden boat.

    The power of the web: Anyone can post anything on the web
    The weakness of the web: Anyone can post anything on the web.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •