Makes Sense to Me (The End)

The most important item first. I promised bonus points to anyone who could tell us the author of the song “My Whole World Lies Waiting Behind Door Number 3.” If you answered Jimmy Buffet, then you are partially correct (and a Parrot Head). The complete answer is that the co-writer was Steve Goodman (no relation to John).

And now to the question: Should I switch from door number one to door number three? To recap: You were offered three doors to chose from, only one has a big prize, you pick door one, door two is revealed to show that it does not contain the big prize. Should you change from door one to door three?  (You can see the details in my previous post.)
There has been a raging debate over this for a number of years. In fact, you can see many of the proofs if you Google “Monty Hall Paradox.” And, in spite of the messages your inner genius may be giving you, the correct answer is to switch doors. The reasoning is very similar to that we used for the three prisoners dilemma discussed in the previous blog. (People seem to have an easier job accepting the conclusion of the three prisoners dilemma over that of the Monty Hall Paradox — hence, it is sometimes easier to start there.) Most people looking at the situation say that it doesn’t matter if you change — there are two doors available and that means there is a 50/50 chance that one of the doors is a winner. But, just as with the prisoners, the combined likelihood of door two and three being the winner stays at 66 percent, which means that there is a two out of three chance that door three is now the winner. Another way to think of it: If you change and you made the correct choice the first time, then one-third of the time you will change and lose. If you change and you made the wrong choice the first time, then two-thirds of the time you will change and win.
Okay, there is a chance you are still in the mental tunnel that says it is a 50/50 chance. And this usually comes from those who argue that what happened in the past has no effect on the fact that there are now two choices and an equal chance for each to hold the big prize. The assumption that the prior actions have no effect is false, and that is at the core of the mental tunnel in which people trap themselves. If you are still unconvinced, I would suggest you experiment. Try this with friends, set up your own game show, write your own computer program. (That was my personal “Aha!” moment. To prove that there was no benefit in changing, I wrote a program. And, while building the logic, I suddenly saw why you should always change.) If you actually play this through, you will see that switching provides the best opportunities.
If you would still like to plead your case, feel free to use the comments sections below. But I would also suggest you peruse the net and see some of the better articulated discussions of this issue.
And so we come to the end of our sojourn through the realms of mental tunnels. If you just joined us, you may want to go back to the beginning and work your way through some of the exercises. If you want to explore more, I refer you to the book I previously suggested: Inevitable Illusions by Massimo Piattelli-Palmarini. Two other very good books I have used while putting this together are Damned Lies and Statistics by Joel Best and How to Lie With Statistics by Darrell Huff and Irving Geiss. (The latter was first published in 1954, but the mere fact it is still in print shows just how valuable it is.)
While I was able to provide some audit examples, it was not my intent to give you specific tools to use. Rather, my purpose was to start you thinking about the assumptions you may be making. Ultimately, this is about reminding each of us to take a look at the suppositions we make, the way we come to conclusions, and the way we misuse statistics and logic.

Posted on Mar 1, 2010 by Mike Jacka

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  1. Here is a fun link -
  1. Here's a link to the site referenced by Mr. Wanner so you don't have to copy and paste.  (Hope this works.)  Of course, the first ten times I tried it, I hit 50%

  1. I have read the arguments and am not convinced. As I see it, when the contestant picks a door, the probability is 50% not 33% - because he knows he can switch and is effectively only choosing between two doors. He is offered three choices, but knows one will be removed so there are really only two choices.

    The logic of the argument for 1/3 is also unsound. When it says that opening one door does not change the odds that the first choice is correct. Extend that to opening a second door - the odds change for sure.

  1. Let me try another approach/explanation.  Let's assume doors one and two have the zonk and three is the winner.  There are three options:

    1)  The contestant choses door three.  Either door one or two is revealed to show a zonk.  The contestant changes doors and loses.

    2)  The contest choses door two.  Door one is revealed to show the zonk.  The contestant changes doors and wins.

    3)  The contestant choses door one.  Door two is revealed to show a zonk.  The contestant changes doors and wins.

    Each of the above has a 1/3 probability of occurring.  Two out of three times the contestant wins; one out of three the contestant loses.

    Now, the more I look at these explanations I realize it is just different ways of saying the same thing.  Ultimately, we do not want to accept this explanation because we believe the first actions have no effect on our choice.  The mental tunnel that exists is our belief that these are independent actions.  But they are not.  The previous choices and revelations DO impact the the odds in the subsequent choices.  Again, I think the best approach is to get a couple of people together and run some tests.  This will consistently prove that, over the long haul, switching is the best approach.

  1. I disagreed, and to continue my argument, I submit this:  There are NOT just 3 options . . . there are SIX options --

    4)  The contestant sticks with door #3 (doesn't switch) and WINS

    5)  The contestant sticks with door #2 (doesn't switch) and LOSES

    6)  The contestant sticks with door #1 (doesn't switch) and LOSES

    Six choices, 2 outcomes (Win or Lose), 3 winners and 3 losers -- 50/50 overall.

  1. Mike,

    Just want you to know I'm still going back to compare this to your original statement to see what holds true . . . As I believe I said in my first post, enlighten me if I'm waaaaaaaaay off base on this.


  1. OK, I thought about this a few more minutes and here's my final opinion:

    Because of the second-chance "switch" option, the initial odds (considering the second-chance) are effectively 1/2 or 50%. 

    Two choices = Stay or Switch

    Two outcomes = Win or Lose

    1 Winning Door, 1 Losing Door

    50/50 overall

    In other words, the third-door zonk is of no regard, and the main decision is not which of 3 doors, but whether to stay with your initial choice or switch away from your initial choice. 

  1. My argument can be furthered by these simplistic statements of options / outcomes:

    If the door you choose is the Prize, and you stay, you win

    If the door you choose is the Prize, and you switch, you lose

    If the door you choose is the Zonk, and you stay, you lose

    If the door you choose is the Zonk, and you switch, you win

    2 possibilities (chose the prize vs. chose a zonk), 2 actions (stay vs. switch), 2 outcomes (win vs. lose)

    Overall 50/50

  1. I should have been very specific in my last response . . . when I say "If the door you choose is . . . "  I'm referring to the door you INITIALLY chose.

  1. For those naysayers out there that still think the percentage would be 50-50, here is another way of thinking about the percentages in the “dilemma” puzzles. A three person internal audit department decides to have a raffle to win a ticket to go to a fraud seminar. To make if fair, the CFO of the company is enlisted to help with the raffle. The CFO places the ticket in one of the envelopes, numbers the envelopes and seals them closed. The CFO knows which envelope contains the ticket. Each employee then draws an envelope out of a hat. One employee decides she will not participate after discovering the seminar is on her birthday and she gives her envelope to the newly hired junior auditor. At this point, the employee with one envelope has a 33 1/3 chance of winning and the other employee, the junior auditor with two envelopes, has a 66 2/3 chance of winning. The CFO decides to make things interesting and instructs the junior auditor to open the envelope number “3”. The CFO knows this is one of the empty envelopes. After this is revealed, the junior auditor feels uneasy about having the additional chance of winning and offers the other employee the chance to trade envelopes. Should the other employee accept the offer. YES! The junior auditors chances, with the two original envelopes are 66 2/3 percent (a 2 out of 3 shot!) and this percentage would remain after envelope 3 was revealed to not contain the ticket. Therefore, the other person should accept the offer to get a better shot at attending the seminar!

  1. Post (Part 2)

    I understand how the “Monty Hall Dilemma” works and agree with the percentages. However, I could not convince a co-worker of the percentages and the reason why she would want to switch boxes in the scenario you have presented. She contends the percentage is 50-50 (because of the additional information we learned by Monty revealing one of the Zonks). Following the logic you provided, this is the opposite conclusion she should have made – because of the additional information we would want to switch boxes to get the higher odds of winning. I suggested she run this past her engineer husband to see what  his conclusion would be to this puzzling dilemma and was met with a unique, “He’s in Canada” response.

    Try this one on for size. You select a box and secretly make a “second choice” selection in your mind during the grand finale of Let’s Make a Deal. Monty then proceeds to reveal a box containing a donkey carrying a cart full of bananas that is obviously one of the “Zonks”! If this box was your “second choice” selection, I would maintain you would want to keep your original choice. If Monty reveled the other box (the box that was not your first or second choice), then you would want to switch choices. Do you agree or disagree with this?

  1. Gentlemen:

    One more thought . . . let's get away from the 3 door example, and use another raffle example . . . 10 tickets are sold and one employee buys only one ticket, another employee buys the remaining 9.  There is only one winner, so the initial chances of any ticket winning is 1/10.  Initially, because the winning ticket "number" is not known, the employee that purchased the nine tickets has a 90% chance of winning the prize.  HOWEVER, if the contest moderator steps in (just to make it interesting) and takes 8 definite-loser envelopes from the contestant holding the 9 tickets -- NOW there are only TWO tickets, one is a loser, and one is a winner.   If "Switching" was an option, why would anyone with an initial 90% chance want to switch?  This negates the "it's better to switch" postulate that you are putting out there for all of us to accept. 

  1. In the 10 ticket raffle - the individual that had the 9 ticket (90% chance) would NOT want to switch - it would be the person that had the 1 ticket originally that would want to switch to get the higher odds.




  1. Please help me think this adaptation of the Monty Hall Dilemma through:
    Let's say there are 3 boxes and 2 players and each gets to select a box. There are two zonks and one grand prize in the boxes. 

    Player 1 has a 33 1/3 percent chance of winning. The other player + the "house" has a 66 2/3 percent chance of getting the grand prize. 

    The same for player 2 (33 1/3 percent chance of getting the grand prize) and the combined chance for the other two boxes is 66 2/3 percent.

    If Monty opens the other box that either player has not selected - it will either be the grand prize or a zonk. Let's say it is a zonk (otherwise the game would end). Monty then offers the players a chance to switch if they both agree. What should the players do?

    Here is the mindbender. Using Mr. Jacka's logic from the posts related to this issue, wouldn't both players want to switch as they would each have a 66 2/3 percent chance of winning? Adding these together, the combined percetnage is greater than 100%. Maybe I'm missing something, or maybe that is where the switching argument falls apart.

    Here are the scenarios:
    Player 1 would have a 33% chance without switching. However, if he opted to switch he would get the 67% chance (of the combined odds of the other two boxes - now collapsed into one box).

    However, player two would have the same exact scenario. A 67% chance of winning by switching.

    Seemingly, this creates a paradox.  How can each have a 67% chance of winning if they both switched. Am a missing something with the logic? 50-50 at this point sounds logical.





  1. Let's see.  The score is one response for Norman, two for me (three if you count this one), four for Greg, and the Academy Award for most fired up goes to Ben with six responses. 

    I'm not going to try and work through this any more.  Beyond experimenting with this yourself, I'd also suggest exploring what's out on the internet.  There are a lot of explanations, and some go into the actual math, so take a wander out there and see what there is to see.

    However, Greg raised a question that had me wondering for a while.  Here's the reason his thinking doesn't work.  In the original Monty Hall problem, with only one contestant, the reveal is always a Zonk.  (On the real show, when there were two contestants, Monty would always reveal one contestant's choice which was a loser.)  In the problem as posed by Greg above, there will be a one in three chance that the unchosen door will be the top prize.  That effectively changes all odds and percentages, resulting in a switch being a 50/50 chance.   One of the keys to the original Monty Hall problem is that a zonk will always be revealed first.

    Now, on with the arguments.

  1. Basically, here is a summation of my "Academy Award" responses (lol, I'm a good sport, so no offense taken).  Using the premises that you outlined, the odds that you INITIALLY chose the winner are simply 1 in 3 or 33.3%.  When one door is eliminated (as is always the case per the show and your example), the contestant now enters PHASE 2 (we'll call it) of the whole ordeal and has a CHOICE TO STAY OR SWITCH, and considering there are NOW only 2 doors [one of which is the grand prize winner] the odds to the contestant between switching and staying (hence, possibly choosing the winner as the result of one action) is 1 in 2 or 50%.  The initial hypothesis of "switching doubles your odds" or however it was worded is not true.  Because of the peripheral action by Monty revealing a zonk and giving the contestant a Phase 2 choice of "switching or staying", the altered odds have increased from 33.3% to 50%, but it isn't strictly due to making a switch.  The most correct statement would be "giving the contestant a 'second-chance' to stay or switch after elimination of a zonk increases the odds to 50/50."  THAT IS MY POINT.  After all, regardless of the first action, the elimination of a zonk and granting of a second-chance renders the whole contest as a 1 in 2 guess as to whether you INITIALLY selected a zonk (and switch or stay) or you INITIALLY selected the GP (and switch or stay).  [Also, my basis for making the statement that the issue becomes one of not WHICH DOOR you chose, but whether you will SWITCH or STAY.]

  1. Sorry to come in late - but I agree with Ben. Given three choices, when the contestant picks one there is a 100% chance that one of the two doors NOT chosen is a Zonk. Given that at least one of those two doors is not a winner how can it change the odds when that certainty is revealed?

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