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Deal or No Deal

  1. Sep 15, 2009 #1
    I am trying to settle a score on probability and odds, and I just wanted to see what everybody thought.

    This springs from the fact that when you watch "Deal or No Deal" and cases are removed one at a time, Howie frequently announces that there is a 15% chance that the case holds a million dollars, then later it is a 25% chance, then 35% and so on, until (if it gets there) a 50% chance.

    The question is, do the odds really improve?

    My stance is that since the there were 26 cases present when the players case was chosen, that the odds will stay frozen at 1/26 up until the point that the player gets to "reguess" (if he gets that far) so essentially if there are two cases left, his case still has a 1/26 chance of being a winner, but if he is allowed to reguess with the new information it becomes 1/2.

    Here is a better example.

    I will randomly choose a card from a regular deck and proclaim that it is the Ace of Spades without showing it. The odds of me being right are of course 1/52. As far as I can see, there are only two ways of determining if I am right:

    A: Flip over my card and check

    B: Flip over all of the other cards, and if the Ace of Spades is not there, it must be my card

    Do you agree so far?

    Would either one of these give me a better chance at getting the Ace of Spades? NO!!!!!!

    Now, lets take a closer look at situation B. Lets say that instead of just flipping over all of the cards, I decide to flip them over one at a time, very slowly. This is ABSOLUTELY NO DIFFERENT FROM JUST FLIPPING THEM ALL OVER! My odds do not get any better as more cards are flipped over. My chances are still 1/52. If I happen to flip over the Ace, I know that I lost, but that has nothing to do with my odds.

    Now, if somewhere in there you give me a chance to choose cards, then my odds become
    1/however many cards remain.
  2. jcsd
  3. Sep 15, 2009 #2


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    Of course your odds improved. If the $1 box is opened, you now know that YOU don't have the $1 box. So your box is either the $2, $5, etc. boxes. Since there's one less value on the list of potential boxes, your odds of having any given one are greater.
  4. Sep 15, 2009 #3
    Well then by that logic, when you have 1 case left and you know that it has to be the X dollar case, you still have a 1/26 chance of it being that case...
  5. Sep 15, 2009 #4
    I think you are missing the point.

    If I guess a case, it is 1/26. If I look at it and there is no million dollars, oh well, it was still a 1/26 chance. Right?

    What is the difference between looking in my case, or determining my case through deduction by looking at the other cases.


    Think of it this way, if I chose a case and then walked out of the room, I would have a 1/26 chance of winning a million. Now that I am out of the room, I let Howie open the case for me to see if I got the million. I don't care how he determined that I had a million or not (either opening my case or opening all the rest) when he comes in the hallway he is either going to tell me that I got the million or I didn't. his methods for finding out what is in my case did not improve my odds.
  6. Sep 15, 2009 #5


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    Yes, there is a 1/26 chance of getting the million in your case, but the point of the game is to try and maximize what the banker offers by weeding out the lower values.
  7. Sep 15, 2009 #6
    No. This reminds me of an old problem called the "Monty Hall problem." In that problem, what you said holds true - there is a 1/3 chance at the end that the contestant will win with the door that they chose because the probability remains the same as in the beginning. But there is a fundamental difference in that problem and in this problem. In the Monty problem, the person choosing the doors (or cases) knows where the prize is at and is purposely not choosing it. In this problem, everything is completely random. Let's say that Howie knew where he million dollar case is and he was choosing the cases for you, purposely not choosing the million dollar case, then at the end, you would have a 1/26 chance of having the million dollar case and a 25/26 chance of not having it. Therefore, if you could, you would want to switch.

    Let me put it another way. You're correct in saying that you have a 1/26 chance of having the million dollar case at the beginning... Just like you have a 1/26 chance of having the 1 dollar case, a 1/26 chance of having the 500 case, etc. So at the end, if it's down to just the 1 dollar case and the 1 million dollar case, you'll have a 50/50 shot no matter what because 1/26=1/26.
    Last edited: Sep 15, 2009
  8. Sep 15, 2009 #7
    Sure they do. Howie may not be a mathematician, but a major network isn't going to get simple probability like this wrong and embarrass themselves.

    I Googled "Deal or no deal" probability. The following came form the second hit:


    Huh? You're saying if he doesn't get a chance to switch the chances of him having the winning case is 1/26 but if he is given the choice his chances magically go to 1/2? When the contestant is down to two choices, his chances of winning $1,000,000 is 1/2 whether or not he is given a chance to switch.

    Of course they do. If you're lucky enough to never flip over the Ace of Spades when down to two cards, you have a 50/50 chance of having originally picked the Ace of Spades. Exactly what the wiki article I linked to above said about Deal or No Deal.
    Last edited: Sep 15, 2009
  9. Sep 15, 2009 #8

    It's you that missed Whatever123's point. If your probability stays the same no matter how many cases are revealed, then by your own logic if it were to get narrowed down to one box, your probability of winning $1,000,000 is 1/26. You realize this is incorrect, right?
  10. Sep 16, 2009 #9
    DavidSnider, thanks for seeing this the correct way. I understand the point of the game is to narrow down possibilities, but my question was based on the fact that the "odds" of having the million dollar case seem to chane every time a non million dollar case is chosen, and as you stated this is not the case.

    For everyone else, my second example is a better one.

    If you think that the odds change, please please please PLEASE explain to me how flipping over cards one at a time is any different than flipping them all over at once, and how either of these would improve my chances of picking the ace of spades over simply just picking a card and looking at it. There is no difference. There were 52 cards at the time I picked the card, and unless I somehow use the new information I gain to make a new decision (i.e. change cards in the middle of the flips, or at least get a chance to and decide to stay with my original) the odds do not change.
  11. Sep 16, 2009 #10


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    It would help if you very precisely stated the example you want to consider.

    But pay attention to this -- you have to analyze all cases if you want the odds to remain the same.

    The odds that you flip over the ace of spades on the 52nd card is 1 in 52.

    However, if you look only at those games where you haven't flipped over the ace of spades in the first 50 flips, then the probability of flipping over the ace of spades on the 52nd card is 1 in 2.

    This is tempered by the fact that if you have already flipped over the ace of spades in the first 50 flips, the probability of flipping over the ace of spades on the 52nd card is zero, and this situation is much more likely.
  12. Sep 16, 2009 #11


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    Flipping over cards one at a time is only different than flipping them all over at the same time because, well, there's time in between. Which means instead of one instantaneous event that vastly changes the chances of having the 1,000,0000 dollars, there's a lot of events that change it slightly (unless it's revealed of course). If I have a deck of 52 cards and you pick one at random:

    I flip over every card except for one left in the deck for you to look at. None of them are the ace of spades. There was a 1/52 chance that your card is the ace of spades, and a 1/52 chance the bottom card of the deck was the ace of spades. So since those are the only two cards left, there's a 1/2 chance of each card being the ace of spades. You seem to think you still only have a 1/52 chance of having the ace of spades. This means the odds of the bottom card of the deck being the ace of spades jumps to 51/52 by me revealing the other cards in the deck.

    So let's play a new game. I give you a card. I flip over every card in the deck except for the bottom one. If the bottom card is then the ace of spades, you win. According to you, if the ace of spades does not show up in any of the cards that I flipped over, there's a 51/52 chance of you winning. But how is this game really any different from the one above? In one game you won if the top card was the ace of spades, in one game you won if the bottom card was the ace of spades. In both games I showed you every card besides the top and the bottom card
  13. Sep 16, 2009 #12
    It is not more information that affects my decision on which card to choose, which was done at 1/52.

    Think of it this way, if I were in a drawing with 51 other people, I have a 1/52 chance of winning. If a winner is randomly drawn, and then the LOSING names were called of one by one instead of simply calling the winner, and I happen to be one of the last two people, my odds are not 1/2 to win the drawing!!!! they are still 1/52!

    Just because names are being called as losers doesn't change the fact that the woinner has already been determined, and eliminating any amount of non wining people will not change the odds whatsoever unless a new winner is drawn after the fact.

    How can you not see that?
  14. Sep 16, 2009 #13
    The odds you PICKED the million dollar case stay the same -- 1/26. The odds you HAVE the million dollar case change -- they improve each time a non-million dollar case is exposed. "Picked" and "have" are different problems.

    This applies to your cards example too.
  15. Sep 16, 2009 #14


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    What if your name was already called as a loser? Do you still have a 1/52 chance of winning?
  16. Sep 16, 2009 #15
    Now, here is the hole in your game. If we made a bet on that game at the very BEGINNING, the odds of me winning are one in 52. The act of you flipping over the cards simply delays us knowing the outcome of the game.

    One the other hand, if you flip over a card and ask me to rewager, then you change the game completely because I am making a new decisin based on new information.

    Again, I don't see how flipping over one card at a time gives me any more chances of winning than just shuffling the deck and looking at the bottom card. I am still only going to win this game with you 1/52 of the time regardless, giving me odds of 1/52.
  17. Sep 16, 2009 #16
    caprirs302, you are ignoring questions asked and points made and don't seem a bit interested in having your point of view changed. I'll play along a little longer and see if you're open to actually having a discussion and not just waving away any explanations that don't support your conclusion.

    Lets do this with three cards instead of 52 for simplicity's sake. Take two random cards from a deck and the Ace of Spades. Mix up these three cards and throw them in a hat. Take one out. One out of every three times you do this (on average), you will have picked the Ace of Spades.

    Now take the same cards, mix them up and throw them in a hat. Take one out and put it face down without ever seeing what card you've chosen. Remove another card from the hat and look at it. If it's not the Ace of Spades what are the chances that the card you have face down is the Ace of Spades. It's 1/2. Don't believe it? You can perform this experiment multiple times and you'll find that if every time you did not pick the the Ace of Spades the time you removed a card and peeked at it, your face down card will be the Ace of Spades about half the time.

    You ask why? Because new information changes probability. Choose one of the 26 cases and you have 1/26 of having the $1,000,000 case. Peek at the insides of 24 of the remaining 25 cases and see that the $1,000,000 case is not in any of those and you have a 1/2 chance of having the $1,000,000 case. Peek inside the other case and see it isn't the $1,000,000 case and you believe your case still only has a 1/26 chance of containing $1,000,000? No, you can be very excited because you have a 1/1 chance of having the $1,000,000 case. You're a guaranteed winner.
  18. Sep 16, 2009 #17
  19. Sep 16, 2009 #18
    It doesn't just simply delay the game; it provides new information. If we flip over the first card and it's the Ace of Spades, the game is over. If it's not, your chances of winning just got a little better. If we flip over 50 of them and none of them were the Ace of Spades, you wouldn't be happy because your odds of winning are significantly better than before you had information about the top 50 cards? You still think you only have a 1/52 chance of winning? No, if you perform this experiment enough times to get to the point where the top 50 cards weren't the Ace of Spades, you'll win 1/2 of the time. If you perform this experiment enough times to get to the point where the top 51 cards weren't the Ace of Spades, you'll win 100% of the time.
  20. Sep 16, 2009 #19
    Again, by that logic I would have to not include all of the times where I was flipping cards and the ace did not come up in the last two (25/26) I cannot simply eliminate those from the set!!! They count too!!!

    Yes, flipping cards over gives me new information, but if I do not get a chance to USE that information in making a decision it cannot affect my odds.

    Again, going back to my original post:

    If I simply pick a card and flip it over, the odds of it being an Ace of spades is 1/52

    If I use deduction to see if I got the ace by flipping over the other 51 cards, my odds are still 1/52

    If I use deduction by flipping over the other 51 cards one at a time you are telling me that i will get better odds of winning? This is awesome! I am going to go buy a lottery ticket right now, and have all of the losers called off one at a time, that way I have a better chance of winning!
  21. Sep 16, 2009 #20
    Exactly! Because the Ace didn't come out those times, your odds improve.

    You said this in your OP:
    The odds really improved because Howie didn't reveal a million dollar case!

    This is irrelevant to every scenario you posted. You brought up scenarios in what the probability would be if the winning case/card, etc. wasn't revealed and what the new probability would be.

    Yes, there is a reason to dismiss these situations. The reason is that those situations didn't play out. Once Howie removes 24 cases and none of them are the $1,000,000 case, we can dismiss them. There are two cases left and both have an equal chance of being the $1,000,000 case. The chances that it's the one you originally chose is 1/2.

    How does changing your guess change the frequency in which your card will be the Ace of Spades? Explain how this works. The card will be the Ace of Spades about 1/2 the time whether or not you speak any words.
    Last edited: Sep 16, 2009
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