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Probability, Deal or no deal

  • 15-12-2011 2:32am
    #1
    yenoah
    Closed Accounts Posts: 159 ✭✭


    Hi folks,

    If any of you are familiar with the TV show, Deal or no Deal; can you tell me, how would I work out the probability of the 250,000 box being chosen by the player ?

    There are 22 boxes, so I assume for one game the probability is 1/22. What about 6 consecutive games? would that be 6/22?


Comments

  • #2
    Victor
    Registered Users, Registered Users 2 Posts: 78,576 ✭✭✭✭Victor


    yenoah wrote: »
    Hi folks,

    If any of you are familiar with the TV show, Deal or no Deal; can you tell me, how would I work out the probability of the 250,000 box being chosen by the player ?

    There are 22 boxes, so I assume for one game the probability is 1/22. What about 6 consecutive games? would that be 6/22?

    No familiar with the game, but if you get the choice of picking 6 boxes out of 22, with the prize only being behind one door, the probability is higher than 6/22 (27.2%). If you think about it, the first one you open it is 1/22, the next one is 1/21 and so on.

    It would be (1/22)+(1/21)+(1/20)+(1/19)+(1/18)+(1/17)=31.0%


  • #3
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    No, you see its reset each day. Each day, the 26 boxes are jumbled up and a new contestant picks one at random.

    so day one is 1/22, then a reset, then day 2 is 1/22 etc or at least I think. So is it 6/22 ????


  • #4
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    Model it as a binomial distribution, with p(success) = 1/22.

    What is the probability of 1 success in 6 trials?

    6!/((5!)(1!)) * (1/22) * (21/22)^5 = 0.2161283
    (I can't get LaTex to work for me, sorry for cruddy notation above)

    That seems counterintuitive. Would the playing stats indicate that the 250K box is chosen by the player roughly 1 week in 5 (assuming 6 games a week)?

    (Of course the big prize being chosen doesn't mean it's won; what with all the shenannigans the Banker gets up to :))


  • #5
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    Yakuza wrote: »
    Model it as a binomial distribution, with p(success) = 1/22.

    What is the probability of 1 success in 6 trials?

    6!/((5!)(1!) * (1/22) * ((21/22))^5 = 0.2161283
    (I can't get LaTex to work for me, sorry for cruddy notation above)

    That seems counterintuitive. Would the playing stats indicate that the 250K box is chosen by the player roughly 1 week in 5 (assuming 6 games a week)?

    (Of course the big prize being chosen doesn't mean it's won; what with all the shenannigans the Banker gets up to :))

    Crikey I don't think this is it. It must be simpler.

    Each day theres a 1/22 chance. So how do I calculate for 6 days?


  • #6
    equivariant
    Registered Users, Registered Users 2 Posts: 966 ✭✭✭equivariant


    yenoah wrote: »
    Hi folks,

    If any of you are familiar with the TV show, Deal or no Deal; can you tell me, how would I work out the probability of the 250,000 box being chosen by the player ?

    There are 22 boxes, so I assume for one game the probability is 1/22. What about 6 consecutive games? would that be 6/22?

    You need to state your question more precisely OP. Do you mean

    A. What is the probability that the 250K box is chosen at least once by the player during a run of 6 consecutive games?

    or

    B. What is the the probability that the 250K box is chosen by the player in each of 6 consecutive games?


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  • #7
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    equivariant wrote: »
    You need to state your question more precisely OP. Do you mean

    A. What is the probability that the 250K box is chosen at least once by the player during a run of 6 consecutive games?

    or

    B. What is the the probability that the 250K box is chosen by the player in each of 6 consecutive games?


    B. What is the the probability that the 250K box is chosen by the player in each of 6 consecutive games?


  • #8
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    yenoah wrote: »
    B. What is the the probability that the 250K box is chosen by the player in each of 6 consecutive games?

    Ok, that's a different question to the one I answered (The probability of exactly one player picking the box in 6 shows).

    If the probability of picking the box in one show is 1/22, then (as each show is independent), the probability of two players picking the box in two consecutive shows is (1/22) * (1/22), for 3 it's (1/22) * (1/22) * (1/22) or (1/22)^3 etc...

    For 6 shows, it's (1/22)^6 or 1 in 113,379,904.

    Oddly enough (see what I did there? :)), those odds are roughly the same as those of getting the top prize in the Euromillions, as it turns out (116,531,800).

    I wouldn't hold my breath.


  • #9
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    Yakuza wrote: »
    Ok, that's a different question to the one I answered (The probability of exactly one player picking the box in 6 shows).

    If the probability of picking the box in one show is 1/22, then (as each show is independent), the probability of two players picking the box in two consecutive shows is (1/22) * (1/22), for 3 it's (1/22) * (1/22) * (1/22) or (1/22)^3 etc...

    For 6 shows, it's (1/22)^6 or 1 in 113,379,904.

    Oddly enough (see what I did there? :)), those odds are roughly the same as those of getting the top prize in the Euromillions, as it turns out (116,531,800).

    I wouldn't hold my breath.

    Many thanks,

    So how would I then calculate the odds of the box being drawn once in those 6 games. For one game it's 1/22. So if we increase the opportunity, 6 games, would that be 6/22?


  • #10
    equivariant
    Registered Users, Registered Users 2 Posts: 966 ✭✭✭equivariant


    yenoah wrote: »
    Many thanks,

    So how would I then calculate the odds of the box being drawn once in those 6 games. For one game it's 1/22. So if we increase the opportunity, 6 games, would that be 6/22?

    Do you mean

    A. Exactly once

    or

    B. At least once

    ?

    The answers are different for these two questions. Presuming that you mean B, then one way to go is to the find the prob of the box not being picked at all in the 6 games and then compute 1 - that. You end up with 1 - (21/22)^6.


  • #11
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    equivariant wrote: »
    Do you mean

    A. Exactly once

    or

    B. At least once

    ?

    The answers are different for these two questions. Presuming that you mean B, then one way to go is to the find the prob of the box not being picked at all in the 6 games and then compute 1 - that. You end up with 1 - (21/22)^6.

    what about Exactly once (A)?


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  • #12
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    My first post answers that; roughly 21% or 1/5

    It is a binomially-distributed variable. The box is either picked or it isn't. With six trials getting exactly one success can happen six ways:
    1 2 3 4 5 6
    p q q q q q, or
    q p q q q q, or
    q q p q q q, or
    q q q p q q, or
    q q q q p q, or
    q q q q q p.

    where p indicates success and q failure

    Each of these can happen with probability (1/22) * (21/22) * (21/22) * (21/22) * (21/22) * (21/22)

    Thus, the total probability is 6 * (1/22) * (21/22) * (21/22) * (21/22) * (21/22) * (21/22) = 0.21612830083186523072025180053072 ;)


  • #13
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    Very greatful for all the help. The reason I started this thread was that I found a web page where someone posted an article on probability in Deal or No Deal. I am sure it's riddled with errors however.

    Have alook at the past paragraph "Interesting statistics" for example
    http://www.jesusevidence.org/deal.html

    I sent him an email asking, but I'd say its a very old page and email address so possibly not working.
    No I'm not that nerdy. It's just that I have started to learn probability again and need literature to make sense otherwise I am sent into a spin. What do you think, is he wrong?


  • #14
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    yenoah wrote: »
    Very greatful for all the help. The reason I started this thread was that I found a web page where someone posted an article on probability in Deal or No Deal. I am sure it's riddled with errors however.

    Have alook at the past paragraph "Interesting statistics" for example
    http://www.jesusevidence.org/deal.html

    I sent him an email asking, but I'd say its a very old page and email address so possibly not working.
    No I'm not that nerdy. It's just that I have started to learn probability again and need literature to make sense otherwise I am sent into a spin. What do you think, is he wrong?

    I haven't checked them all, but the ones stating "[FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica][FONT=arial,helvetica]The odds of the £250,000 appearing at least once in the next 6 shows = 24.4%[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]" etc are all correct.

    What he's stating there is not the same as what you've been asking.

    The odds of the 250k box appearing *at least* once in 6 shows is not the same as the odds of it appearing *exacty* once on 6 shows.

    To calculate the odds of an "at least once" type event, the easiest thing is to work out the probability of the opposite thing happening (ie not happening at all), and take that figure from 1.

    So, the odds of zero sucesses in 6 trials (this can only happen one way) is : (21/22) ^ 6 = 0.756449053. Thus, the probabilty of *at least* one appearance is 1 minus this, or approx 24.4%, as he states. Lashing the other figures into Excel (BINOMDIST function is v handy) corroborates the others.

    The long-winded way to work out the probabilty of at least one success would be to work out the probabilty of 1 (done in a post above by me), 2, 3, 4, 5 and 6 and add them up; which is easier? :)


  • #15
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    Would the long winded way be
    1/22 + 1/22 + 1/22 + 1/22 + 1/22 + 1/22 = 27%

    The other way, you describe above is 1 - the probability of it not happening in 6 attempts thus
    1 - 21/22 ^6 = 24%

    I'm really confused, very new to this and trying to get my head around it. Why have I two different answers here?


  • #16
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    Probability of 2 sucesses:
    There are 15 ways of getting 2 sucesses from 6 trials:
    ppqqqq pqpqqq pqqpqq pqqqpq pqqqqp
    qppqqq qpqpqq qpqqpq qpqqqp
    qqppqq qqpqpq qqpqqp
    qqqppq qqqpqp
    qqqqpp

    The probability of any of the above is (1/22)^2 * (21/22)^4, so the probability of getting two sucesses is 15 * (1/22)^2 * (21/22)^4 = 0.02572956

    There are 20 ways of getting 3 sucesses from , 15 ways of getting 4, 6 ways of getting 5 and 1 way of getting 6 . I'm not going to type in all the permutations here (check out http://stattrek.com/Tables/Binomial.aspx and look up Binomial Distribution on Wikipedia) but the results are:

    P(1): 0.2161283008
    P(2): 0.0257295596
    P(3): 0.0016336228
    P(4): 0.0000583437
    P(5): 0.0000011113
    P(6): 0.0000000088

    Total 0.2435509471

    I repeat, which is easier? :)


  • #17
    yenoah
    Closed Accounts Posts: 159 ✭✭yenoah


    Yakuza wrote: »
    Probability of 2 sucesses:
    There are 15 ways of getting 2 sucesses from 6 trials:
    ppqqqq pqpqqq pqqpqq pqqqpq pqqqqp
    qppqqq qpqpqq qpqqpq qpqqqp
    qqppqq qqpqpq qqpqqp
    qqqppq qqqpqp
    qqqqpp

    The probability of any of the above is (1/22)^2 * (21/22)^4, so the probability of getting two sucesses is 15 * (1/22)^2 * (21/22)^4 = 0.02572956

    There are 20 ways of getting 3 sucesses from , 15 ways of getting 4, 6 ways of getting 5 and 1 way of getting 6 . I'm not going to type in all the permutations here (check out http://stattrek.com/Tables/Binomial.aspx and look up Binomial Distribution on Wikipedia) but the results are:

    P(1): 0.2161283008
    P(2): 0.0257295596
    P(3): 0.0016336228
    P(4): 0.0000583437
    P(5): 0.0000011113
    P(6): 0.0000000088

    Total 0.2435509471

    I repeat, which is easier? :)

    Im very grateful for your answers. Unfortunately you are conmpletely blinding me and as the thread progresses I am getting more confused. Lets end it here!

    As a newcomer to probability I need some initial hand holding and teaching as though I was 6 years old using visuals and simple examples, simple language etc.

    Again many thanks for your efforts, I just need to consume it in a different way!


  • #18
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    yenoah wrote: »
    Im very grateful for your answers. Unfortunately you are conmpletely blinding me and as the thread progresses I am getting more confused. Lets end it here!

    As a newcomer to probability I need some initial hand holding and teaching as though I was 6 years old using visuals and simple examples, simple language etc.

    Again many thanks for your efforts, I just need to consume it in a different way!
    No problem, I know I'm not the best teacher! If you google "Binomial Distribution" there are lots of links to tutorials to start. Perhaps start with easier probabilties like getting x heads when throwing a coin y times.


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