deal or no deal

Discussion in 'Python' started by rbt, Dec 22, 2005.

  1. rbt

    rbt Guest

    The house almost always wins or are my assumptions wrong...

    import random

    amounts = [.01, 1, 5, 10, 25, 50, 75, 100, 200, 300, 400, 500, 750,
    1000, 5000, 10000, 25000, 50000, 75000, 100000, 200000,
    300000, 400000, 500000, 750000, 1000000]

    results = []

    count = 0
    while count < 10:
    count = count + 1
    pick = random.choice(amounts)
    if pick < 100000:
    results.append("NBC won... Your briefcase contains $%s" %pick)
    else:
    results.append("You won... Your briefcase contains $%s" %pick)

    results.sort()
    print "Here are 10 random picks: "
    for result in results:
    print result
    rbt, Dec 22, 2005
    #1
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  2. On Thu, 22 Dec 2005 09:29:49 -0500, rbt <> wrote:

    >The house almost always wins or are my assumptions wrong...
    >
    >import random
    >
    >amounts = [.01, 1, 5, 10, 25, 50, 75, 100, 200, 300, 400, 500, 750,
    > 1000, 5000, 10000, 25000, 50000, 75000, 100000, 200000,
    > 300000, 400000, 500000, 750000, 1000000]
    >
    >results = []
    >
    >count = 0
    >while count < 10:
    > count = count + 1
    > pick = random.choice(amounts)
    > if pick < 100000:
    > results.append("NBC won... Your briefcase contains $%s" %pick)
    > else:
    > results.append("You won... Your briefcase contains $%s" %pick)
    >
    >results.sort()
    >print "Here are 10 random picks: "
    >for result in results:
    > print result


    I don't know what you are doing, but 10 is a small sample to draw confident
    conclusions from. E.g., try counting how many times pick<100000 out of
    a larger number, e.g.,

    (BTW, the expression x<y has the value True or False, and True and False are
    int subclass instances with int values 1 and 0 respectively, so the sum below
    is adding 1 whenever random.choice(amounts)<100000 and 0 otherwise)

    >>> import random
    >>> amounts = [.01, 1, 5, 10, 25, 50, 75, 100, 200, 300, 400, 500, 750,

    ... 1000, 5000, 10000, 25000, 50000, 75000, 100000, 200000,
    ... 300000, 400000, 500000, 750000, 1000000]
    >>> sum(random.choice(amounts)<100000 for _ in xrange(1000))

    736
    >>> sum(random.choice(amounts)<100000 for _ in xrange(1000))

    734
    >>> sum(random.choice(amounts)<100000 for _ in xrange(1000))

    730
    >>> sum(random.choice(amounts)<100000 for _ in xrange(1000))

    721
    >>> sum(random.choice(amounts)<100000 for _ in xrange(1000))

    753

    What should the sum be?

    >>> amounts.index(100000)

    19
    is the number of numbers < 100000 in amounts

    >>> len(amounts)

    26
    is the total number of numbers

    So if you choose with uniform probability, you'd expect
    >>> 19./26

    0.73076923076923073

    to be the fraction satisfying pick<100000

    With a larger sample, the fraction should show better, e.g.,

    >>> sum(random.choice(amounts)<100000 for _ in xrange(1000000))

    730983

    Does that help?

    Regards,
    Bengt Richter
    Bengt Richter, Dec 22, 2005
    #2
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  3. rbt

    rbt Guest

    Bengt Richter wrote:
    > On Thu, 22 Dec 2005 09:29:49 -0500, rbt <> wrote:
    >
    >> The house almost always wins or are my assumptions wrong...
    >>
    >> import random
    >>
    >> amounts = [.01, 1, 5, 10, 25, 50, 75, 100, 200, 300, 400, 500, 750,
    >> 1000, 5000, 10000, 25000, 50000, 75000, 100000, 200000,
    >> 300000, 400000, 500000, 750000, 1000000]
    >>
    >> results = []
    >>
    >> count = 0
    >> while count < 10:
    >> count = count + 1
    >> pick = random.choice(amounts)
    >> if pick < 100000:
    >> results.append("NBC won... Your briefcase contains $%s" %pick)
    >> else:
    >> results.append("You won... Your briefcase contains $%s" %pick)
    >>
    >> results.sort()
    >> print "Here are 10 random picks: "
    >> for result in results:
    >> print result

    >
    > I don't know what you are doing, but 10 is a small sample to draw confident
    > conclusions from. E.g., try counting how many times pick<100000 out of
    > a larger number, e.g.,


    The TV show on NBC in the USA running this week during primetime (Deal
    or No Deal). I figure there are roughly 10, maybe 15 contestants. They
    pick a briefcase that has between 1 penny and 1 million bucks and then
    play this silly game where NBC tries to buy the briefcase from them
    while amounts of money are taken away from the list of possibilities.
    The contestant's hope is that they've picked a briefcase with a lot of
    money and that when an amount is removed from the list that it is small
    amount of money not a large amount (I categorize a large amount to be
    more than 100,000)

    NBC tries to give the least amount of money possible away. The
    contestants try to get the most.

    > What should the sum be?


    100,000 or above unless you're already rich ;)
    rbt, Dec 22, 2005
    #3
  4. rbt wrote:

    > The TV show on NBC in the USA running this week during primetime (Deal
    > or No Deal). I figure there are roughly 10, maybe 15 contestants. They
    > pick a briefcase that has between 1 penny and 1 million bucks and then
    > play this silly game where NBC tries to buy the briefcase from them
    > while amounts of money are taken away from the list of possibilities.
    > The contestant's hope is that they've picked a briefcase with a lot of
    > money and that when an amount is removed from the list that it is small
    > amount of money not a large amount (I categorize a large amount to be
    > more than 100,000)


    Well, if the contestants' choices are truly random, and they stick with
    their first choice all the way to the end, each contestant wins, on
    average, $131 477.54 (sum(amounts)/len(amounts)).

    Assuming that the buyout offer is always less than (or equal to) the
    average of the still-available amounts, NBC will (on average) never have
    to pay out more than ~$132k per contestant. Likely they'll pay out less,
    because most people will get nervous before the very end, and will take
    the low ball offer NBC is fronting.

    What I would really like to know, is how they calculate the offer.
    Obviously, they set the upper limit at the average of the still standing
    offers, but I wonder if and how they take subsequent rounds into
    consideration. Is there a "Monty Hall"
    (http://en.wikipedia.org/wiki/Monty_Hall_problem) type consideration
    that needs to be taken into effect as cases are eliminated?
    Rocco Moretti, Dec 22, 2005
    #4
  5. rbt

    Duncan Smith Guest

    Rocco Moretti wrote:
    > rbt wrote:
    >
    >> The TV show on NBC in the USA running this week during primetime (Deal
    >> or No Deal). I figure there are roughly 10, maybe 15 contestants. They
    >> pick a briefcase that has between 1 penny and 1 million bucks and then
    >> play this silly game where NBC tries to buy the briefcase from them
    >> while amounts of money are taken away from the list of possibilities.
    >> The contestant's hope is that they've picked a briefcase with a lot of
    >> money and that when an amount is removed from the list that it is
    >> small amount of money not a large amount (I categorize a large amount
    >> to be more than 100,000)

    >
    >
    > Well, if the contestants' choices are truly random, and they stick with
    > their first choice all the way to the end, each contestant wins, on
    > average, $131 477.54 (sum(amounts)/len(amounts)).
    >
    > Assuming that the buyout offer is always less than (or equal to) the
    > average of the still-available amounts, NBC will (on average) never have
    > to pay out more than ~$132k per contestant. Likely they'll pay out less,
    > because most people will get nervous before the very end, and will take
    > the low ball offer NBC is fronting.
    >
    > What I would really like to know, is how they calculate the offer.
    > Obviously, they set the upper limit at the average of the still standing
    > offers, but I wonder if and how they take subsequent rounds into
    > consideration.


    I've seen them offer more than the average towards the end of the game
    (UK version).

    Duncan
    Duncan Smith, Dec 25, 2005
    #5
  6. rbt

    Guest

    [snip]
    >
    > What I would really like to know, is how they calculate the offer.
    > Obviously, they set the upper limit at the average of the still standing
    > offers, but I wonder if and how they take subsequent rounds into
    > consideration. Is there a "Monty Hall"
    > (http://en.wikipedia.org/wiki/Monty_Hall_problem) type consideration
    > that needs to be taken into effect as cases are eliminated?


    I believe not; the Monty Hall problem is biased by the fact that the
    presenter knows where the prize is, and eliminates one box accordingly.
    Where boxes are eliminated at random, it's impossible for any given
    box to have a higher probability of containing any given amount of
    money than another. And for the contestants box to be worth more or
    less than the mean, it must have a higher probability of containing a
    certain amount.

    Like another member of the group, I've seen them offer more than the
    average on the UK version, which puzzled me quite a lot.

    James M
    , Dec 26, 2005
    #6
  7. rbt

    Chip Turner Guest

    On 2005-12-26 15:05:21 -0500, said:

    > I believe not; the Monty Hall problem is biased by the fact that the
    > presenter knows where the prize is, and eliminates one box accordingly.
    > Where boxes are eliminated at random, it's impossible for any given
    > box to have a higher probability of containing any given amount of
    > money than another. And for the contestants box to be worth more or
    > less than the mean, it must have a higher probability of containing a
    > certain amount.


    Agreed -- unless the presenter takes away a case based on knowledge he
    has about the contents, then Monty Hall doesn't enter into it. Deal or
    No Deal seems to be a purely chance based game. However, that doesn't
    mean there aren't strategies beyond strictly expecting the average
    payout.

    > Like another member of the group, I've seen them offer more than the
    > average on the UK version, which puzzled me quite a lot.


    I imagine it is about risks. Many gameshows take out insurance
    policies against the larger payoffs to protect the show and network
    from big winners. I believe Who Wants to be a Millionaire actually had
    some difficulty with their insurance when they were paying out too
    often, or something. Perhaps the UK Deal or No Deal didn't want to
    risk increasing their premium :)

    But even the contestant has a reason to not just play the average,
    thereby bringing psychology into the game. It comes down to the odd
    phenomenon that the value of money isn't linear to the amount of money
    in question. If you're playing the game, and only two briefcases are
    left -- 1,000,000 and 0.01, and the house offers you 400,000... take
    it! On average you'll win around 500,000, but half the time, you'll
    get a penny. Averages break down when you try to apply them to a
    single instance. On the flip side, if you think about how much
    difference 500,000 will make in your life vs, say, 750,000, then taking
    a risk to get 750,000 is probably worth it; sure, you might lose
    250,000 but on top of 500,000, the impact of the loss you would suffer
    is significantly lessened. In the end, it comes down to what the money
    on the table means to *you* and how willing you are to lose the
    guaranteed amount to take risks.

    It's similar to the old game of coin flipping to double your money.
    Put a dollar on the table. Flip a coin. Heads, you double your bet,
    tails you lose it all. You can stop any time you want. The expected
    outcome is infinite money (1 * 1/2 + 2 * 1/4 + 4 * 1/8 ...), but a
    human playing it would do well to stop before the inevitable tails
    comes along, even though mathematically the house pays out an expected
    infinite number of dollars over time. Exponential growth in winnings
    doesn't offset exponential risk in taking a loss because, once you hit
    a certain point, the money on the table is worth more than the 50%
    chance of having twice as much.

    Chip

    --
    Chip Turner
    Chip Turner, Dec 26, 2005
    #7
  8. rbt

    Duncan Smith Guest

    Chip Turner wrote:
    > On 2005-12-26 15:05:21 -0500, said:
    >
    >> I believe not; the Monty Hall problem is biased by the fact that the
    >> presenter knows where the prize is, and eliminates one box accordingly.
    >> Where boxes are eliminated at random, it's impossible for any given
    >> box to have a higher probability of containing any given amount of
    >> money than another. And for the contestants box to be worth more or
    >> less than the mean, it must have a higher probability of containing a
    >> certain amount.

    >
    >
    > Agreed -- unless the presenter takes away a case based on knowledge he
    > has about the contents, then Monty Hall doesn't enter into it. Deal or
    > No Deal seems to be a purely chance based game. However, that doesn't
    > mean there aren't strategies beyond strictly expecting the average payout.
    >
    >> Like another member of the group, I've seen them offer more than the
    >> average on the UK version, which puzzled me quite a lot.

    >
    >
    > I imagine it is about risks. Many gameshows take out insurance policies
    > against the larger payoffs to protect the show and network from big
    > winners. I believe Who Wants to be a Millionaire actually had some
    > difficulty with their insurance when they were paying out too often, or
    > something. Perhaps the UK Deal or No Deal didn't want to risk
    > increasing their premium :)
    >
    > But even the contestant has a reason to not just play the average,
    > thereby bringing psychology into the game. It comes down to the odd
    > phenomenon that the value of money isn't linear to the amount of money
    > in question. If you're playing the game, and only two briefcases are
    > left -- 1,000,000 and 0.01, and the house offers you 400,000... take
    > it! On average you'll win around 500,000, but half the time, you'll get
    > a penny. Averages break down when you try to apply them to a single
    > instance. On the flip side, if you think about how much difference
    > 500,000 will make in your life vs, say, 750,000, then taking a risk to
    > get 750,000 is probably worth it; sure, you might lose 250,000 but on
    > top of 500,000, the impact of the loss you would suffer is significantly
    > lessened. In the end, it comes down to what the money on the table
    > means to *you* and how willing you are to lose the guaranteed amount to
    > take risks.
    >
    > It's similar to the old game of coin flipping to double your money. Put
    > a dollar on the table. Flip a coin. Heads, you double your bet, tails
    > you lose it all. You can stop any time you want. The expected outcome
    > is infinite money (1 * 1/2 + 2 * 1/4 + 4 * 1/8 ...), but a human playing
    > it would do well to stop before the inevitable tails comes along, even
    > though mathematically the house pays out an expected infinite number of
    > dollars over time. Exponential growth in winnings doesn't offset
    > exponential risk in taking a loss because, once you hit a certain point,
    > the money on the table is worth more than the 50% chance of having twice
    > as much.
    >
    > Chip
    >


    As you say, it depends on the player's utility function. But it's not a
    straightforward question of comparing the offer to the expected values
    of the remaining boxes, even for a risk-neutral player. At most stages
    of the game refusing an offer means that there will be a future offer,
    and, later in the game, these tend to be closer to (or even greater
    than) the expected value.

    Duncan
    Duncan Smith, Dec 28, 2005
    #8
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