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Thread: Geoff Hall: A,7 vs 9

  1. #1
    Geoff Hall
    Guest

    Geoff Hall: A,7 vs 9

    I had a friend of mine recently involved in an interesting proposition and he asked me for my advice which unfortunately is more limited now that I do not spend any time teaching math's or stat's.
    The proposition evolved from a disagreement with an obnoxious player who was convinced that soft 18 should never be hit against 9,10 or Ace.
    I told my friend that a hit on soft 18 was more advantageous against a 9 and to use this upcard for the following scenario :-
    200 hands are dealt with soft 18 verses dealer 9 and the total of hands won is tallied, for hitting and standing. The method which wins the most hands gets paid a fixed amount.
    The question I have is that after 200 hands what is the probability that the 'Hit soft 18' will be ahead of the 'stand 18'. I also managed to work out that you would expect to win 90 hands by hitting and 82 hands by standing but I could not calculate what chance my friend had of losing the bet over 200 hands.

    Any ideas gratefully received.

    Best regards

    Geoff Hall

  2. #2
    PaddyBoy
    Guest

    PaddyBoy: A7 v 9

    This is my amatuerish attempt at answering. From p332 of ProBJ.6 decks,S17.Ev stand is -.183.EV hit is -.098.So a gain of .085 for hitting.So 200X.00085=1.7 units.
    This to me means that the person that hits will be ahead 1.7 units on the person that stands over 200 hands.
    I dont know how to work out the percentage of wins to losses as pushes would affect that i suppose.
    Ok Don,Norm how did i do?

  3. #3
    Karel
    Guest

    Karel: Re: A,7 vs 9


    I am not sure I understand correctly. It is not enough to know how many hands are won. It is also necessary to know how many hands are lost (or pushed).

    It is clear that standing on soft 18 will push more hands that hitting soft 18. (For example, standing cannot bust.) Thus, it is not a proper argument to say that hitting wins more hands, even if it is true. Most likely, hitting also loses more hands and pushes fewer.

    If we know the frequency of loss/win/push distribution, it is possible to calculate approximately by using normal distribution. The approximation will be excellent for 200 hands, with negligible difference compared to precise numbers. The calculation is as follows:
    Suppose that frequency $p$ hands win$, $q$ lose, $r=1-p-q$ push. Then the edge is
    $p-q$, and the variance per one game is
    $p + q - (p-q)^2$.

    The distribution after 200 hands will approximately be, with excellent approximation for practical purposes, the normal distribution with mean 200*(p-q) and variance 200*(p+q-(p-q)^2).

    To analyze the difference of results we note that the games are supposed to be independent, so variance of the difference is sum of variances. The difference between games 1 and 2 will have the distribution:

    Normal( 200*((p1-q1)-(p2-q2)), 200*(p1+q1-(p1-q1)^2) + 200*(p2+p2-(p2-q2)^2) ).

    Given the mean and standard deviation (variance) of the normal distribution, it is an easy question to find the probability that the actual result will be more than 0. A function
    NormsDist(edge/standard deviation))
    in Excel gives the result. I can do the calculation of anybody posts p1, q1, p2, q2.

    Regards,

    Karel

    > The question I have is that after 200 hands
    > what is the probability that the 'Hit soft
    > 18' will be ahead of the 'stand 18'. I also
    > managed to work out that you would expect to
    > win 90 hands by hitting and 82 hands by
    > standing but I could not calculate what
    > chance my friend had of losing the bet over
    > 200 hands.

  4. #4
    Don Schlesinger
    Guest

    Don Schlesinger: Re: A,7 vs 9

    > I can do the
    > calculation of anybody posts p1, q1, p2, q2.

    Unfortunately, this probably won't help, unless we stipulate resolved hands and ignore pushes. If so, hitting loses 54.9% and wins 45.1% for a net EV of -.098 (rounded). Standing loses 59.1% and wins 40.9%, for a net EV of -.183 (rounded).

    Don

  5. #5
    Geoff Hall
    Guest

    Geoff Hall: Re: A,7 vs 9

    > Unfortunately, this probably won't help,
    > unless we stipulate resolved hands and
    > ignore pushes. If so, hitting loses 54.9%
    > and wins 45.1% for a net EV of -.098
    > (rounded). Standing loses 59.1% and wins
    > 40.9%, for a net EV of -.183 (rounded).

    > Don

    The above percentages is where I got the 'hands won' numbers i.e. hitting ~ 90 hands won (45.1% x 2) and standing ~ 82 hands won (40.9% x 2).
    So you would expect to be 8 hands ahead after dealing 200 hands with hitting.
    My query is what is the chance of still being behind after just 200 hands. Furthermore, how many hands would needed to be dealt in order for P(hitting) > P(standing) to be greater than 0.9% ?
    If 2 people tossed an unbiased coin then you would expect that both results would be between 86-114 'heads' (1 sd) 67% of the time.
    I have a problem in the fact that E(X) is different for each distribution and can't remember the proper 'tools' for calculating the answer.
    I did consider using a 't' or 'z' test but not sure if I was on the right lines.

    Thanks for your input Don,Karel,Paddyboy.

    Best regards

    Geoff

  6. #6
    Karel
    Guest

    Karel: A rough guess would be okay


    The problem is that we need also the frequency of unresolved hands to be precise. Pretty much, those are the hands that do not help either way. The pushed trials are useless for the result. However, it would be okay to have a very rough idea. For example, if the frequency was 10% in both cases, we would need to use 10% more rounds to play compared to "no pushes", in order to achieve the same confidence.

    Karel

    > Unfortunately, this probably won't help,
    > unless we stipulate resolved hands and
    > ignore pushes. If so, hitting loses 54.9%
    > and wins 45.1% for a net EV of -.098
    > (rounded). Standing loses 59.1% and wins
    > 40.9%, for a net EV of -.183 (rounded).

    > Don

  7. #7
    Cacarulo
    Guest

    Cacarulo: Here is some data that may help

    > I am not sure I understand correctly. It is
    > not enough to know how many hands are won.
    > It is also necessary to know how many hands
    > are lost (or pushed).

    > It is clear that standing on soft 18 will
    > push more hands that hitting soft 18. (For
    > example, standing cannot bust.) Thus, it is
    > not a proper argument to say that hitting
    > wins more hands, even if it is true. Most
    > likely, hitting also loses more hands and
    > pushes fewer.

    > If we know the frequency of loss/win/push
    > distribution, it is possible to calculate
    > approximately by using normal distribution.
    > The approximation will be excellent for 200
    > hands, with negligible difference compared
    > to precise numbers. The calculation is as
    > follows:
    > Suppose that frequency $p$ hands win$, $q$
    > lose, $r=1-p-q$ push. Then the edge is
    > $p-q$, and the variance per one game is
    > $p + q - (p-q)^2$.

    > The distribution after 200 hands will
    > approximately be, with excellent
    > approximation for practical purposes, the
    > normal distribution with mean 200*(p-q) and
    > variance 200*(p+q-(p-q)^2).

    > To analyze the difference of results we note
    > that the games are supposed to be
    > independent, so variance of the difference
    > is sum of variances. The difference between
    > games 1 and 2 will have the distribution:

    > Normal( 200*((p1-q1)-(p2-q2)),
    > 200*(p1+q1-(p1-q1)^2) +
    > 200*(p2+p2-(p2-q2)^2) ).

    > Given the mean and standard deviation
    > (variance) of the normal distribution, it is
    > an easy question to find the probability
    > that the actual result will be more than 0.
    > A function
    > NormsDist(edge/standard deviation))
    > in Excel gives the result. I can do the
    > calculation of anybody posts p1, q1, p2, q2.

    A7v9 (mean, sd)
    ---------------
    S = -0.18263993715641 | 0.92132041168269
    H = -0.09846902159307 | 0.93675669575608

    We can use the random variable X = H-S and find the P(X>0).

    Mean(X) = Mean(H)-Mean(S) = 0.08417091556334
    Var(X) = Var(H)+Var(S) = 1.72634440802701
    Sd(X) = sqrt(Var(X)) = 1.31390426136268

    After 200 trials the mean would be:
    Mean(X')=200*Mean(X) and the variance Var(X')=200*Var(X).

    You can then use U = [0 - Mean(X')]/sqrt(Var(X')

    Hope this helps.

    Sincerely,
    Cacarulo

  8. #8
    Geoff Hall
    Guest

    Geoff Hall: Re: Here is some data that may help

    > A7v9 (mean, sd)
    > ---------------
    > S = -0.18263993715641 | 0.92132041168269
    > H = -0.09846902159307 | 0.93675669575608

    > We can use the random variable X = H-S and
    > find the P(X>0).

    > Mean(X) = Mean(H)-Mean(S) = 0.08417091556334
    > Var(X) = Var(H)+Var(S) = 1.72634440802701
    > Sd(X) = sqrt(Var(X)) = 1.31390426136268

    > After 200 trials the mean would be:
    > Mean(X')=200*Mean(X) and the variance
    > Var(X')=200*Var(X).

    > You can then use U = [0 -
    > Mean(X')]/sqrt(Var(X')

    > Hope this helps.

    > Sincerely,
    > Cacarulo

    Thanks for the analysis Cacarulo - there's a ?10,000 ($15,000) wager involved so I can understand why my friend wants to be reasonably certain before he accepts the bet.

    Best regards

    Geoff Hall

  9. #9
    Don Schlesinger
    Guest

    Don Schlesinger: Re: A,7 vs 9

    "If 2 people tossed an unbiased coin [200 times] then you would expect that both results would be between 86-114 'heads' (1 sd) 67% of the time."

    No, that's incorrect. Leaving out, for the moment, that the normal distribution is only an approximation here to the binomial, and that, for relativley small numbers of trials (200), the precise binomial probabilities are higher than the normal ones, your SD is twice what it should be.

    Square root of 200 is, roughly, 14. You need to divide by 2 for the SD, when moving left or right from the mean of 100. So, the one-SD interval is 100 plus or minus 7 (actually 7.07), not 14. The interval is 93-107, and that's what should occur 68.3% of the time (not 67%, by the way).

    In point of fact, just to be precise, since this is binomial, it actually occurs 71.09% of the time.

    Don

  10. #10
    Don Schlesinger
    Guest

    Don Schlesinger: Re: Here is some data that may help

    > Thanks for the analysis Cacarulo - there's a
    > ?10,000 ($15,000) wager involved so I can
    > understand why my friend wants to be
    > reasonably certain before he accepts the
    > bet.

    Let's put it this way: there is a HUGE probability that hitting will win. So, if the wager is for even money, just go ahead and accept it! :-)

    Don

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