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Thread: On Average How Many Blackjack Hands Have to be Played to Win 5 in a Row

  1. #27


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    Quote Originally Posted by aceside View Post
    That is impossible in only 14 rounds. However, let me give you an example of having both 5 head and 3 head in a 24-round flip.
    Let us define x=head, o=tail.

    xox,xoo,oxx,oxo,xxx,xxo,oox,ooo

    This represents a set of 8 3-round flips, and there is one 3-head flip, xxx; however, there is also one 5-head flip, xxxxx.
    It all depends on how many flips we are going to do, there will be 3-head, 4-head, 5-head ... and so on.

    The average 14 rounds required is based on below rules :-

    1) You flip a fair(50/50) coin continuosly.
    2) once you get 3 head in a row, you STOP immediately and record total no of flips since first flip, including the last three flips, say R.
    3) You repeat 1) and 2) for X times.
    4) Average rounds to have a 3 head in a row = (R1 + R2 + R3 + . . ...Rx)/X = 14. !

    If involve tie/push(break or not break the streak), as per Midwest PlayerBJ promo, it will be more complex, you have to solve 5 simultaneous equations as describe in post #9. For Midwest Player BJ promo, the average rounds required to have a 5 winning streak(pushes not break the streak) is 84.28441, with p=0.4332, q=0.4788,r = 0.0880.

    If p=0.4332, q=0.5668,r = 0, average rounds required to have a 5 winning streak is 113.88099.
    If p=0.475, q=0.525,r = 0, average rounds required to have a 5 winning streak is 76.86738.

    When r = 0, it does not matter pushes break the streak or not.
    Last edited by James989; 11-21-2021 at 09:18 PM.

  2. #28


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    Quote Originally Posted by Midwest Player View Post
    However, on the other hand I'm still having my best year ever. I'm presently Diamond status and only need less than 20,000 points out of 1,000,000 to reach the top tier of Onyx by the end of the year. Maybe I will even get better promotions then.
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  3. #29


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    One point that no one has made is that, when you do qualify for this bonus, your last five non-push results are guaranteed to be wins, so your average EV in qualifying will be somewhat higher than the EV for the average number of rounds before hitting the streak.

  4. #30
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    Quote Originally Posted by James989 View Post
    The average 14 rounds required is based on below rules :-

    1) You flip a fair(50/50) coin continuosly.
    2) once you get 3 head in a row, you STOP immediately and record total no of flips since first flip, including the last three flips, say R.
    3) You repeat 1) and 2) for X times.
    4) Average rounds to have a 3 head in a row = (R1 + R2 + R3 + . . ...Rx)/X = 14. !
    .
    This becomes clearer for me now. Can you demonstrate the calculation for the average number of required flips for a 3 head streak? It seems to me the above number X is infinite and thus it is impossible to calculate. If I can understand this number 14, I should be able to understand your blackjack streak calculator.

  5. #31


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    Quote Originally Posted by aceside View Post
    It seems to me the above number X is infinite and thus it is impossible to calculate
    James989 has provided a methodology for computing the number by simulation. X is the number of times you run the simulation. You choose X. The higher X is, the more accurate the result will be (i.e. more correct decimal places).

  6. #32
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    Quote Originally Posted by Gronbog View Post
    James989 has provided a methodology for computing the number by simulation. X is the number of times you run the simulation. You choose X. The higher X is, the more accurate the result will be (i.e. more correct decimal places).
    But this number 14 is a magic number. It may not need the recursive method of solving a set of three equations. My concern is still about if it is exactly a 3-head steak or a 3-or-more-head streak.

  7. #33


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    Quote Originally Posted by aceside View Post
    But this number 14 is a magic number. It may not need the recursive method of solving a set of three equations. My concern is still about if it is exactly a 3-head steak or a 3-or-more-head streak.
    It's exactly a three-head streak, because you stop as soon as you get three heads! Then you record that number. You do this over and over again and take the arithmetic average of the results. That number is 14. Stop beating this to death. There's nothing more to be said about it.

    Don

  8. #34
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    0 out of 2 members found this post helpful. Did you find this post helpful? Yes | No
    Isn’t it cute to write the number 14 as the sum of an infinite series? Based on this process. This is fun!

  9. #35


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    Quote Originally Posted by aceside View Post
    Isn’t it cute to write the number 14 as the sum of an infinite series? Based on this process. This is fun!
    Suppose Rn = Additional rounds required after you get n winning streak(with target N winning streak, where n <= N). p= win probability, q = loss probability, r = tie probability and PUSHES DO NOT BREAK THE STREAK.
    Generally, Rn = p * Rn+1 + q * R0 + r * Rn + 1

    For coin flipping, r=0
    R3 = 0(When you have a 3 head in a row, you DON'T need to flip further, additional rounds required after you have a 3 head in a row is 0 !)
    R
    2 = p * R3 + q * R0 + 1
    R1 = p * R2 + q * R0 + 1
    R0 = p * R1 + q * R0 + 1

    You can(you are on your own) simplify above four simultaneous equations as below :-

    R0 = (p^2 + p + 1)/(p - p^2q - pq), edited

    when p =q = 0.5, R0 = 14
    Last edited by James989; 11-22-2021 at 05:53 PM.

  10. #36
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    Quote Originally Posted by James989 View Post
    You can(you are on your own) simplify above four simultaneous equations as below :-

    R0 = (pq + p + 1)/(p - p^2q - pq)
    Thank you for your patience. I just derived this myself and found a typo here. The formula should be
    R0 = (p^2 + p + 1)/(p - p^2q - pq)

  11. #37


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    Quote Originally Posted by aceside View Post
    Thank you for your patience. I just derived this myself and found a typo here. The formula should be
    R0 = (p^2 + p + 1)/(p - p^2q - pq)
    I agree, R0 = (p^2 + p + 1)/(p - p^2q - pq)


    So, coin flipping, 5 head in a row, I think(just a wild guess) R0 = (p^4 + p^3 + p^2 + p + 1)/(p - p^4q - p^3q- p^2q - pq) = 62 !
    Last edited by James989; 11-22-2021 at 06:01 PM.

  12. #38
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    Quote Originally Posted by James989 View Post
    I agree, R0 = (p^2 + p + 1)/(p - p^2q - pq)


    So, coin flipping, 5 head in a row, I think(just a wild guess) R0 = (p^4 + p^3 + p^2 + p + 1)/(p - p^4q - p^3q- p^2q - pq) = 62 !
    I have verified with both Excel and Krapstuff calculator that the formula is valid for all N.

    For N successes in a row on the p-side event, R0 = (p^(N-1) + p^(N-2)+... + p^2 + p + 1)/{1 - [p^(N-1) + p^(N-2)+...+ p^2+p+1]q}.

    To simplify, we let P=p^(N-1) + p^(N-2)+... + p^2 + p + 1=(1-p^N)/(1-p), then we have

    R0=P/(1-Pq).

    Also, let me generalize this process to the blackjack streak calculator you developed. Firstly we define: p=win probability, q=loss probability, r=tie probability, then we have the average hands required for an N success in a row on the p-side exclusive of blackjack outcome.

    R0=P/(1-r-Pq),

    where P={1-[p/(1-r)]^N}/{1-[p/(1-r)]}.

    Please verify this formula. (Also, how can we find the average hands required for an N success in a row on the p-side or r-side of blackjack outcome?)

    This result shows that there is a tiny difference between taking into account the tie probability and leaving it out! This is my proof.
    Last edited by aceside; 11-23-2021 at 02:23 PM.

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