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Thread: Odds for best of 3 series.

  1. #1
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    Question Odds for best of 3 series.

    hi all!


    I'm playing around with odds and I would like someone to confirm if I'm correct.
    Let's take imaginary game between Team A and Team B

    They will play a video game and probabilities for winning Map are: Team A 60% Team B 40%
    I want to calculate probabilities and odds for Best of 3.

    BO3

    AA 0.60 * 0.60 = 0.36
    BB 0.40 * 0.40 = 0.16
    ABA 0.60 * 0.40 * 0.60 = 0.144
    BAB 0.40 * 0.60 * 0.40 = 0.096

    If we sum these we get 0.76 - which is less than 100 % so we can't compile odds from that. That's why I think we have to convert it to 100% sports book.

    0.36 / 0.76 = 0.474
    0.16 / 0.76 = 0.211
    0.144 / 0.76 = 0.189
    0.096 / 0.76 = 0.126

    Now when we sum these we get 1.

    When we sum all probabilities for Team A win ( AA, ABA) that gives us 0.66. Team B (BB, BAB) gives us 0.34.
    Now let's convert this to true decimal odds ( without bookie margin):

    1/0.66 = 1.51
    1/0.34 = 2.97


    If I'm correct it mans that probability of Team A to win 2-0 is 47%
    Probability of Team B winning 0-2 is 21%

    Is this approach correct? Could somebody confirm that this conversion to 100% is correct? and also make an example with BO5 ?
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  2. #2
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    Re: Odds for best of 3 series.

    I really don't understand your basic concept but I am sure this problem is much simpler than that!

    A can win by:
    AA
    ABA
    BAA
    Since the probability of A winning any one game is 0.6, the probability of "AA" (A wins 2-0) is (0.6)(0.6)= 0.36, 36%, not 47%, the probability of "ABA" is (0.6)(0.6)(0.4)= 0.144, and the probability of "BAA" is (0.4)(0.6)(0.6)= 0.144 again. The probability of any one of those is 0.36+ 0.144+ 0.144= 0.648.

    Similarly, B can win by
    BB
    BAB
    ABB
    The probability of BB (B wins 0-2) is (0.4)(0.4)= 0.16, 16%, not 21%, the probability of BAB= (0.4)(0.6)(0.4)= 0.096, and the probability of ABB is (0.6)(0.4)(0.6)= 0.096. The probability of any of those is 0.16+ 0.096+ 0.096= 0.352.

    Of course, 0.648+ 0.352= 1.000.
    Last edited by HallsofIvy; Jul 14th 2017 at 04:44 PM.
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  3. #3
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    Re: Odds for best of 3 series.

    Quote Originally Posted by epic View Post
    I'm playing around with odds and I would like someone to confirm if I'm correct.
    Let's take imaginary game between Team A and Team B
    They will play a video game and probabilities for winning Map are: Team A 60% Team B 40%
    I want to calculate probabilities and odds for Best of 3.
    Why are you making this so complicated? Here is the outcome table:
    $\begin{array}{*{20}{c}} A&A&A \\ A&A&B \\ A&B&A \\ A&B&B \\ B&A&A \\ B&A&B \\ B&B&A \\ B&B&B \end{array}$
    See that in four A wins, in four B wins.
    The probability that A wins is $(.6)^3+\dbinom{3}{2}(.6)^2(.4)=0.648$ SEE HERE.

    The probability that B wins is $(.4)^3+\dbinom{3}{2}(.4)^2(.6)=0.352$ SEE HERE.
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  4. #4
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    Re: Odds for best of 3 series.

    For best out of 5:

    A wins 3-0: $\dbinom{2}{0}(0.6)^3(0.4)^0 = 0.216$
    A wins 3-1: $\dbinom{3}{1} (0.6)^3(0.4)^1 = 0.2592$
    A wins 3-2: $\dbinom{4}{2} (0.6)^3(0.4)^2 = 0.20736$
    B wins 2-3: $\dbinom{4}{2} (0.6)^2(0.4)^3 = 0.13824$
    B wins 1-3: $\dbinom{3}{1} (0.6)^1(0.4)^3 = 0.1152$
    B wins 0-3: $\dbinom{2}{0}(0.6)^0(0.4)^3 = 0.064$

    In general, if A wins a best of $2n-1$ with a game record of n to k, then you know A won the last game (the n-th game). Prior to that game, you know A won (n-1) games and B won k games in any order. If team A wins with probability a and team B wins with probability b, the probability of the outcome that A wins n-k is:
    $\dbinom{n-1+k}{k}a^nb^k$
    Last edited by SlipEternal; Jul 15th 2017 at 07:57 AM.
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    Re: Odds for best of 3 series.

    Thank you both for replies I really appreciate - sometimes the most obvious thing can be invisible !
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    Re: Odds for best of 3 series.

    One question though!

    The reason for my confusion is the fact that you take into consideration games which wouldn't take place.
    For example AAB - in best of 3 game. AA would give team A a win 2-0 and game would be over why should we take this game into consideration?


    On the other hand example provided by SleepEternal for 3-1 in best of 5 is ignoring one of the game

    3-1:

    BAAA - 0,0864
    ABAA - 0,0864
    AABA - 0,0864
    AAAB - 0,0864

    = 0,3456


    While in his example he ignores last game AAAB (game over 3-0)

    BAAA - 0,0864
    ABAA - 0,0864
    AABA - 0,0864

    = 0,2592


    That's why it's so confusing for me. Are we taking into consideration games which wouldn't take place?
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  7. #7
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    Re: Odds for best of 3 series.

    I, in my post above, didn't, just using "P(AA)= (0.6)(0.6)= 0.36". Some other responses used both "P(AAB)= (0.6)(0.6)(0.4)= 0.144" and "P(AAA)= (0.6)(0.6)(0.6)= 0.216" so that P(AA)= P(AAB)+ P(AAA)= 0.144+ 0.216= 0.36.
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    Re: Odds for best of 3 series.

    Quote Originally Posted by SlipEternal View Post
    For best out of 5:

    A wins 3-0: $\dbinom{2}{0}(0.6)^3(0.4)^0 = 0.216$
    A wins 3-1: $\dbinom{3}{1} (0.6)^3(0.4)^1 = 0.2592$
    A wins 3-2: $\dbinom{4}{2} (0.6)^3(0.4)^2 = 0.20736$
    B wins 2-3: $\dbinom{4}{2} (0.6)^2(0.4)^3 = 0.13824$
    B wins 1-3: $\dbinom{3}{1} (0.6)^1(0.4)^3 = 0.1152$
    B wins 0-3: $\dbinom{2}{0}(0.6)^0(0.4)^3 = 0.064$

    In general, if A wins a best of $2n-1$ with a game record of n to k, then you know A won the last game (the n-th game). Prior to that game, you know A won (n-1) games and B won k games in any order. If team A wins with probability a and team B wins with probability b, the probability of the outcome that A wins n-k is:
    $\dbinom{n-1+k}{k}a^nb^k$


    I think 3-2 is causing a lot of confusion here for me, could you please check what am I missing:
    ABBAA AABBA BABAA BBAAA ABABA
    0.03456 0.03456 0.03456 0.03456 0.03456

    that's 0.1728 and it's wrong... one combination is missing and I can't find it.


    If I use your formula I get:

    (3-1+2) /2 = 4/2 = 2 so 2*0.60*0.60*0.60*0.40*0.40 = 0.06912

    Thanks in advance.
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  9. #9
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    Re: Odds for best of 3 series.

    Quote Originally Posted by epic View Post
    I think 3-2 is causing a lot of confusion here for me, could you please check what am I missing:
    ABBAA AABBA BABAA BBAAA ABABA
    0.03456 0.03456 0.03456 0.03456 0.03456

    that's 0.1728 and it's wrong... one combination is missing and I can't find it.


    If I use your formula I get:

    (3-1+2) /2 = 4/2 = 2 so 2*0.60*0.60*0.60*0.40*0.40 = 0.06912

    Thanks in advance.
    $\dbinom{n-1+k}{k} = \dfrac{(n-1+k)!}{k!(n-1)!}$

    $\dbinom{4}{2} = \dfrac{4!}{2!2!} = \dfrac{4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1} = 6$

    You can enter this in wolframalpha as: (4 choose 2)*(0.6)^3*(0.4)^2

    Anyway:

    AABBA
    ABABA
    ABBAA
    BAABA
    BABAA
    BBAAA
    Last edited by SlipEternal; Jul 16th 2017 at 08:10 AM.
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  10. #10
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    Re: Odds for best of 3 series.

    Anyone interested in this topic, would do well to read this on the problem of points.
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