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Thread: Streaks & Runs

  1. #1
    Mar 2006

    Streaks & Runs

    Say you have the following distribution of streaks of wins & losses of length N. (A streak is defined by exactly N wins or losses in a row, followed by a new streak of the opposite kind.)

    (Sorry this table did not display properly; replace the spaces w/ tabs & it should be OK.)

    1 206 206 1 179 179
    2 89 178 2 81 162
    3 49 147 3 52 156
    4 8 32 4 19 76
    5 4 20 5 18 90
    6 3 18 6 7 42
    7 1 7 7 4 28
    360 608 360 733

    So 360 winning streaks and 360 losing streaks, with the overall probability of a win of 608 / ( 608 + 733 ) = 45%.

    I would like to create a table that provides the probabilities of a streak continuing/ not continuing once it has reached length N. For example, what is the probability that a single win will be followed by another win? that two consecutive losses will be followed by exactly one loss, or more than one loss?

    Here's my thinking so far:

    Assume you have 1 win: You could go on to have another win, thus starting a longer winning streak, or you could have a loss. So probability of a loss following the win = 360 / ( 360 + (360 - 206) ) = 70%, i.e., the number of losing streaks divided by the total number of streaks, excluding the streak of 1 win.

    Assume you have 6 wins in a row: Probability of another win = 1 / (1 + 360) = 0.28%

    Am I on the right track here?

    Thanks for any help!
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  2. #2
    Mar 2006
    I do understand that you may be disappointed by this solution of the problem, but Iím afraid you will be unable to solve this problem without a deep knowledge of the theory of probabilities.
    Let's try to solve this problem in the way it must be solve using theory of probabilities. First, we must set a model. Our stochastic experiment is a series of games (one experiment stops when a series of wins ends with loss or vice versa). Therefore, the elementary events are the streaks of wings or losses of a certain length. Now we are ready to define the set $\displaystyle \Omega$ a model of the set of elementary events. The elements of $\displaystyle \Omega $ will be streaks of -1 or +1 where for example (11111) denotes a streak of 5 wins and (-1-1-1) Ė a streak of 3 losses.
    Now letís define a probability. As $\displaystyle \Omega $ is discrete, it is enough to define probability only for the elements of $\displaystyle \Omega $. The probability of an elementary event is approximately equal to the number of experiments in which this event happened divided by the number of all experiments. The number of experiments is equal to the total number of streaks that is 720. So we get the following probabilities:
    event probability
    1 0.286111
    11 0.123611
    111 0.068056
    1111 0.011111
    11111 0.005556
    111111 0.004167
    1111111 0.001389
    -1 0.248611
    -1-1 0.1125
    -1-1-1 0.072222
    -1-1-1-1 0.026389
    -1-1-1-1-1 0.025
    -1-1-1-1-1-1 0.009722
    -1-1-1-1-1-1-1 0.005556
    Now, what do you want to find? You want to find the probability of a win if it is known that there has already been a streak of one win (we take length of one for example). This is a conditional probability of the event A that the streak is longer then one under the condition B that it is not less then one. The formula to calculate the conditional probability is
    It is evident that A is a subset of B. So P(A∩B)= P(A)= 0.123611+0.123611+0.068056+0.011111+0.005556+0.004 167+0.001389=0.213889,
    P(B)= 0.286111+0.123611+0.123611+0.068056+0.011111+0.005 556+0.004167+0.001389=0,5
    P(A|B)= 0.213889/0,5=0,427778.
    The probabilities for other length are calculated in the same way.
    Last edited by yourmathhelp; Mar 30th 2006 at 05:16 AM.
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  3. #3
    Mar 2006
    Thank you for the response.

    My approach was definitely not headed down the right track.

    I'll have to study your answer carefully & get back to you if I have any questions.

    Many thanks again.
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