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Thread: Game probability problem

  1. #31
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    Re: Game probability problem




    OK thanks. And is it otherwise correct?

  2. #32
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    Re: Game probability problem

    Quote Originally Posted by BGM View Post
    Let a_k be the number of attackers survived just before the k-th battle (if exist), so it is including those singleton attacker stayed in the territory. And a_1 is the initial number of the attackers. Note that the actual number of attackers entering the the k-th battle is a_k - k.

    Let d_k be the number of defenders entering the k-th battle.

    Let p(\cdot|(a, d)) be the probability mass function of the number of attackers left after a battle given that the number of attackers and defenders entering the battle are (a, d)

    Therefore the probability of winning is

    \sum_{a_2 -1 = 2}^{a_1-1} \sum_{a_3 -2= 1}^{a_2 - 2} p(a_2-1|(a_1-1, d_1))p(a_3-2|(a_2-2, d_2))

    Sorry if I count it wrong
    Please help, I need to know if my 3-battle solution is correct so I can generalize. (See post #29)

    Furthermore, could you show how to find the probability for the 2-battle scenario of there being m attackers surviving and placed on the last conquered territory at the end (in terms of the pm functions we already know how to calculate), for 1 \leq m \leq a_0-2 where a_0 is the number of attackers that originally started (maximum m is a_0-2 because the first battle involves a_0-1 attackers, the second involves maximum a_0-2 attackers and if these attackers all win there will be a_0-2 surviving attackers on the last conquered territory).
    Last edited by GameGod; 07-06-2013 at 10:54 AM.

  3. #33
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    Re: Game probability problem

    I think #29 is alright (at least the same as my idea but maybe I am wrong). You may try to verify it by simulation.

    For the latter problem you want to ask if there is exactly m attackers survive and placed on the last territory.

    Note that in the formula \sum_{a_3 - 2 = 1}^{a_2 - 2} means we consider all the scenario with the attackers from 1 to a_2 - 2 so that you are winning the last battle. (Actually a_3 - 2 is the m you considered here).

    Therefore you just need to modify it as

    \sum_{a_2 - 1 = m+1}^{a_1 - 1} p(a_2 - 1|(a_1 - 1, d_1))p(m|(a_2 - 2, d_2)

    Note that you require m + 1 survived after the second last battle.

  4. #34
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    Re: Game probability problem

    Thanks. So what are the two summation bounds for the 3-battle case?

    \sum \sum p(a_2-1|(a_1-1, d_1))\cdot p(a_3-2|(a_2-2, d_2)) \cdot p(m|(a_3-3, d_3))

    But what are the summation bounds ...

  5. #35
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    Re: Game probability problem


    Actually there is nothing special/mysterious. Just need to include all the cases, and ensure that it is possible to led to m survivors at last.

    \sum_{a_2 -1 = m+2}^{a_1-1} \sum_{a_3 -2= m+1}^{a_2 - 2} 
p(a_2-1|(a_1-1, d_1))p(a_3-2|(a_2-2, d_2))p(m|(a_3-3, d_3))

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