Variation of Polya’s urn problem?











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I was asked this question at an interview and was stumped by it. The problem is as follows:




Alice is playing an online game. She wins the first round and loses the second. The probability she then wins subsequent rounds is proportional to the number of wins. Calculate the probability that after 100 games, the number of wins and losses are equal.




I am unsure of how to proceed. At the interview, I was originally thinking of using a simple sum of



$$sum_i alpha w_i delta_i,$$



where $alpha$ is normalisation constant, $w_i$ is the number of wins at the $i$th game and $delta_i$ is either 0 or 1. However, I doubt this is a correct approach, and $w_i$ is dependent on the $delta$’s for $j<i$.



I was discussing with a mathematician friend who said that this is a possible variation of Polya’s urn, but I can’t see the connection. Could anyone help me with this visualisation and provide hints on how to move forward? Thank you!










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  • 1




    Saying the probability in subsequent rounds is proportional to the number of wins is not sufficient. It might be fixed at $0$. I think this is usually asked with the probability in each round being the fraction of wins up to that point, so for round $3$ she has $frac 12$ chance to win and if she wins in round $3$ she has $frac 23$ chance to win round $4$.
    – Ross Millikan
    Nov 18 at 19:55










  • @RossMillikan I see – could you talk a bit more about that? I’m curious!
    – user107224
    Nov 18 at 19:57










  • This is exactly the Polya urn model. en.wikipedia.org/wiki/P%C3%B3lya_urn_model See the 2nd paragraph. Then interpret a black ball as WIN and a white ball as LOSE. $Prob(WIN) = $ exactly the fraction of black balls = fraction of previous wins. Having said this, I don't know how to solve this model but I suspect some googling would reveal quite a lot of resources.
    – antkam
    Nov 18 at 20:14















up vote
2
down vote

favorite
1












I was asked this question at an interview and was stumped by it. The problem is as follows:




Alice is playing an online game. She wins the first round and loses the second. The probability she then wins subsequent rounds is proportional to the number of wins. Calculate the probability that after 100 games, the number of wins and losses are equal.




I am unsure of how to proceed. At the interview, I was originally thinking of using a simple sum of



$$sum_i alpha w_i delta_i,$$



where $alpha$ is normalisation constant, $w_i$ is the number of wins at the $i$th game and $delta_i$ is either 0 or 1. However, I doubt this is a correct approach, and $w_i$ is dependent on the $delta$’s for $j<i$.



I was discussing with a mathematician friend who said that this is a possible variation of Polya’s urn, but I can’t see the connection. Could anyone help me with this visualisation and provide hints on how to move forward? Thank you!










share|cite|improve this question




















  • 1




    Saying the probability in subsequent rounds is proportional to the number of wins is not sufficient. It might be fixed at $0$. I think this is usually asked with the probability in each round being the fraction of wins up to that point, so for round $3$ she has $frac 12$ chance to win and if she wins in round $3$ she has $frac 23$ chance to win round $4$.
    – Ross Millikan
    Nov 18 at 19:55










  • @RossMillikan I see – could you talk a bit more about that? I’m curious!
    – user107224
    Nov 18 at 19:57










  • This is exactly the Polya urn model. en.wikipedia.org/wiki/P%C3%B3lya_urn_model See the 2nd paragraph. Then interpret a black ball as WIN and a white ball as LOSE. $Prob(WIN) = $ exactly the fraction of black balls = fraction of previous wins. Having said this, I don't know how to solve this model but I suspect some googling would reveal quite a lot of resources.
    – antkam
    Nov 18 at 20:14













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I was asked this question at an interview and was stumped by it. The problem is as follows:




Alice is playing an online game. She wins the first round and loses the second. The probability she then wins subsequent rounds is proportional to the number of wins. Calculate the probability that after 100 games, the number of wins and losses are equal.




I am unsure of how to proceed. At the interview, I was originally thinking of using a simple sum of



$$sum_i alpha w_i delta_i,$$



where $alpha$ is normalisation constant, $w_i$ is the number of wins at the $i$th game and $delta_i$ is either 0 or 1. However, I doubt this is a correct approach, and $w_i$ is dependent on the $delta$’s for $j<i$.



I was discussing with a mathematician friend who said that this is a possible variation of Polya’s urn, but I can’t see the connection. Could anyone help me with this visualisation and provide hints on how to move forward? Thank you!










share|cite|improve this question















I was asked this question at an interview and was stumped by it. The problem is as follows:




Alice is playing an online game. She wins the first round and loses the second. The probability she then wins subsequent rounds is proportional to the number of wins. Calculate the probability that after 100 games, the number of wins and losses are equal.




I am unsure of how to proceed. At the interview, I was originally thinking of using a simple sum of



$$sum_i alpha w_i delta_i,$$



where $alpha$ is normalisation constant, $w_i$ is the number of wins at the $i$th game and $delta_i$ is either 0 or 1. However, I doubt this is a correct approach, and $w_i$ is dependent on the $delta$’s for $j<i$.



I was discussing with a mathematician friend who said that this is a possible variation of Polya’s urn, but I can’t see the connection. Could anyone help me with this visualisation and provide hints on how to move forward? Thank you!







probability probability-theory statistics game-theory polya-urn-model






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edited Nov 18 at 19:34

























asked Nov 18 at 10:22









user107224

378214




378214








  • 1




    Saying the probability in subsequent rounds is proportional to the number of wins is not sufficient. It might be fixed at $0$. I think this is usually asked with the probability in each round being the fraction of wins up to that point, so for round $3$ she has $frac 12$ chance to win and if she wins in round $3$ she has $frac 23$ chance to win round $4$.
    – Ross Millikan
    Nov 18 at 19:55










  • @RossMillikan I see – could you talk a bit more about that? I’m curious!
    – user107224
    Nov 18 at 19:57










  • This is exactly the Polya urn model. en.wikipedia.org/wiki/P%C3%B3lya_urn_model See the 2nd paragraph. Then interpret a black ball as WIN and a white ball as LOSE. $Prob(WIN) = $ exactly the fraction of black balls = fraction of previous wins. Having said this, I don't know how to solve this model but I suspect some googling would reveal quite a lot of resources.
    – antkam
    Nov 18 at 20:14














  • 1




    Saying the probability in subsequent rounds is proportional to the number of wins is not sufficient. It might be fixed at $0$. I think this is usually asked with the probability in each round being the fraction of wins up to that point, so for round $3$ she has $frac 12$ chance to win and if she wins in round $3$ she has $frac 23$ chance to win round $4$.
    – Ross Millikan
    Nov 18 at 19:55










  • @RossMillikan I see – could you talk a bit more about that? I’m curious!
    – user107224
    Nov 18 at 19:57










  • This is exactly the Polya urn model. en.wikipedia.org/wiki/P%C3%B3lya_urn_model See the 2nd paragraph. Then interpret a black ball as WIN and a white ball as LOSE. $Prob(WIN) = $ exactly the fraction of black balls = fraction of previous wins. Having said this, I don't know how to solve this model but I suspect some googling would reveal quite a lot of resources.
    – antkam
    Nov 18 at 20:14








1




1




Saying the probability in subsequent rounds is proportional to the number of wins is not sufficient. It might be fixed at $0$. I think this is usually asked with the probability in each round being the fraction of wins up to that point, so for round $3$ she has $frac 12$ chance to win and if she wins in round $3$ she has $frac 23$ chance to win round $4$.
– Ross Millikan
Nov 18 at 19:55




Saying the probability in subsequent rounds is proportional to the number of wins is not sufficient. It might be fixed at $0$. I think this is usually asked with the probability in each round being the fraction of wins up to that point, so for round $3$ she has $frac 12$ chance to win and if she wins in round $3$ she has $frac 23$ chance to win round $4$.
– Ross Millikan
Nov 18 at 19:55












@RossMillikan I see – could you talk a bit more about that? I’m curious!
– user107224
Nov 18 at 19:57




@RossMillikan I see – could you talk a bit more about that? I’m curious!
– user107224
Nov 18 at 19:57












This is exactly the Polya urn model. en.wikipedia.org/wiki/P%C3%B3lya_urn_model See the 2nd paragraph. Then interpret a black ball as WIN and a white ball as LOSE. $Prob(WIN) = $ exactly the fraction of black balls = fraction of previous wins. Having said this, I don't know how to solve this model but I suspect some googling would reveal quite a lot of resources.
– antkam
Nov 18 at 20:14




This is exactly the Polya urn model. en.wikipedia.org/wiki/P%C3%B3lya_urn_model See the 2nd paragraph. Then interpret a black ball as WIN and a white ball as LOSE. $Prob(WIN) = $ exactly the fraction of black balls = fraction of previous wins. Having said this, I don't know how to solve this model but I suspect some googling would reveal quite a lot of resources.
– antkam
Nov 18 at 20:14










1 Answer
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Let $W_i$ and $L_i$ denote the number of wins and losses after the $i$-th rounds. Then, assuming that the probability of winning the $(i+1)$-th round is $W_i/(W_i+L_i)$,
begin{align}
&mathsf{P}(W_{100}=L_{100}mid W_2=L_2) \
&qquad=binom{98}{49}frac{W_2(W_2+1)cdots (W_2+48)times L_2(L_2+1)cdots (L_2+48)}{(W_2+L_2)(W_2+L_2+1)cdots(W_2+L_2+97)} \
&qquad=binom{98}{49}frac{(1cdot 2cdots 48cdot 49)^2}{2cdot 3cdots 98cdot 99}=frac{1}{99}.
end{align}






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    up vote
    4
    down vote



    accepted










    Let $W_i$ and $L_i$ denote the number of wins and losses after the $i$-th rounds. Then, assuming that the probability of winning the $(i+1)$-th round is $W_i/(W_i+L_i)$,
    begin{align}
    &mathsf{P}(W_{100}=L_{100}mid W_2=L_2) \
    &qquad=binom{98}{49}frac{W_2(W_2+1)cdots (W_2+48)times L_2(L_2+1)cdots (L_2+48)}{(W_2+L_2)(W_2+L_2+1)cdots(W_2+L_2+97)} \
    &qquad=binom{98}{49}frac{(1cdot 2cdots 48cdot 49)^2}{2cdot 3cdots 98cdot 99}=frac{1}{99}.
    end{align}






    share|cite|improve this answer



























      up vote
      4
      down vote



      accepted










      Let $W_i$ and $L_i$ denote the number of wins and losses after the $i$-th rounds. Then, assuming that the probability of winning the $(i+1)$-th round is $W_i/(W_i+L_i)$,
      begin{align}
      &mathsf{P}(W_{100}=L_{100}mid W_2=L_2) \
      &qquad=binom{98}{49}frac{W_2(W_2+1)cdots (W_2+48)times L_2(L_2+1)cdots (L_2+48)}{(W_2+L_2)(W_2+L_2+1)cdots(W_2+L_2+97)} \
      &qquad=binom{98}{49}frac{(1cdot 2cdots 48cdot 49)^2}{2cdot 3cdots 98cdot 99}=frac{1}{99}.
      end{align}






      share|cite|improve this answer

























        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        Let $W_i$ and $L_i$ denote the number of wins and losses after the $i$-th rounds. Then, assuming that the probability of winning the $(i+1)$-th round is $W_i/(W_i+L_i)$,
        begin{align}
        &mathsf{P}(W_{100}=L_{100}mid W_2=L_2) \
        &qquad=binom{98}{49}frac{W_2(W_2+1)cdots (W_2+48)times L_2(L_2+1)cdots (L_2+48)}{(W_2+L_2)(W_2+L_2+1)cdots(W_2+L_2+97)} \
        &qquad=binom{98}{49}frac{(1cdot 2cdots 48cdot 49)^2}{2cdot 3cdots 98cdot 99}=frac{1}{99}.
        end{align}






        share|cite|improve this answer














        Let $W_i$ and $L_i$ denote the number of wins and losses after the $i$-th rounds. Then, assuming that the probability of winning the $(i+1)$-th round is $W_i/(W_i+L_i)$,
        begin{align}
        &mathsf{P}(W_{100}=L_{100}mid W_2=L_2) \
        &qquad=binom{98}{49}frac{W_2(W_2+1)cdots (W_2+48)times L_2(L_2+1)cdots (L_2+48)}{(W_2+L_2)(W_2+L_2+1)cdots(W_2+L_2+97)} \
        &qquad=binom{98}{49}frac{(1cdot 2cdots 48cdot 49)^2}{2cdot 3cdots 98cdot 99}=frac{1}{99}.
        end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 18 at 22:59

























        answered Nov 18 at 20:17









        d.k.o.

        8,189527




        8,189527






























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