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Fair gambler's ruin problem intuition


Probability of Gambler's Ruin with Unequal Gain/LossAdaptive gambler's ruin problemGambler's Ruin with no set target for winGambler's ruin problem - unsure about the number of roundsEffect of Gambler's Ruin Bet Size on DurationGambler's ruin: verifying Markov propertyComparison of duration of two gambler's ruin gamesGambler's Ruin - Probability of Losing in t StepsGambler's Ruin: Win 2 dollars, Lose 1 dollarGambler's ruin Markov chain













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$begingroup$


In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_{k-1} = p_{k-1} - p_{k-2} = ... = p_1 - p_0 = p_1$.



Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



Is there an intuitive reason why this is the case?










share|cite









$endgroup$

















    1












    $begingroup$


    In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



    In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_{k-1} = p_{k-1} - p_{k-2} = ... = p_1 - p_0 = p_1$.



    Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



    Is there an intuitive reason why this is the case?










    share|cite









    $endgroup$















      1












      1








      1





      $begingroup$


      In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



      In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_{k-1} = p_{k-1} - p_{k-2} = ... = p_1 - p_0 = p_1$.



      Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



      Is there an intuitive reason why this is the case?










      share|cite









      $endgroup$




      In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



      In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_{k-1} = p_{k-1} - p_{k-2} = ... = p_1 - p_0 = p_1$.



      Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



      Is there an intuitive reason why this is the case?







      probability






      share|cite













      share|cite











      share|cite




      share|cite










      asked 2 hours ago









      platypus17platypus17

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      366






















          2 Answers
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          active

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          2












          $begingroup$

          Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



          $$q_{k-1} - q_k = q_{k-2} - q_{k - 1} = ldots = q_1 - q_2 = q_0 - q_1 tag{1}label{eq1}$$



          Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



          Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



          $$p_i = frac{1}{2}p_{i - 1} + frac{1}{2}p_{i + 1} tag{2}label{eq2}$$



          based on the probabilities of either winning or losing the first time. Summing eqref{eq2} for $i$ from $1$ to $k - 1$ gives



          $$sum_{i=1}^{k-1} p_i = frac{1}{2}sum_{i=1}^{k-1} p_{i - 1} + frac{1}{2}sum_{i=1}^{k-1} p_{i + 1} tag{3}label{eq3}$$



          Having the summations only include the common terms on both sides gives



          $$p_1 + sum_{i=2}^{k - 2} p_i + p_{k-1} = frac{1}{2}p_0 + frac{1}{2}p_1 + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}p_{k-1} + frac{1}{2}p_k tag{4}label{eq4}$$



          Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_{k-1}$ term on the left to the RHS, eqref{eq4} becomes



          $$frac{1}{2}p_1 - frac{1}{2}p_0 = frac{1}{2}p_k - frac{1}{2}p_{k-1} tag{5}label{eq5}$$



          Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqref{eq2} to get that $p_{i+1} - p_{i} = p_{i} - p_{i-1}$, like John Doe's answer states.






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_{k-1}+p_{k+1})$$ Rearranging this gives $$2p_k=p_{k-1}+p_{k+1}\p_k-p_{k-1}=p_{k+1}-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






            share|cite|improve this answer











            $endgroup$














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              $begingroup$

              Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



              $$q_{k-1} - q_k = q_{k-2} - q_{k - 1} = ldots = q_1 - q_2 = q_0 - q_1 tag{1}label{eq1}$$



              Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



              Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



              $$p_i = frac{1}{2}p_{i - 1} + frac{1}{2}p_{i + 1} tag{2}label{eq2}$$



              based on the probabilities of either winning or losing the first time. Summing eqref{eq2} for $i$ from $1$ to $k - 1$ gives



              $$sum_{i=1}^{k-1} p_i = frac{1}{2}sum_{i=1}^{k-1} p_{i - 1} + frac{1}{2}sum_{i=1}^{k-1} p_{i + 1} tag{3}label{eq3}$$



              Having the summations only include the common terms on both sides gives



              $$p_1 + sum_{i=2}^{k - 2} p_i + p_{k-1} = frac{1}{2}p_0 + frac{1}{2}p_1 + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}p_{k-1} + frac{1}{2}p_k tag{4}label{eq4}$$



              Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_{k-1}$ term on the left to the RHS, eqref{eq4} becomes



              $$frac{1}{2}p_1 - frac{1}{2}p_0 = frac{1}{2}p_k - frac{1}{2}p_{k-1} tag{5}label{eq5}$$



              Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqref{eq2} to get that $p_{i+1} - p_{i} = p_{i} - p_{i-1}$, like John Doe's answer states.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



                $$q_{k-1} - q_k = q_{k-2} - q_{k - 1} = ldots = q_1 - q_2 = q_0 - q_1 tag{1}label{eq1}$$



                Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



                Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



                $$p_i = frac{1}{2}p_{i - 1} + frac{1}{2}p_{i + 1} tag{2}label{eq2}$$



                based on the probabilities of either winning or losing the first time. Summing eqref{eq2} for $i$ from $1$ to $k - 1$ gives



                $$sum_{i=1}^{k-1} p_i = frac{1}{2}sum_{i=1}^{k-1} p_{i - 1} + frac{1}{2}sum_{i=1}^{k-1} p_{i + 1} tag{3}label{eq3}$$



                Having the summations only include the common terms on both sides gives



                $$p_1 + sum_{i=2}^{k - 2} p_i + p_{k-1} = frac{1}{2}p_0 + frac{1}{2}p_1 + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}p_{k-1} + frac{1}{2}p_k tag{4}label{eq4}$$



                Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_{k-1}$ term on the left to the RHS, eqref{eq4} becomes



                $$frac{1}{2}p_1 - frac{1}{2}p_0 = frac{1}{2}p_k - frac{1}{2}p_{k-1} tag{5}label{eq5}$$



                Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqref{eq2} to get that $p_{i+1} - p_{i} = p_{i} - p_{i-1}$, like John Doe's answer states.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



                  $$q_{k-1} - q_k = q_{k-2} - q_{k - 1} = ldots = q_1 - q_2 = q_0 - q_1 tag{1}label{eq1}$$



                  Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



                  Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



                  $$p_i = frac{1}{2}p_{i - 1} + frac{1}{2}p_{i + 1} tag{2}label{eq2}$$



                  based on the probabilities of either winning or losing the first time. Summing eqref{eq2} for $i$ from $1$ to $k - 1$ gives



                  $$sum_{i=1}^{k-1} p_i = frac{1}{2}sum_{i=1}^{k-1} p_{i - 1} + frac{1}{2}sum_{i=1}^{k-1} p_{i + 1} tag{3}label{eq3}$$



                  Having the summations only include the common terms on both sides gives



                  $$p_1 + sum_{i=2}^{k - 2} p_i + p_{k-1} = frac{1}{2}p_0 + frac{1}{2}p_1 + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}p_{k-1} + frac{1}{2}p_k tag{4}label{eq4}$$



                  Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_{k-1}$ term on the left to the RHS, eqref{eq4} becomes



                  $$frac{1}{2}p_1 - frac{1}{2}p_0 = frac{1}{2}p_k - frac{1}{2}p_{k-1} tag{5}label{eq5}$$



                  Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqref{eq2} to get that $p_{i+1} - p_{i} = p_{i} - p_{i-1}$, like John Doe's answer states.






                  share|cite|improve this answer











                  $endgroup$



                  Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



                  $$q_{k-1} - q_k = q_{k-2} - q_{k - 1} = ldots = q_1 - q_2 = q_0 - q_1 tag{1}label{eq1}$$



                  Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



                  Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



                  $$p_i = frac{1}{2}p_{i - 1} + frac{1}{2}p_{i + 1} tag{2}label{eq2}$$



                  based on the probabilities of either winning or losing the first time. Summing eqref{eq2} for $i$ from $1$ to $k - 1$ gives



                  $$sum_{i=1}^{k-1} p_i = frac{1}{2}sum_{i=1}^{k-1} p_{i - 1} + frac{1}{2}sum_{i=1}^{k-1} p_{i + 1} tag{3}label{eq3}$$



                  Having the summations only include the common terms on both sides gives



                  $$p_1 + sum_{i=2}^{k - 2} p_i + p_{k-1} = frac{1}{2}p_0 + frac{1}{2}p_1 + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}sum_{i=2}^{k - 2} p_i + frac{1}{2}p_{k-1} + frac{1}{2}p_k tag{4}label{eq4}$$



                  Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_{k-1}$ term on the left to the RHS, eqref{eq4} becomes



                  $$frac{1}{2}p_1 - frac{1}{2}p_0 = frac{1}{2}p_k - frac{1}{2}p_{k-1} tag{5}label{eq5}$$



                  Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqref{eq2} to get that $p_{i+1} - p_{i} = p_{i} - p_{i-1}$, like John Doe's answer states.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 50 mins ago

























                  answered 2 hours ago









                  John OmielanJohn Omielan

                  4,5362215




                  4,5362215























                      3












                      $begingroup$

                      The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_{k-1}+p_{k+1})$$ Rearranging this gives $$2p_k=p_{k-1}+p_{k+1}\p_k-p_{k-1}=p_{k+1}-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






                      share|cite|improve this answer











                      $endgroup$


















                        3












                        $begingroup$

                        The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_{k-1}+p_{k+1})$$ Rearranging this gives $$2p_k=p_{k-1}+p_{k+1}\p_k-p_{k-1}=p_{k+1}-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






                        share|cite|improve this answer











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_{k-1}+p_{k+1})$$ Rearranging this gives $$2p_k=p_{k-1}+p_{k+1}\p_k-p_{k-1}=p_{k+1}-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






                          share|cite|improve this answer











                          $endgroup$



                          The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_{k-1}+p_{k+1})$$ Rearranging this gives $$2p_k=p_{k-1}+p_{k+1}\p_k-p_{k-1}=p_{k+1}-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 1 hour ago

























                          answered 2 hours ago









                          John DoeJohn Doe

                          11.5k11239




                          11.5k11239






























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