Fair gambler's ruin problem intuitionProbability 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

ssTTsSTtRrriinInnnnNNNIiinngg

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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













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












    $begingroup$


    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

      366




      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



          $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



          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



          $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



          based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



          $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



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



          $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



          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, eqrefeq4 becomes



          $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



          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 eqrefeq2 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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              2 Answers
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              2 Answers
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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



              $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



              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



              $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



              based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



              $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



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



              $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



              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, eqrefeq4 becomes



              $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



              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 eqrefeq2 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



                $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



                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



                $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



                based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



                $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



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



                $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



                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, eqrefeq4 becomes



                $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



                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 eqrefeq2 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



                  $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



                  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



                  $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



                  based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



                  $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



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



                  $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



                  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, eqrefeq4 becomes



                  $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



                  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 eqrefeq2 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



                  $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



                  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



                  $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



                  based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



                  $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



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



                  $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



                  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, eqrefeq4 becomes



                  $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



                  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 eqrefeq2 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 44 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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