Fair gambler's ruin problem intuitionProbability of Gambler's Ruin with Unequal Gain/LossAdaptive gambler's...
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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
$begingroup$
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?
probability
asked 3 hours ago
platypus17platypus17
416
$endgroup$
add a comment |
$begingroup$
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?
probability
asked 3 hours ago
platypus17platypus17
416
$endgroup$
add a comment |
$begingroup$
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?
probability
asked 3 hours ago
platypus17platypus17
416
$endgroup$
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
probability
asked 3 hours ago
platypus17platypus17
416
asked 3 hours ago
platypus17platypus17
416
asked 3 hours ago
platypus17platypus17
416
asked 3 hours ago
platypus17platypus17
416
asked 3 hours ago
platypus17platypus17
416
416
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
$endgroup$
add a comment |
$begingroup$
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, $p_0=0$.
answered 3 hours ago
John DoeJohn Doe
11.5k11239
$endgroup$
add a comment |
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2 Answers
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2 Answers
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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.
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
$endgroup$
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.
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
edited 1 hour ago
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
answered 2 hours ago
John OmielanJohn Omielan
4,5362215
4,5362215
add a comment |
add a comment |
$begingroup$
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, $p_0=0$.
answered 3 hours ago
John DoeJohn Doe
11.5k11239
$endgroup$
add a comment |
$begingroup$
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, $p_0=0$.
answered 3 hours ago
John DoeJohn Doe
11.5k11239
$endgroup$
add a comment |
$begingroup$
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, $p_0=0$.
answered 3 hours ago
John DoeJohn Doe
11.5k11239
$endgroup$
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, $p_0=0$.
answered 3 hours ago
John DoeJohn Doe
11.5k11239
edited 2 hours ago
answered 3 hours ago
John DoeJohn Doe
11.5k11239
answered 3 hours ago
John DoeJohn Doe
11.5k11239
answered 3 hours ago
John DoeJohn Doe
11.5k11239
11.5k11239
add a comment |
add a comment |
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