Mixing
Gambler’s Ruin in finite time - solve by Markov chains?
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Consider the Gambler's Ruin Problem: an infinite Markov process of steps with binary outcomes of plus or minus, each of equal probability. At each step of the process there may be a play of the game.
At each play of the game starting with step 1, the gambler's initial fortune, B, increases by one unit with probability 1/2 or decreases by one unit with probability 1/2. The game ends at step N or sooner (after fewer plays) if the gambler's profit first reaches a Target of +A or a Ruin (total loss) of -B units anytime during the process (which continues beyond N steps).
There are three profit outcomes after step N:
-B, +A, or Neither.
Ruin is defined by profit reaching -B for the first time during N plays without first reaching +A.
Success is defined by profit reaching +A for the first time during N plays without first reaching -B.
Problem: What is the probability of a. Ruin or b. Success or c. Neither, in terms of A, B, N?
I cannot solve this except by path-counting. Can a.,b.,c. be solved in a closed form by Markov chains?
This problem is also called a random walk of a particle in finite time with two absorbing barriers.
probability
asked Jan 2 at 16:24
artbenisartbenis
112
$endgroup$
add a comment |
$begingroup$
Consider the Gambler's Ruin Problem: an infinite Markov process of steps with binary outcomes of plus or minus, each of equal probability. At each step of the process there may be a play of the game.
At each play of the game starting with step 1, the gambler's initial fortune, B, increases by one unit with probability 1/2 or decreases by one unit with probability 1/2. The game ends at step N or sooner (after fewer plays) if the gambler's profit first reaches a Target of +A or a Ruin (total loss) of -B units anytime during the process (which continues beyond N steps).
There are three profit outcomes after step N:
-B, +A, or Neither.
Ruin is defined by profit reaching -B for the first time during N plays without first reaching +A.
Success is defined by profit reaching +A for the first time during N plays without first reaching -B.
Problem: What is the probability of a. Ruin or b. Success or c. Neither, in terms of A, B, N?
I cannot solve this except by path-counting. Can a.,b.,c. be solved in a closed form by Markov chains?
This problem is also called a random walk of a particle in finite time with two absorbing barriers.
probability
asked Jan 2 at 16:24
artbenisartbenis
112
$endgroup$
$begingroup$
He will be ruined
$endgroup$
– AkatsukiMaliki
Jan 2 at 17:10
$begingroup$
@AkatsukiMaliki. Not so. For example with A = B= N=2, P of Neither = 1/2.
$endgroup$
– artbenis
Jan 2 at 18:04
add a comment |
$begingroup$
Consider the Gambler's Ruin Problem: an infinite Markov process of steps with binary outcomes of plus or minus, each of equal probability. At each step of the process there may be a play of the game.
At each play of the game starting with step 1, the gambler's initial fortune, B, increases by one unit with probability 1/2 or decreases by one unit with probability 1/2. The game ends at step N or sooner (after fewer plays) if the gambler's profit first reaches a Target of +A or a Ruin (total loss) of -B units anytime during the process (which continues beyond N steps).
There are three profit outcomes after step N:
-B, +A, or Neither.
Ruin is defined by profit reaching -B for the first time during N plays without first reaching +A.
Success is defined by profit reaching +A for the first time during N plays without first reaching -B.
Problem: What is the probability of a. Ruin or b. Success or c. Neither, in terms of A, B, N?
I cannot solve this except by path-counting. Can a.,b.,c. be solved in a closed form by Markov chains?
This problem is also called a random walk of a particle in finite time with two absorbing barriers.
probability
asked Jan 2 at 16:24
artbenisartbenis
112
$endgroup$
Consider the Gambler's Ruin Problem: an infinite Markov process of steps with binary outcomes of plus or minus, each of equal probability. At each step of the process there may be a play of the game.
At each play of the game starting with step 1, the gambler's initial fortune, B, increases by one unit with probability 1/2 or decreases by one unit with probability 1/2. The game ends at step N or sooner (after fewer plays) if the gambler's profit first reaches a Target of +A or a Ruin (total loss) of -B units anytime during the process (which continues beyond N steps).
There are three profit outcomes after step N:
-B, +A, or Neither.
Ruin is defined by profit reaching -B for the first time during N plays without first reaching +A.
Success is defined by profit reaching +A for the first time during N plays without first reaching -B.
Problem: What is the probability of a. Ruin or b. Success or c. Neither, in terms of A, B, N?
I cannot solve this except by path-counting. Can a.,b.,c. be solved in a closed form by Markov chains?
This problem is also called a random walk of a particle in finite time with two absorbing barriers.
probability
probability
asked Jan 2 at 16:24
artbenisartbenis
112
asked Jan 2 at 16:24
artbenisartbenis
112
edited Jan 2 at 18:16
artbenis
asked Jan 2 at 16:24
artbenisartbenis
112
asked Jan 2 at 16:24
artbenisartbenis
112
asked Jan 2 at 16:24
artbenisartbenis
112
112
$begingroup$
He will be ruined
$endgroup$
– AkatsukiMaliki
Jan 2 at 17:10
$begingroup$
@AkatsukiMaliki. Not so. For example with A = B= N=2, P of Neither = 1/2.
$endgroup$
– artbenis
Jan 2 at 18:04
add a comment |
$begingroup$
He will be ruined
$endgroup$
– AkatsukiMaliki
Jan 2 at 17:10
$begingroup$
@AkatsukiMaliki. Not so. For example with A = B= N=2, P of Neither = 1/2.
$endgroup$
– artbenis
Jan 2 at 18:04
$begingroup$
He will be ruined
$endgroup$
– AkatsukiMaliki
Jan 2 at 17:10
$begingroup$
He will be ruined
$endgroup$
– AkatsukiMaliki
Jan 2 at 17:10
$begingroup$
@AkatsukiMaliki. Not so. For example with A = B= N=2, P of Neither = 1/2.
$endgroup$
– artbenis
Jan 2 at 18:04
$begingroup$
@AkatsukiMaliki. Not so. For example with A = B= N=2, P of Neither = 1/2.
$endgroup$
– artbenis
Jan 2 at 18:04
add a comment |
1 Answer
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Would leave as comment, but too long:
Since each path corresponds to a 0-1 bit string on Length $N$, you must implicitly compute the sum of binomial coefficients to determine your final probabilities. I think this is what you are already doing (my interpretation of path counting?). Unfortunately, summing binomial coefficients is infeasible for large values of $N$, so this is the best you are going to be able to do. For large $N$, Determining exact values for the sum of binomial coefficients is not possible (It would make my research a lot easier if it was :)), we only know of good approximations (again, this is assuming N is very large, otherwise you can just sum up the binomial coefficients you're interested in via some math program). Depending on the relationship between $N$ $B$ and $A$, you may be able to find asymptotically tight bounds for the desired probabilities. I would recommend checking out Stirling's approximation, Entropy bounds for binomial coefficients, and related literature. This may be a useful resource if you are interested in the "large N" regime.
EDIT: As a side note, this does not become feasible even when the probability of win/loss is not 1/2. In this regime, you can still get useful bounds using KL-Divergence based estimates of binomial coefficients. This generalizes the entropy bounds mentioned earlier (See this math stack post if you are interested).
answered Jan 2 at 18:17
mm8511mm8511
54328
$endgroup$
$begingroup$
Thanks for your reply. A quick point on your reference to very large N values: then there are well-known closed form solutions to a., b., and c. they are a. B/( A+ B) , b. A/(A + B) , c. zero.
$endgroup$
– artbenis
Jan 3 at 19:13
$begingroup$
There are also approximate solutions for them e.g. for c. (A+B - 3)/N.
$endgroup$
– artbenis
Jan 3 at 19:21
add a comment |
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$begingroup$
Would leave as comment, but too long:
Since each path corresponds to a 0-1 bit string on Length $N$, you must implicitly compute the sum of binomial coefficients to determine your final probabilities. I think this is what you are already doing (my interpretation of path counting?). Unfortunately, summing binomial coefficients is infeasible for large values of $N$, so this is the best you are going to be able to do. For large $N$, Determining exact values for the sum of binomial coefficients is not possible (It would make my research a lot easier if it was :)), we only know of good approximations (again, this is assuming N is very large, otherwise you can just sum up the binomial coefficients you're interested in via some math program). Depending on the relationship between $N$ $B$ and $A$, you may be able to find asymptotically tight bounds for the desired probabilities. I would recommend checking out Stirling's approximation, Entropy bounds for binomial coefficients, and related literature. This may be a useful resource if you are interested in the "large N" regime.
EDIT: As a side note, this does not become feasible even when the probability of win/loss is not 1/2. In this regime, you can still get useful bounds using KL-Divergence based estimates of binomial coefficients. This generalizes the entropy bounds mentioned earlier (See this math stack post if you are interested).
answered Jan 2 at 18:17
mm8511mm8511
54328
$endgroup$
$begingroup$
Thanks for your reply. A quick point on your reference to very large N values: then there are well-known closed form solutions to a., b., and c. they are a. B/( A+ B) , b. A/(A + B) , c. zero.
$endgroup$
– artbenis
Jan 3 at 19:13
$begingroup$
There are also approximate solutions for them e.g. for c. (A+B - 3)/N.
$endgroup$
– artbenis
Jan 3 at 19:21
add a comment |
$begingroup$
Would leave as comment, but too long:
Since each path corresponds to a 0-1 bit string on Length $N$, you must implicitly compute the sum of binomial coefficients to determine your final probabilities. I think this is what you are already doing (my interpretation of path counting?). Unfortunately, summing binomial coefficients is infeasible for large values of $N$, so this is the best you are going to be able to do. For large $N$, Determining exact values for the sum of binomial coefficients is not possible (It would make my research a lot easier if it was :)), we only know of good approximations (again, this is assuming N is very large, otherwise you can just sum up the binomial coefficients you're interested in via some math program). Depending on the relationship between $N$ $B$ and $A$, you may be able to find asymptotically tight bounds for the desired probabilities. I would recommend checking out Stirling's approximation, Entropy bounds for binomial coefficients, and related literature. This may be a useful resource if you are interested in the "large N" regime.
EDIT: As a side note, this does not become feasible even when the probability of win/loss is not 1/2. In this regime, you can still get useful bounds using KL-Divergence based estimates of binomial coefficients. This generalizes the entropy bounds mentioned earlier (See this math stack post if you are interested).
answered Jan 2 at 18:17
mm8511mm8511
54328
$endgroup$
$begingroup$
Thanks for your reply. A quick point on your reference to very large N values: then there are well-known closed form solutions to a., b., and c. they are a. B/( A+ B) , b. A/(A + B) , c. zero.
$endgroup$
– artbenis
Jan 3 at 19:13
$begingroup$
There are also approximate solutions for them e.g. for c. (A+B - 3)/N.
$endgroup$
– artbenis
Jan 3 at 19:21
add a comment |
$begingroup$
Would leave as comment, but too long:
Since each path corresponds to a 0-1 bit string on Length $N$, you must implicitly compute the sum of binomial coefficients to determine your final probabilities. I think this is what you are already doing (my interpretation of path counting?). Unfortunately, summing binomial coefficients is infeasible for large values of $N$, so this is the best you are going to be able to do. For large $N$, Determining exact values for the sum of binomial coefficients is not possible (It would make my research a lot easier if it was :)), we only know of good approximations (again, this is assuming N is very large, otherwise you can just sum up the binomial coefficients you're interested in via some math program). Depending on the relationship between $N$ $B$ and $A$, you may be able to find asymptotically tight bounds for the desired probabilities. I would recommend checking out Stirling's approximation, Entropy bounds for binomial coefficients, and related literature. This may be a useful resource if you are interested in the "large N" regime.
EDIT: As a side note, this does not become feasible even when the probability of win/loss is not 1/2. In this regime, you can still get useful bounds using KL-Divergence based estimates of binomial coefficients. This generalizes the entropy bounds mentioned earlier (See this math stack post if you are interested).
answered Jan 2 at 18:17
mm8511mm8511
54328
$endgroup$
Would leave as comment, but too long:
Since each path corresponds to a 0-1 bit string on Length $N$, you must implicitly compute the sum of binomial coefficients to determine your final probabilities. I think this is what you are already doing (my interpretation of path counting?). Unfortunately, summing binomial coefficients is infeasible for large values of $N$, so this is the best you are going to be able to do. For large $N$, Determining exact values for the sum of binomial coefficients is not possible (It would make my research a lot easier if it was :)), we only know of good approximations (again, this is assuming N is very large, otherwise you can just sum up the binomial coefficients you're interested in via some math program). Depending on the relationship between $N$ $B$ and $A$, you may be able to find asymptotically tight bounds for the desired probabilities. I would recommend checking out Stirling's approximation, Entropy bounds for binomial coefficients, and related literature. This may be a useful resource if you are interested in the "large N" regime.
EDIT: As a side note, this does not become feasible even when the probability of win/loss is not 1/2. In this regime, you can still get useful bounds using KL-Divergence based estimates of binomial coefficients. This generalizes the entropy bounds mentioned earlier (See this math stack post if you are interested).
answered Jan 2 at 18:17
mm8511mm8511
54328
edited Jan 2 at 18:22
answered Jan 2 at 18:17
mm8511mm8511
54328
answered Jan 2 at 18:17
mm8511mm8511
54328
answered Jan 2 at 18:17
mm8511mm8511
54328
54328
$begingroup$
Thanks for your reply. A quick point on your reference to very large N values: then there are well-known closed form solutions to a., b., and c. they are a. B/( A+ B) , b. A/(A + B) , c. zero.
$endgroup$
– artbenis
Jan 3 at 19:13
$begingroup$
There are also approximate solutions for them e.g. for c. (A+B - 3)/N.
$endgroup$
– artbenis
Jan 3 at 19:21
add a comment |
$begingroup$
Thanks for your reply. A quick point on your reference to very large N values: then there are well-known closed form solutions to a., b., and c. they are a. B/( A+ B) , b. A/(A + B) , c. zero.
$endgroup$
– artbenis
Jan 3 at 19:13
$begingroup$
There are also approximate solutions for them e.g. for c. (A+B - 3)/N.
$endgroup$
– artbenis
Jan 3 at 19:21
$begingroup$
Thanks for your reply. A quick point on your reference to very large N values: then there are well-known closed form solutions to a., b., and c. they are a. B/( A+ B) , b. A/(A + B) , c. zero.
$endgroup$
– artbenis
Jan 3 at 19:13
$begingroup$
Thanks for your reply. A quick point on your reference to very large N values: then there are well-known closed form solutions to a., b., and c. they are a. B/( A+ B) , b. A/(A + B) , c. zero.
$endgroup$
– artbenis
Jan 3 at 19:13
$begingroup$
There are also approximate solutions for them e.g. for c. (A+B - 3)/N.
$endgroup$
– artbenis
Jan 3 at 19:21
$begingroup$
There are also approximate solutions for them e.g. for c. (A+B - 3)/N.
$endgroup$
– artbenis
Jan 3 at 19:21
add a comment |
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$begingroup$
He will be ruined
$endgroup$
– AkatsukiMaliki
Jan 2 at 17:10
$begingroup$
@AkatsukiMaliki. Not so. For example with A = B= N=2, P of Neither = 1/2.
$endgroup$
– artbenis
Jan 2 at 18:04