Mixing
Gambler's ruin problem variant with unfair coin and different starting funds?
$begingroup$
Someone wants to play the Gambler’s Ruin game with you, where you pay
them $1$ dollar if the coin comes up heads and they pay you $1$ if it comes up
tails. You discover that a biased coin will be used
that has probability 2/3 of coming up heads. Even though the other player seems to have the advantage, you get to start with 2 dollars and they will start with 1. Whoever gets all $3$ dollars first will win the game.
Do you still have an advantage, even though you use a biased coin?
Find out by computing $p_1$, where $p_i$ is the probability that the other player wins the game starting with i dollars, for 0 ≤ i ≤ 3. Give an exact value for $p_1$ as a simplified fraction.
How do I go about solving this problem?
I understand the general formula for conditional probability, and in this case I would start with P($E_i$ | H)P(H) + P($E_i$ | T)P(T)
But since "i" doesn't seem to have an upper bound, I don't really know how to compute $p_i$, let alone $p_1$.
Any help is appreciated. Thanks!
probability combinatorics discrete-mathematics conditional-probability
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
$endgroup$
add a comment |
$begingroup$
Someone wants to play the Gambler’s Ruin game with you, where you pay
them $1$ dollar if the coin comes up heads and they pay you $1$ if it comes up
tails. You discover that a biased coin will be used
that has probability 2/3 of coming up heads. Even though the other player seems to have the advantage, you get to start with 2 dollars and they will start with 1. Whoever gets all $3$ dollars first will win the game.
Do you still have an advantage, even though you use a biased coin?
Find out by computing $p_1$, where $p_i$ is the probability that the other player wins the game starting with i dollars, for 0 ≤ i ≤ 3. Give an exact value for $p_1$ as a simplified fraction.
How do I go about solving this problem?
I understand the general formula for conditional probability, and in this case I would start with P($E_i$ | H)P(H) + P($E_i$ | T)P(T)
But since "i" doesn't seem to have an upper bound, I don't really know how to compute $p_i$, let alone $p_1$.
Any help is appreciated. Thanks!
probability combinatorics discrete-mathematics conditional-probability
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
$endgroup$
add a comment |
$begingroup$
Someone wants to play the Gambler’s Ruin game with you, where you pay
them $1$ dollar if the coin comes up heads and they pay you $1$ if it comes up
tails. You discover that a biased coin will be used
that has probability 2/3 of coming up heads. Even though the other player seems to have the advantage, you get to start with 2 dollars and they will start with 1. Whoever gets all $3$ dollars first will win the game.
Do you still have an advantage, even though you use a biased coin?
Find out by computing $p_1$, where $p_i$ is the probability that the other player wins the game starting with i dollars, for 0 ≤ i ≤ 3. Give an exact value for $p_1$ as a simplified fraction.
How do I go about solving this problem?
I understand the general formula for conditional probability, and in this case I would start with P($E_i$ | H)P(H) + P($E_i$ | T)P(T)
But since "i" doesn't seem to have an upper bound, I don't really know how to compute $p_i$, let alone $p_1$.
Any help is appreciated. Thanks!
probability combinatorics discrete-mathematics conditional-probability
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
$endgroup$
Someone wants to play the Gambler’s Ruin game with you, where you pay
them $1$ dollar if the coin comes up heads and they pay you $1$ if it comes up
tails. You discover that a biased coin will be used
that has probability 2/3 of coming up heads. Even though the other player seems to have the advantage, you get to start with 2 dollars and they will start with 1. Whoever gets all $3$ dollars first will win the game.
Do you still have an advantage, even though you use a biased coin?
Find out by computing $p_1$, where $p_i$ is the probability that the other player wins the game starting with i dollars, for 0 ≤ i ≤ 3. Give an exact value for $p_1$ as a simplified fraction.
How do I go about solving this problem?
I understand the general formula for conditional probability, and in this case I would start with P($E_i$ | H)P(H) + P($E_i$ | T)P(T)
But since "i" doesn't seem to have an upper bound, I don't really know how to compute $p_i$, let alone $p_1$.
Any help is appreciated. Thanks!
probability combinatorics discrete-mathematics conditional-probability
probability combinatorics discrete-mathematics conditional-probability
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
asked Jan 31 at 2:17
digiHarmoniousdigiHarmonious
303
303
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given the instructions, do the following:
Let $p_i$ be the probability you lose if you are playing this game and currently you have $3-i$ dollars and they have $i$. Call this "state" $i$.
Then note that $p_0=0$: since they have $0$, they have already lost the game.
The interesting cases are the middle ones, where you can write a recursive relationship. Each round of the game the opponent either wins $1$ (with probability 2/3) or loses $1$ (with probability 1/3). So, after one round you move from state $i$ to either $i+1$ or $i-1$.
By definition of $p_i$ (the winning probability, given that we started at state $i$) we therefore have
$$p_1=frac{2}{3}p_2+frac{1}{3}p_0$$
You can solve for all the $p_i$s from here.
answered Jan 31 at 2:33
obscuransobscurans
1,152311
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given the instructions, do the following:
Let $p_i$ be the probability you lose if you are playing this game and currently you have $3-i$ dollars and they have $i$. Call this "state" $i$.
Then note that $p_0=0$: since they have $0$, they have already lost the game.
The interesting cases are the middle ones, where you can write a recursive relationship. Each round of the game the opponent either wins $1$ (with probability 2/3) or loses $1$ (with probability 1/3). So, after one round you move from state $i$ to either $i+1$ or $i-1$.
By definition of $p_i$ (the winning probability, given that we started at state $i$) we therefore have
$$p_1=frac{2}{3}p_2+frac{1}{3}p_0$$
You can solve for all the $p_i$s from here.
answered Jan 31 at 2:33
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
Given the instructions, do the following:
Let $p_i$ be the probability you lose if you are playing this game and currently you have $3-i$ dollars and they have $i$. Call this "state" $i$.
Then note that $p_0=0$: since they have $0$, they have already lost the game.
The interesting cases are the middle ones, where you can write a recursive relationship. Each round of the game the opponent either wins $1$ (with probability 2/3) or loses $1$ (with probability 1/3). So, after one round you move from state $i$ to either $i+1$ or $i-1$.
By definition of $p_i$ (the winning probability, given that we started at state $i$) we therefore have
$$p_1=frac{2}{3}p_2+frac{1}{3}p_0$$
You can solve for all the $p_i$s from here.
answered Jan 31 at 2:33
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
Given the instructions, do the following:
Let $p_i$ be the probability you lose if you are playing this game and currently you have $3-i$ dollars and they have $i$. Call this "state" $i$.
Then note that $p_0=0$: since they have $0$, they have already lost the game.
The interesting cases are the middle ones, where you can write a recursive relationship. Each round of the game the opponent either wins $1$ (with probability 2/3) or loses $1$ (with probability 1/3). So, after one round you move from state $i$ to either $i+1$ or $i-1$.
By definition of $p_i$ (the winning probability, given that we started at state $i$) we therefore have
$$p_1=frac{2}{3}p_2+frac{1}{3}p_0$$
You can solve for all the $p_i$s from here.
answered Jan 31 at 2:33
obscuransobscurans
1,152311
$endgroup$
Given the instructions, do the following:
Let $p_i$ be the probability you lose if you are playing this game and currently you have $3-i$ dollars and they have $i$. Call this "state" $i$.
Then note that $p_0=0$: since they have $0$, they have already lost the game.
The interesting cases are the middle ones, where you can write a recursive relationship. Each round of the game the opponent either wins $1$ (with probability 2/3) or loses $1$ (with probability 1/3). So, after one round you move from state $i$ to either $i+1$ or $i-1$.
By definition of $p_i$ (the winning probability, given that we started at state $i$) we therefore have
$$p_1=frac{2}{3}p_2+frac{1}{3}p_0$$
You can solve for all the $p_i$s from here.
answered Jan 31 at 2:33
obscuransobscurans
1,152311
answered Jan 31 at 2:33
obscuransobscurans
1,152311
answered Jan 31 at 2:33
obscuransobscurans
1,152311
answered Jan 31 at 2:33
obscuransobscurans
1,152311
1,152311
add a comment |
add a comment |
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