Calculating probability of 2 people reaching together The Next CEO of Stack OverflowExact Probability of Collision of Two Independent Random Walkers After N StepsExact Probability of Collision of Two Independent Random Walkers After N StepsIndependent, Uniform, Random Variable:The probability of a drunk person/random walkProbability 2 Random Walkers Will Meet (1D)Result of a $2D$ random walk with position dependent probabilities of stepResult of a $2D$ random walk with position dependent probabilitiesProbability - A drunk man walking along the axisProbability of reaching a pathcalculating probability of drunk man walking on a lineDrunken man walking on an axis
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Calculating probability of 2 people reaching together
The Next CEO of Stack OverflowExact Probability of Collision of Two Independent Random Walkers After N StepsExact Probability of Collision of Two Independent Random Walkers After N StepsIndependent, Uniform, Random Variable:The probability of a drunk person/random walkProbability 2 Random Walkers Will Meet (1D)Result of a $2D$ random walk with position dependent probabilities of stepResult of a $2D$ random walk with position dependent probabilitiesProbability - A drunk man walking along the axisProbability of reaching a pathcalculating probability of drunk man walking on a lineDrunken man walking on an axis
$begingroup$
How do I solve given problem? Or how do I approach this question?
Two drunks start out together at the origin, each having equal probability of making a step to the left or right along the $x$ axis. Find the probability that they meet again after $N$ steps. It is understood that the men make their steps simultaneously. (It may be helpful to consider their relative motion.)
probability probability-theory probability-distributions
edited 2 days ago
Matti P.
2,2811514
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
$endgroup$
add a comment |
$begingroup$
How do I solve given problem? Or how do I approach this question?
Two drunks start out together at the origin, each having equal probability of making a step to the left or right along the $x$ axis. Find the probability that they meet again after $N$ steps. It is understood that the men make their steps simultaneously. (It may be helpful to consider their relative motion.)
probability probability-theory probability-distributions
edited 2 days ago
Matti P.
2,2811514
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
$endgroup$
$begingroup$
Possible duplicate of Exact Probability of Collision of Two Independent Random Walkers After N Steps
$endgroup$
– Yanior Weg
yesterday
add a comment |
$begingroup$
How do I solve given problem? Or how do I approach this question?
Two drunks start out together at the origin, each having equal probability of making a step to the left or right along the $x$ axis. Find the probability that they meet again after $N$ steps. It is understood that the men make their steps simultaneously. (It may be helpful to consider their relative motion.)
probability probability-theory probability-distributions
edited 2 days ago
Matti P.
2,2811514
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
$endgroup$
How do I solve given problem? Or how do I approach this question?
Two drunks start out together at the origin, each having equal probability of making a step to the left or right along the $x$ axis. Find the probability that they meet again after $N$ steps. It is understood that the men make their steps simultaneously. (It may be helpful to consider their relative motion.)
probability probability-theory probability-distributions
probability probability-theory probability-distributions
edited 2 days ago
Matti P.
2,2811514
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
edited 2 days ago
Matti P.
2,2811514
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
edited 2 days ago
Matti P.
2,2811514
edited 2 days ago
Matti P.
2,2811514
edited 2 days ago
Matti P.
2,2811514
2,2811514
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
asked Mar 26 at 4:31
Abhi7731756Abhi7731756
74
74
$begingroup$
Possible duplicate of Exact Probability of Collision of Two Independent Random Walkers After N Steps
$endgroup$
– Yanior Weg
yesterday
add a comment |
$begingroup$
Possible duplicate of Exact Probability of Collision of Two Independent Random Walkers After N Steps
$endgroup$
– Yanior Weg
yesterday
$begingroup$
Possible duplicate of Exact Probability of Collision of Two Independent Random Walkers After N Steps
$endgroup$
– Yanior Weg
yesterday
$begingroup$
Possible duplicate of Exact Probability of Collision of Two Independent Random Walkers After N Steps
$endgroup$
– Yanior Weg
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $X_n$ is the coordinate of the first man after $n$ steps, and $Y$ - of the second one. Suppose $Z_n = X_n - Y_n$. Then your problem is equivalent to finding $P(Z_N = 0)$.
We see, that $$Z_n - Z_n-1 = begincases 2 & quad text with probability frac14 text (the probability that the first one makes a step right and the other a step left) \ 0 & quad text with probability frac12 text (the probability that they move in the same direction) \ -2 & quad text with probability frac14 text (the probability that the first one makes a step left and the other a step right) endcases$$
And each step is assumed to be made independently.
Suppose $N_1$ is the number of the occurrences of the first case, $N_2$ - of the second case, and $N_3$ - of the third case. Then, in case $Z_N = 0$ we have $N = N_1 + N_2 + N_3$ and $N_1 = N_3$. There are $C_N^N_2C_N - N_2^fracN - N_22$ such configurations for any fixed $N_2$, such that $N equiv N_2 (textmod 2)$ (and $0$ otherwise). And the probability of each of them is $frac12^N_2frac14^N - N_2 = frac12^2N - N_2$. Thus, we have $$P(Z_N = 0) = begincases fracSigma_i = 0^fracN2 C_N^2iC_N - 2i^fracN2 - i4^N - i & quad N text is even \ fracSigma_i = 0^fracN-12 C_N^2i+1C_N - 2i-1^fracN-12 - i2^2N - 2i - 1 & quad N text is oddendcases$$
That is your answer.
answered 2 days ago
Yanior WegYanior Weg
2,78211346
$endgroup$
$begingroup$
Shouldn't $N_1=N_3 as oppose to N_1=N_2$?
$endgroup$
– Abhi7731756
yesterday
$begingroup$
@Abhi7731756, yes, it should. $N_1 = N_2$ was a typo.
$endgroup$
– Yanior Weg
yesterday
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $X_n$ is the coordinate of the first man after $n$ steps, and $Y$ - of the second one. Suppose $Z_n = X_n - Y_n$. Then your problem is equivalent to finding $P(Z_N = 0)$.
We see, that $$Z_n - Z_n-1 = begincases 2 & quad text with probability frac14 text (the probability that the first one makes a step right and the other a step left) \ 0 & quad text with probability frac12 text (the probability that they move in the same direction) \ -2 & quad text with probability frac14 text (the probability that the first one makes a step left and the other a step right) endcases$$
And each step is assumed to be made independently.
Suppose $N_1$ is the number of the occurrences of the first case, $N_2$ - of the second case, and $N_3$ - of the third case. Then, in case $Z_N = 0$ we have $N = N_1 + N_2 + N_3$ and $N_1 = N_3$. There are $C_N^N_2C_N - N_2^fracN - N_22$ such configurations for any fixed $N_2$, such that $N equiv N_2 (textmod 2)$ (and $0$ otherwise). And the probability of each of them is $frac12^N_2frac14^N - N_2 = frac12^2N - N_2$. Thus, we have $$P(Z_N = 0) = begincases fracSigma_i = 0^fracN2 C_N^2iC_N - 2i^fracN2 - i4^N - i & quad N text is even \ fracSigma_i = 0^fracN-12 C_N^2i+1C_N - 2i-1^fracN-12 - i2^2N - 2i - 1 & quad N text is oddendcases$$
That is your answer.
answered 2 days ago
Yanior WegYanior Weg
2,78211346
$endgroup$
$begingroup$
Shouldn't $N_1=N_3 as oppose to N_1=N_2$?
$endgroup$
– Abhi7731756
yesterday
$begingroup$
@Abhi7731756, yes, it should. $N_1 = N_2$ was a typo.
$endgroup$
– Yanior Weg
yesterday
add a comment |
$begingroup$
Suppose $X_n$ is the coordinate of the first man after $n$ steps, and $Y$ - of the second one. Suppose $Z_n = X_n - Y_n$. Then your problem is equivalent to finding $P(Z_N = 0)$.
We see, that $$Z_n - Z_n-1 = begincases 2 & quad text with probability frac14 text (the probability that the first one makes a step right and the other a step left) \ 0 & quad text with probability frac12 text (the probability that they move in the same direction) \ -2 & quad text with probability frac14 text (the probability that the first one makes a step left and the other a step right) endcases$$
And each step is assumed to be made independently.
Suppose $N_1$ is the number of the occurrences of the first case, $N_2$ - of the second case, and $N_3$ - of the third case. Then, in case $Z_N = 0$ we have $N = N_1 + N_2 + N_3$ and $N_1 = N_3$. There are $C_N^N_2C_N - N_2^fracN - N_22$ such configurations for any fixed $N_2$, such that $N equiv N_2 (textmod 2)$ (and $0$ otherwise). And the probability of each of them is $frac12^N_2frac14^N - N_2 = frac12^2N - N_2$. Thus, we have $$P(Z_N = 0) = begincases fracSigma_i = 0^fracN2 C_N^2iC_N - 2i^fracN2 - i4^N - i & quad N text is even \ fracSigma_i = 0^fracN-12 C_N^2i+1C_N - 2i-1^fracN-12 - i2^2N - 2i - 1 & quad N text is oddendcases$$
That is your answer.
answered 2 days ago
Yanior WegYanior Weg
2,78211346
$endgroup$
$begingroup$
Shouldn't $N_1=N_3 as oppose to N_1=N_2$?
$endgroup$
– Abhi7731756
yesterday
$begingroup$
@Abhi7731756, yes, it should. $N_1 = N_2$ was a typo.
$endgroup$
– Yanior Weg
yesterday
add a comment |
$begingroup$
Suppose $X_n$ is the coordinate of the first man after $n$ steps, and $Y$ - of the second one. Suppose $Z_n = X_n - Y_n$. Then your problem is equivalent to finding $P(Z_N = 0)$.
We see, that $$Z_n - Z_n-1 = begincases 2 & quad text with probability frac14 text (the probability that the first one makes a step right and the other a step left) \ 0 & quad text with probability frac12 text (the probability that they move in the same direction) \ -2 & quad text with probability frac14 text (the probability that the first one makes a step left and the other a step right) endcases$$
And each step is assumed to be made independently.
Suppose $N_1$ is the number of the occurrences of the first case, $N_2$ - of the second case, and $N_3$ - of the third case. Then, in case $Z_N = 0$ we have $N = N_1 + N_2 + N_3$ and $N_1 = N_3$. There are $C_N^N_2C_N - N_2^fracN - N_22$ such configurations for any fixed $N_2$, such that $N equiv N_2 (textmod 2)$ (and $0$ otherwise). And the probability of each of them is $frac12^N_2frac14^N - N_2 = frac12^2N - N_2$. Thus, we have $$P(Z_N = 0) = begincases fracSigma_i = 0^fracN2 C_N^2iC_N - 2i^fracN2 - i4^N - i & quad N text is even \ fracSigma_i = 0^fracN-12 C_N^2i+1C_N - 2i-1^fracN-12 - i2^2N - 2i - 1 & quad N text is oddendcases$$
That is your answer.
answered 2 days ago
Yanior WegYanior Weg
2,78211346
$endgroup$
Suppose $X_n$ is the coordinate of the first man after $n$ steps, and $Y$ - of the second one. Suppose $Z_n = X_n - Y_n$. Then your problem is equivalent to finding $P(Z_N = 0)$.
We see, that $$Z_n - Z_n-1 = begincases 2 & quad text with probability frac14 text (the probability that the first one makes a step right and the other a step left) \ 0 & quad text with probability frac12 text (the probability that they move in the same direction) \ -2 & quad text with probability frac14 text (the probability that the first one makes a step left and the other a step right) endcases$$
And each step is assumed to be made independently.
Suppose $N_1$ is the number of the occurrences of the first case, $N_2$ - of the second case, and $N_3$ - of the third case. Then, in case $Z_N = 0$ we have $N = N_1 + N_2 + N_3$ and $N_1 = N_3$. There are $C_N^N_2C_N - N_2^fracN - N_22$ such configurations for any fixed $N_2$, such that $N equiv N_2 (textmod 2)$ (and $0$ otherwise). And the probability of each of them is $frac12^N_2frac14^N - N_2 = frac12^2N - N_2$. Thus, we have $$P(Z_N = 0) = begincases fracSigma_i = 0^fracN2 C_N^2iC_N - 2i^fracN2 - i4^N - i & quad N text is even \ fracSigma_i = 0^fracN-12 C_N^2i+1C_N - 2i-1^fracN-12 - i2^2N - 2i - 1 & quad N text is oddendcases$$
That is your answer.
answered 2 days ago
Yanior WegYanior Weg
2,78211346
edited yesterday
answered 2 days ago
Yanior WegYanior Weg
2,78211346
answered 2 days ago
Yanior WegYanior Weg
2,78211346
answered 2 days ago
Yanior WegYanior Weg
2,78211346
2,78211346
$begingroup$
Shouldn't $N_1=N_3 as oppose to N_1=N_2$?
$endgroup$
– Abhi7731756
yesterday
$begingroup$
@Abhi7731756, yes, it should. $N_1 = N_2$ was a typo.
$endgroup$
– Yanior Weg
yesterday
add a comment |
$begingroup$
Shouldn't $N_1=N_3 as oppose to N_1=N_2$?
$endgroup$
– Abhi7731756
yesterday
$begingroup$
@Abhi7731756, yes, it should. $N_1 = N_2$ was a typo.
$endgroup$
– Yanior Weg
yesterday
$begingroup$
Shouldn't $N_1=N_3 as oppose to N_1=N_2$?
$endgroup$
– Abhi7731756
yesterday
$begingroup$
Shouldn't $N_1=N_3 as oppose to N_1=N_2$?
$endgroup$
– Abhi7731756
yesterday
$begingroup$
@Abhi7731756, yes, it should. $N_1 = N_2$ was a typo.
$endgroup$
– Yanior Weg
yesterday
$begingroup$
@Abhi7731756, yes, it should. $N_1 = N_2$ was a typo.
$endgroup$
– Yanior Weg
yesterday
add a comment |
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$begingroup$
Possible duplicate of Exact Probability of Collision of Two Independent Random Walkers After N Steps
$endgroup$
– Yanior Weg
yesterday