Mixing
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Expected number of rolls to cover all stops
1
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Suppose there is a board game, the map is a circle with 70 stops on it. The game uses a 6 faces fair die. Assume there is only 1 player, every time the player rolls a die, s/he will move counter clockwise based on the number shows on the die.
What is the expected number of rolls to get on every stop at least once on the map?
Any suggestion would be appreciated!
Thanks
probability statistics optimization combinations
asked Jan 24 at 21:31
Y KY K
62
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add a comment |
1
$begingroup$
Suppose there is a board game, the map is a circle with 70 stops on it. The game uses a 6 faces fair die. Assume there is only 1 player, every time the player rolls a die, s/he will move counter clockwise based on the number shows on the die.
What is the expected number of rolls to get on every stop at least once on the map?
Any suggestion would be appreciated!
Thanks
probability statistics optimization combinations
asked Jan 24 at 21:31
Y KY K
62
$endgroup$
$begingroup$
Suggestion: Imagine a probability of the player being in position $i$ at time $n$: $P_i^n$ (vector of 70 components for each time) and each roll disperses this probability with a discrete convolution with $(0,1/6,1/6,1/6,1/6,1/6,0,0,0,0,ldots)$. to get $P_i^{n+1}$. This part cen be simplified with fourier transform (though probably not in closed form). Then you have to express probability of not having landed on any of them yet (you'll have to take into account all previous times including the current one). This will be a function of $n$, then you can take expected value of $n$.
$endgroup$
– orion
Jan 24 at 21:41
$begingroup$
For reference, simulation puts the average at $approx335.8$ if you don't count the initial position until the first return.
$endgroup$
– Daniel Mathias
Jan 25 at 14:29
add a comment |
1
1
1
$begingroup$
Suppose there is a board game, the map is a circle with 70 stops on it. The game uses a 6 faces fair die. Assume there is only 1 player, every time the player rolls a die, s/he will move counter clockwise based on the number shows on the die.
What is the expected number of rolls to get on every stop at least once on the map?
Any suggestion would be appreciated!
Thanks
probability statistics optimization combinations
asked Jan 24 at 21:31
Y KY K
62
$endgroup$
Suppose there is a board game, the map is a circle with 70 stops on it. The game uses a 6 faces fair die. Assume there is only 1 player, every time the player rolls a die, s/he will move counter clockwise based on the number shows on the die.
What is the expected number of rolls to get on every stop at least once on the map?
Any suggestion would be appreciated!
Thanks
probability statistics optimization combinations
probability statistics optimization combinations
asked Jan 24 at 21:31
Y KY K
62
asked Jan 24 at 21:31
Y KY K
62
asked Jan 24 at 21:31
Y KY K
62
asked Jan 24 at 21:31
Y KY K
62
asked Jan 24 at 21:31
Y KY K
62
62
$begingroup$
Suggestion: Imagine a probability of the player being in position $i$ at time $n$: $P_i^n$ (vector of 70 components for each time) and each roll disperses this probability with a discrete convolution with $(0,1/6,1/6,1/6,1/6,1/6,0,0,0,0,ldots)$. to get $P_i^{n+1}$. This part cen be simplified with fourier transform (though probably not in closed form). Then you have to express probability of not having landed on any of them yet (you'll have to take into account all previous times including the current one). This will be a function of $n$, then you can take expected value of $n$.
$endgroup$
– orion
Jan 24 at 21:41
$begingroup$
For reference, simulation puts the average at $approx335.8$ if you don't count the initial position until the first return.
$endgroup$
– Daniel Mathias
Jan 25 at 14:29
add a comment |
$begingroup$
Suggestion: Imagine a probability of the player being in position $i$ at time $n$: $P_i^n$ (vector of 70 components for each time) and each roll disperses this probability with a discrete convolution with $(0,1/6,1/6,1/6,1/6,1/6,0,0,0,0,ldots)$. to get $P_i^{n+1}$. This part cen be simplified with fourier transform (though probably not in closed form). Then you have to express probability of not having landed on any of them yet (you'll have to take into account all previous times including the current one). This will be a function of $n$, then you can take expected value of $n$.
$endgroup$
– orion
Jan 24 at 21:41
$begingroup$
For reference, simulation puts the average at $approx335.8$ if you don't count the initial position until the first return.
$endgroup$
– Daniel Mathias
Jan 25 at 14:29
$begingroup$
Suggestion: Imagine a probability of the player being in position $i$ at time $n$: $P_i^n$ (vector of 70 components for each time) and each roll disperses this probability with a discrete convolution with $(0,1/6,1/6,1/6,1/6,1/6,0,0,0,0,ldots)$. to get $P_i^{n+1}$. This part cen be simplified with fourier transform (though probably not in closed form). Then you have to express probability of not having landed on any of them yet (you'll have to take into account all previous times including the current one). This will be a function of $n$, then you can take expected value of $n$.
$endgroup$
– orion
Jan 24 at 21:41
$begingroup$
Suggestion: Imagine a probability of the player being in position $i$ at time $n$: $P_i^n$ (vector of 70 components for each time) and each roll disperses this probability with a discrete convolution with $(0,1/6,1/6,1/6,1/6,1/6,0,0,0,0,ldots)$. to get $P_i^{n+1}$. This part cen be simplified with fourier transform (though probably not in closed form). Then you have to express probability of not having landed on any of them yet (you'll have to take into account all previous times including the current one). This will be a function of $n$, then you can take expected value of $n$.
$endgroup$
– orion
Jan 24 at 21:41
$begingroup$
For reference, simulation puts the average at $approx335.8$ if you don't count the initial position until the first return.
$endgroup$
– Daniel Mathias
Jan 25 at 14:29
$begingroup$
For reference, simulation puts the average at $approx335.8$ if you don't count the initial position until the first return.
$endgroup$
– Daniel Mathias
Jan 25 at 14:29
add a comment |
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$begingroup$
Suggestion: Imagine a probability of the player being in position $i$ at time $n$: $P_i^n$ (vector of 70 components for each time) and each roll disperses this probability with a discrete convolution with $(0,1/6,1/6,1/6,1/6,1/6,0,0,0,0,ldots)$. to get $P_i^{n+1}$. This part cen be simplified with fourier transform (though probably not in closed form). Then you have to express probability of not having landed on any of them yet (you'll have to take into account all previous times including the current one). This will be a function of $n$, then you can take expected value of $n$.
$endgroup$
– orion
Jan 24 at 21:41
$begingroup$
For reference, simulation puts the average at $approx335.8$ if you don't count the initial position until the first return.
$endgroup$
– Daniel Mathias
Jan 25 at 14:29