Calculating if I'm going to have my gift swapped
$begingroup$
My company is doing something...strange for the christmas party. Its sort of a yankee swap, but with a twist.
Roll The Dice Swap!
Roll a 1 – keep your gift
Roll a 2 – steal anyone’s gift
Roll a 3 – everyone passes their gift to the left
Roll a 4 – everyone passes their gift to the right
Roll a 5 – keep your gift
Roll a 6 – steal anyone’s gift
I'm trying to figure out the probability of your gift (assuming you have one already) getting swapped out.
It looks like keeping your gift is 33%, a gift being stolen is 33% and gifts shifting is 33%. What I'd like to do is find out, each round, what the chance is that the gift you have is either shifted away or stolen.
It seems like it would be 1/3 + integral of (dn/(N-n)) for each person after you n out of N people have drawn and assuming axiom of choice of the gift (with respect to motivation to swap) but I can't get the numbers to work out.
Can someone help me specifically understand how to calculate this sort of nonsense?
EDIT: I'm not sure what the plan is. Before we've done an ordinary Yankee Swap where we draw lots for order, then each person either takes a present from the pool or takes someone else's. I would think it works that way with presents shifting only to those who already have presents (effectively the pool of present-holders grows incrementally). Although they could also decide to just randomly assign the presents and then each person in turn rolls the dice. In that case I think the probability would just be 1/3+1/N?
EDIT 2: Lets say that no person ends up with more than one gift.
EDIT 3: For simplicity, I'm assuming each present is equally desirable (knowing my company, its probably booze) and that each person gets one roll of the die. Further, once they get "steal a gift" they get to chose which gift to take.
probability combinatorics
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
$endgroup$
|
show 3 more comments
$begingroup$
My company is doing something...strange for the christmas party. Its sort of a yankee swap, but with a twist.
Roll The Dice Swap!
Roll a 1 – keep your gift
Roll a 2 – steal anyone’s gift
Roll a 3 – everyone passes their gift to the left
Roll a 4 – everyone passes their gift to the right
Roll a 5 – keep your gift
Roll a 6 – steal anyone’s gift
I'm trying to figure out the probability of your gift (assuming you have one already) getting swapped out.
It looks like keeping your gift is 33%, a gift being stolen is 33% and gifts shifting is 33%. What I'd like to do is find out, each round, what the chance is that the gift you have is either shifted away or stolen.
It seems like it would be 1/3 + integral of (dn/(N-n)) for each person after you n out of N people have drawn and assuming axiom of choice of the gift (with respect to motivation to swap) but I can't get the numbers to work out.
Can someone help me specifically understand how to calculate this sort of nonsense?
EDIT: I'm not sure what the plan is. Before we've done an ordinary Yankee Swap where we draw lots for order, then each person either takes a present from the pool or takes someone else's. I would think it works that way with presents shifting only to those who already have presents (effectively the pool of present-holders grows incrementally). Although they could also decide to just randomly assign the presents and then each person in turn rolls the dice. In that case I think the probability would just be 1/3+1/N?
EDIT 2: Lets say that no person ends up with more than one gift.
EDIT 3: For simplicity, I'm assuming each present is equally desirable (knowing my company, its probably booze) and that each person gets one roll of the die. Further, once they get "steal a gift" they get to chose which gift to take.
probability combinatorics
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
$endgroup$
$begingroup$
For clarity: Under the "steal anyone's gift" rolls, do you steal from everyone with equal probability? Can you steal your own gift? Also, are you assuming gifts are initially given to all $N$ people, and we proceed sequentially with the rolling thereafter? Another variation would be that the gifts are in the middle and we progressively pick gifts (so we can only steal from people who have picked before us), but there is no dice roll for "pick a gift from the middle."
$endgroup$
– Michael
Dec 20 '18 at 18:28
$begingroup$
If you steal a gift, do you have two gifts, or do you swap your gift with theirs?
$endgroup$
– zahbaz
Dec 20 '18 at 18:32
$begingroup$
I'm going to say when someone steals your gift, they give you what they had before.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:35
$begingroup$
Related: Assuming we pick a gift from the middle before each roll, but then we roll a "shift left" it means the next person (who has not yet picked from the middle) gets a free gift and a chance to pick from the middle, meaning that person will have two gifts before his/her roll of the dice.
$endgroup$
– Michael
Dec 20 '18 at 18:36
$begingroup$
I'm starting to think that the powers that be may not have thought through this scheme.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:39
|
show 3 more comments
$begingroup$
My company is doing something...strange for the christmas party. Its sort of a yankee swap, but with a twist.
Roll The Dice Swap!
Roll a 1 – keep your gift
Roll a 2 – steal anyone’s gift
Roll a 3 – everyone passes their gift to the left
Roll a 4 – everyone passes their gift to the right
Roll a 5 – keep your gift
Roll a 6 – steal anyone’s gift
I'm trying to figure out the probability of your gift (assuming you have one already) getting swapped out.
It looks like keeping your gift is 33%, a gift being stolen is 33% and gifts shifting is 33%. What I'd like to do is find out, each round, what the chance is that the gift you have is either shifted away or stolen.
It seems like it would be 1/3 + integral of (dn/(N-n)) for each person after you n out of N people have drawn and assuming axiom of choice of the gift (with respect to motivation to swap) but I can't get the numbers to work out.
Can someone help me specifically understand how to calculate this sort of nonsense?
EDIT: I'm not sure what the plan is. Before we've done an ordinary Yankee Swap where we draw lots for order, then each person either takes a present from the pool or takes someone else's. I would think it works that way with presents shifting only to those who already have presents (effectively the pool of present-holders grows incrementally). Although they could also decide to just randomly assign the presents and then each person in turn rolls the dice. In that case I think the probability would just be 1/3+1/N?
EDIT 2: Lets say that no person ends up with more than one gift.
EDIT 3: For simplicity, I'm assuming each present is equally desirable (knowing my company, its probably booze) and that each person gets one roll of the die. Further, once they get "steal a gift" they get to chose which gift to take.
probability combinatorics
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
$endgroup$
My company is doing something...strange for the christmas party. Its sort of a yankee swap, but with a twist.
Roll The Dice Swap!
Roll a 1 – keep your gift
Roll a 2 – steal anyone’s gift
Roll a 3 – everyone passes their gift to the left
Roll a 4 – everyone passes their gift to the right
Roll a 5 – keep your gift
Roll a 6 – steal anyone’s gift
I'm trying to figure out the probability of your gift (assuming you have one already) getting swapped out.
It looks like keeping your gift is 33%, a gift being stolen is 33% and gifts shifting is 33%. What I'd like to do is find out, each round, what the chance is that the gift you have is either shifted away or stolen.
It seems like it would be 1/3 + integral of (dn/(N-n)) for each person after you n out of N people have drawn and assuming axiom of choice of the gift (with respect to motivation to swap) but I can't get the numbers to work out.
Can someone help me specifically understand how to calculate this sort of nonsense?
EDIT: I'm not sure what the plan is. Before we've done an ordinary Yankee Swap where we draw lots for order, then each person either takes a present from the pool or takes someone else's. I would think it works that way with presents shifting only to those who already have presents (effectively the pool of present-holders grows incrementally). Although they could also decide to just randomly assign the presents and then each person in turn rolls the dice. In that case I think the probability would just be 1/3+1/N?
EDIT 2: Lets say that no person ends up with more than one gift.
EDIT 3: For simplicity, I'm assuming each present is equally desirable (knowing my company, its probably booze) and that each person gets one roll of the die. Further, once they get "steal a gift" they get to chose which gift to take.
probability combinatorics
probability combinatorics
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
edited Dec 20 '18 at 18:40
Kevin Milner
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
asked Dec 20 '18 at 18:24
Kevin MilnerKevin Milner
1255
1255
$begingroup$
For clarity: Under the "steal anyone's gift" rolls, do you steal from everyone with equal probability? Can you steal your own gift? Also, are you assuming gifts are initially given to all $N$ people, and we proceed sequentially with the rolling thereafter? Another variation would be that the gifts are in the middle and we progressively pick gifts (so we can only steal from people who have picked before us), but there is no dice roll for "pick a gift from the middle."
$endgroup$
– Michael
Dec 20 '18 at 18:28
$begingroup$
If you steal a gift, do you have two gifts, or do you swap your gift with theirs?
$endgroup$
– zahbaz
Dec 20 '18 at 18:32
$begingroup$
I'm going to say when someone steals your gift, they give you what they had before.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:35
$begingroup$
Related: Assuming we pick a gift from the middle before each roll, but then we roll a "shift left" it means the next person (who has not yet picked from the middle) gets a free gift and a chance to pick from the middle, meaning that person will have two gifts before his/her roll of the dice.
$endgroup$
– Michael
Dec 20 '18 at 18:36
$begingroup$
I'm starting to think that the powers that be may not have thought through this scheme.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:39
|
show 3 more comments
$begingroup$
For clarity: Under the "steal anyone's gift" rolls, do you steal from everyone with equal probability? Can you steal your own gift? Also, are you assuming gifts are initially given to all $N$ people, and we proceed sequentially with the rolling thereafter? Another variation would be that the gifts are in the middle and we progressively pick gifts (so we can only steal from people who have picked before us), but there is no dice roll for "pick a gift from the middle."
$endgroup$
– Michael
Dec 20 '18 at 18:28
$begingroup$
If you steal a gift, do you have two gifts, or do you swap your gift with theirs?
$endgroup$
– zahbaz
Dec 20 '18 at 18:32
$begingroup$
I'm going to say when someone steals your gift, they give you what they had before.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:35
$begingroup$
Related: Assuming we pick a gift from the middle before each roll, but then we roll a "shift left" it means the next person (who has not yet picked from the middle) gets a free gift and a chance to pick from the middle, meaning that person will have two gifts before his/her roll of the dice.
$endgroup$
– Michael
Dec 20 '18 at 18:36
$begingroup$
I'm starting to think that the powers that be may not have thought through this scheme.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:39
$begingroup$
For clarity: Under the "steal anyone's gift" rolls, do you steal from everyone with equal probability? Can you steal your own gift? Also, are you assuming gifts are initially given to all $N$ people, and we proceed sequentially with the rolling thereafter? Another variation would be that the gifts are in the middle and we progressively pick gifts (so we can only steal from people who have picked before us), but there is no dice roll for "pick a gift from the middle."
$endgroup$
– Michael
Dec 20 '18 at 18:28
$begingroup$
For clarity: Under the "steal anyone's gift" rolls, do you steal from everyone with equal probability? Can you steal your own gift? Also, are you assuming gifts are initially given to all $N$ people, and we proceed sequentially with the rolling thereafter? Another variation would be that the gifts are in the middle and we progressively pick gifts (so we can only steal from people who have picked before us), but there is no dice roll for "pick a gift from the middle."
$endgroup$
– Michael
Dec 20 '18 at 18:28
$begingroup$
If you steal a gift, do you have two gifts, or do you swap your gift with theirs?
$endgroup$
– zahbaz
Dec 20 '18 at 18:32
$begingroup$
If you steal a gift, do you have two gifts, or do you swap your gift with theirs?
$endgroup$
– zahbaz
Dec 20 '18 at 18:32
$begingroup$
I'm going to say when someone steals your gift, they give you what they had before.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:35
$begingroup$
I'm going to say when someone steals your gift, they give you what they had before.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:35
$begingroup$
Related: Assuming we pick a gift from the middle before each roll, but then we roll a "shift left" it means the next person (who has not yet picked from the middle) gets a free gift and a chance to pick from the middle, meaning that person will have two gifts before his/her roll of the dice.
$endgroup$
– Michael
Dec 20 '18 at 18:36
$begingroup$
Related: Assuming we pick a gift from the middle before each roll, but then we roll a "shift left" it means the next person (who has not yet picked from the middle) gets a free gift and a chance to pick from the middle, meaning that person will have two gifts before his/her roll of the dice.
$endgroup$
– Michael
Dec 20 '18 at 18:36
$begingroup$
I'm starting to think that the powers that be may not have thought through this scheme.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:39
$begingroup$
I'm starting to think that the powers that be may not have thought through this scheme.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:39
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The odds that you'll lose your current item in a given roll are in $(1/3,2/3]$ with the upper bound hit for $n=2$ players.
Assume $nge2$ players. On any given turn, there is a $frac13$ chance of losing your gift through a rotation. Assuming that on a stealing roll a player is forced to exchange their gift with that of another, and assuming this decision is uniformly drawn from the remaining $n-1$ players, then there is a $frac{1}{3}left(frac{1}{n-1}right)$ chance of losing through a steal every round. So the odds you lose your current gift each round is the sum
$$frac13left(frac{n}{n-1}right)$$.
I'd say the next step is to consider the odds of losing your gift after one entire cycle of dice rolls (i.e. everyone in the game's circle gets to roll the dice once). This seems like combining a drunken walk problem with a bit of teleportation due to the stealing roll.
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The odds that you'll lose your current item in a given roll are in $(1/3,2/3]$ with the upper bound hit for $n=2$ players.
Assume $nge2$ players. On any given turn, there is a $frac13$ chance of losing your gift through a rotation. Assuming that on a stealing roll a player is forced to exchange their gift with that of another, and assuming this decision is uniformly drawn from the remaining $n-1$ players, then there is a $frac{1}{3}left(frac{1}{n-1}right)$ chance of losing through a steal every round. So the odds you lose your current gift each round is the sum
$$frac13left(frac{n}{n-1}right)$$.
I'd say the next step is to consider the odds of losing your gift after one entire cycle of dice rolls (i.e. everyone in the game's circle gets to roll the dice once). This seems like combining a drunken walk problem with a bit of teleportation due to the stealing roll.
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
$endgroup$
add a comment |
$begingroup$
The odds that you'll lose your current item in a given roll are in $(1/3,2/3]$ with the upper bound hit for $n=2$ players.
Assume $nge2$ players. On any given turn, there is a $frac13$ chance of losing your gift through a rotation. Assuming that on a stealing roll a player is forced to exchange their gift with that of another, and assuming this decision is uniformly drawn from the remaining $n-1$ players, then there is a $frac{1}{3}left(frac{1}{n-1}right)$ chance of losing through a steal every round. So the odds you lose your current gift each round is the sum
$$frac13left(frac{n}{n-1}right)$$.
I'd say the next step is to consider the odds of losing your gift after one entire cycle of dice rolls (i.e. everyone in the game's circle gets to roll the dice once). This seems like combining a drunken walk problem with a bit of teleportation due to the stealing roll.
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
$endgroup$
add a comment |
$begingroup$
The odds that you'll lose your current item in a given roll are in $(1/3,2/3]$ with the upper bound hit for $n=2$ players.
Assume $nge2$ players. On any given turn, there is a $frac13$ chance of losing your gift through a rotation. Assuming that on a stealing roll a player is forced to exchange their gift with that of another, and assuming this decision is uniformly drawn from the remaining $n-1$ players, then there is a $frac{1}{3}left(frac{1}{n-1}right)$ chance of losing through a steal every round. So the odds you lose your current gift each round is the sum
$$frac13left(frac{n}{n-1}right)$$.
I'd say the next step is to consider the odds of losing your gift after one entire cycle of dice rolls (i.e. everyone in the game's circle gets to roll the dice once). This seems like combining a drunken walk problem with a bit of teleportation due to the stealing roll.
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
$endgroup$
The odds that you'll lose your current item in a given roll are in $(1/3,2/3]$ with the upper bound hit for $n=2$ players.
Assume $nge2$ players. On any given turn, there is a $frac13$ chance of losing your gift through a rotation. Assuming that on a stealing roll a player is forced to exchange their gift with that of another, and assuming this decision is uniformly drawn from the remaining $n-1$ players, then there is a $frac{1}{3}left(frac{1}{n-1}right)$ chance of losing through a steal every round. So the odds you lose your current gift each round is the sum
$$frac13left(frac{n}{n-1}right)$$.
I'd say the next step is to consider the odds of losing your gift after one entire cycle of dice rolls (i.e. everyone in the game's circle gets to roll the dice once). This seems like combining a drunken walk problem with a bit of teleportation due to the stealing roll.
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
answered Dec 20 '18 at 18:51
zahbazzahbaz
8,44921938
8,44921938
add a comment |
add a comment |
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$begingroup$
For clarity: Under the "steal anyone's gift" rolls, do you steal from everyone with equal probability? Can you steal your own gift? Also, are you assuming gifts are initially given to all $N$ people, and we proceed sequentially with the rolling thereafter? Another variation would be that the gifts are in the middle and we progressively pick gifts (so we can only steal from people who have picked before us), but there is no dice roll for "pick a gift from the middle."
$endgroup$
– Michael
Dec 20 '18 at 18:28
$begingroup$
If you steal a gift, do you have two gifts, or do you swap your gift with theirs?
$endgroup$
– zahbaz
Dec 20 '18 at 18:32
$begingroup$
I'm going to say when someone steals your gift, they give you what they had before.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:35
$begingroup$
Related: Assuming we pick a gift from the middle before each roll, but then we roll a "shift left" it means the next person (who has not yet picked from the middle) gets a free gift and a chance to pick from the middle, meaning that person will have two gifts before his/her roll of the dice.
$endgroup$
– Michael
Dec 20 '18 at 18:36
$begingroup$
I'm starting to think that the powers that be may not have thought through this scheme.
$endgroup$
– Kevin Milner
Dec 20 '18 at 18:39