How can the DM most simply and fairly choose 1 out of an odd number of players to be targeted by an attack or...





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When DMing and randomly determining a character to take damage or an effect, it's easy to roll a die if you have an even number of players. But what about if you have an odd number of players?



What is the simplest method for randomly choosing 1 out of an odd number of players, while ensuring that each player's probability of getting chosen is equal or very close to equal?










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  • $begingroup$
    How many players do you have?
    $endgroup$
    – KorvinStarmast
    yesterday






  • 4




    $begingroup$
    As it stands this question is just asking for an open-ended list or poll of methods that people use without any way to judge any that are better or worse. Can you add some guidelines that you are looking for so that the question can fit the format better?
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    (I made an edit that tries to steer it further away from just asking for an unbound list, but I think further revisions may need to be made.)
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    "Most effective" might be arguably still opinion-based, as there's no criterion for effectiveness. A more straightforward and answerable version might be "What is the simplest method [...] while ensuring equal probabilities for each result?" (I've edited the question accordingly and voted to reopen.)
    $endgroup$
    – V2Blast
    5 hours ago




















11












$begingroup$


When DMing and randomly determining a character to take damage or an effect, it's easy to roll a die if you have an even number of players. But what about if you have an odd number of players?



What is the simplest method for randomly choosing 1 out of an odd number of players, while ensuring that each player's probability of getting chosen is equal or very close to equal?










share|improve this question











$endgroup$












  • $begingroup$
    How many players do you have?
    $endgroup$
    – KorvinStarmast
    yesterday






  • 4




    $begingroup$
    As it stands this question is just asking for an open-ended list or poll of methods that people use without any way to judge any that are better or worse. Can you add some guidelines that you are looking for so that the question can fit the format better?
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    (I made an edit that tries to steer it further away from just asking for an unbound list, but I think further revisions may need to be made.)
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    "Most effective" might be arguably still opinion-based, as there's no criterion for effectiveness. A more straightforward and answerable version might be "What is the simplest method [...] while ensuring equal probabilities for each result?" (I've edited the question accordingly and voted to reopen.)
    $endgroup$
    – V2Blast
    5 hours ago
















11












11








11


1



$begingroup$


When DMing and randomly determining a character to take damage or an effect, it's easy to roll a die if you have an even number of players. But what about if you have an odd number of players?



What is the simplest method for randomly choosing 1 out of an odd number of players, while ensuring that each player's probability of getting chosen is equal or very close to equal?










share|improve this question











$endgroup$




When DMing and randomly determining a character to take damage or an effect, it's easy to roll a die if you have an even number of players. But what about if you have an odd number of players?



What is the simplest method for randomly choosing 1 out of an odd number of players, while ensuring that each player's probability of getting chosen is equal or very close to equal?







gm-techniques dungeons-and-dragons






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share|improve this question













share|improve this question




share|improve this question








edited 5 hours ago









V2Blast

26.2k590160




26.2k590160










asked yesterday









Andrew FranklinAndrew Franklin

4712416




4712416












  • $begingroup$
    How many players do you have?
    $endgroup$
    – KorvinStarmast
    yesterday






  • 4




    $begingroup$
    As it stands this question is just asking for an open-ended list or poll of methods that people use without any way to judge any that are better or worse. Can you add some guidelines that you are looking for so that the question can fit the format better?
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    (I made an edit that tries to steer it further away from just asking for an unbound list, but I think further revisions may need to be made.)
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    "Most effective" might be arguably still opinion-based, as there's no criterion for effectiveness. A more straightforward and answerable version might be "What is the simplest method [...] while ensuring equal probabilities for each result?" (I've edited the question accordingly and voted to reopen.)
    $endgroup$
    – V2Blast
    5 hours ago




















  • $begingroup$
    How many players do you have?
    $endgroup$
    – KorvinStarmast
    yesterday






  • 4




    $begingroup$
    As it stands this question is just asking for an open-ended list or poll of methods that people use without any way to judge any that are better or worse. Can you add some guidelines that you are looking for so that the question can fit the format better?
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    (I made an edit that tries to steer it further away from just asking for an unbound list, but I think further revisions may need to be made.)
    $endgroup$
    – Rubiksmoose
    yesterday










  • $begingroup$
    "Most effective" might be arguably still opinion-based, as there's no criterion for effectiveness. A more straightforward and answerable version might be "What is the simplest method [...] while ensuring equal probabilities for each result?" (I've edited the question accordingly and voted to reopen.)
    $endgroup$
    – V2Blast
    5 hours ago


















$begingroup$
How many players do you have?
$endgroup$
– KorvinStarmast
yesterday




$begingroup$
How many players do you have?
$endgroup$
– KorvinStarmast
yesterday




4




4




$begingroup$
As it stands this question is just asking for an open-ended list or poll of methods that people use without any way to judge any that are better or worse. Can you add some guidelines that you are looking for so that the question can fit the format better?
$endgroup$
– Rubiksmoose
yesterday




$begingroup$
As it stands this question is just asking for an open-ended list or poll of methods that people use without any way to judge any that are better or worse. Can you add some guidelines that you are looking for so that the question can fit the format better?
$endgroup$
– Rubiksmoose
yesterday












$begingroup$
(I made an edit that tries to steer it further away from just asking for an unbound list, but I think further revisions may need to be made.)
$endgroup$
– Rubiksmoose
yesterday




$begingroup$
(I made an edit that tries to steer it further away from just asking for an unbound list, but I think further revisions may need to be made.)
$endgroup$
– Rubiksmoose
yesterday












$begingroup$
"Most effective" might be arguably still opinion-based, as there's no criterion for effectiveness. A more straightforward and answerable version might be "What is the simplest method [...] while ensuring equal probabilities for each result?" (I've edited the question accordingly and voted to reopen.)
$endgroup$
– V2Blast
5 hours ago






$begingroup$
"Most effective" might be arguably still opinion-based, as there's no criterion for effectiveness. A more straightforward and answerable version might be "What is the simplest method [...] while ensuring equal probabilities for each result?" (I've edited the question accordingly and voted to reopen.)
$endgroup$
– V2Blast
5 hours ago












3 Answers
3






active

oldest

votes


















35












$begingroup$

Dice work just fine for the most common numbers of players



I work with probabilities a lot. Dice allow you to represent many common probabilities, but when you want to roll to pick from options, I've discovered the following "law": it stops making sense to people when you have to roll more than a single die at once. Luckily, the most common numbers of players can be chosen from using just a single die roll, or a single die roll that occasionally gets re-rolled when its result is invalid.



If you have a single player, no need to roll.



If you have three players, a d6 will do --- one player gets results 1-2, another 3-4 and the third one 5-6.



Five players, roll a d10. One player gets 1-2, the next 3-4, then 5-6, then 7-8, and the final one 9-10.



There is just a bit of math behind these two options: they work neatly because three and five are factors of six and ten, respectively. That means we can "partition" each dice's possible outcomes into that many equally probable sets, similar to how we can use "odd" and "even" to select from two options using a d6.



Seven players is the first tricky one because it's the least prime number that's not a factor of any of the common die sizes of 4, 6, 8, 10, 12 or 20. This means you cannot resolve a choice between seven players with any bounded number of these die rolls with complete fairness. You can, however, use a d8 with numbers 1-7 corresponding to each of the players and 8 indicating a reroll. The chances of having to roll the die impractically many times for a single decision are minimal.



Nine players or eleven players, similar to seven except use a d10 or d12, respectively. Since nine is divisible by three, you can also use two rolls of a d6: first divide the player into three groups of three, let the first roll decide which group you pick and the second roll which particular player from that group. The d10 method is simpler, though.



You can of course use arbitrarily large dice to cover any amount of players using the "reroll if too high" strategy, but it can get unwieldy for larger numbers.



Drawing from a deck of cards works for many sizes of groups



Using a deck of cards provides a simple solution for groups small and large, without having to remember any complex roll rules. Have each player draw a card from a standard deck of 52 cards. The player who gets the highest card (or lowest, if you prefer) is chosen. Remember to have a set convention on the ordering used (aces high or low and precedence of suits).



There's 52 cards in a normal French-style deck of cards, and each is unique, so you can choose a random player with everyone drawing a single card for groups up to 52 players. That's probably more than enough.



Draw from a bag



Yet another classic method: have each player write their name on a slip of paper. Put them in a box or bag, and resolve the decision by drawing one without looking. My experience is that this is quite awkward especially when table space is limited --- I appreciate the other methods' compactness. However, this method works quite well regardless of the size of the game, assuming you have a big enough box for the name slips and shuffle it well.



Random number list, if you want to keep it secret



If you know the amount of players you'll have in advance, you can create a list of random numbers and bring that to the game with you. When consulting the list, cross over the topmost number on it and select the player based on that number.



This solution is, for most games, over-engineered, but it has an advantage in that unlike die rolls and draws from the deck of cards, it is very subtle and allows invoking the randomness without signaling it to your players. My experience has been, though, that it's best to let the players know when you're actually choosing one of them at random in most cases --- otherwise, they can assume the choice is you being arbitrary.






share|improve this answer











$endgroup$













  • $begingroup$
    Re dice: If you have four or fewer players, I suppose you could use these. Just be very sure you don't accidentally mix them in with the regular d12s, or the DM will not be happy with you ("What do you mean, you rolled a 47 on 1d12?").
    $endgroup$
    – Kevin
    23 hours ago












  • $begingroup$
    If you go with drawing from a bag, I would recommend physical items. Differently-coloured tiddlywinks (or dice of the same size) if exact fairness is more important than immersion; cheap unpainted miniatures if the reverse.
    $endgroup$
    – TimLymington
    18 hours ago










  • $begingroup$
    Yeah, I think this is best. It's not perfect, but this works well, is fair, and rarely makes arguments. Maybe also mention apps or websites as an alternative, if "just getting a number" outweighs "being able to see how" for some people.
    $endgroup$
    – Jack V.
    14 hours ago



















23












$begingroup$

If you have 5 players, roll a d6. If you get a 6, re-roll until you get a number other than 6. And so on, for other odd numbers. Yes, this is completely fair, if the die you're rolling is a fair die.






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$endgroup$









  • 1




    $begingroup$
    You can also form5 players roll a d10 assigning 1-2 to one player, 3-4 to another and so on until all five have a number range that's theirs.
    $endgroup$
    – Sarriesfan
    yesterday










  • $begingroup$
    This is actually much simpler than the accepted answer, and works with any number of player unlike that answer.
    $endgroup$
    – John
    1 hour ago





















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$begingroup$

How I have done this over the years:



It makes for even probabilities.




  1. For three players: roll 1d6, divide result by 2, round up.


  2. For five players: roll 1d10, divide result by 2, round up


  3. For seven players: roll 1d8, re roll any 8.



  4. For nine players: Roll 2d6 of different colors. (Let's say red and
    green, since that is what my bag is filled with)




    • Green die indicates which range of three: 1 or 2 =1-3, 3 or 4 = 4-6, 5 or 6 = 7-9.

    • You now have a "three players" situation

    • Red die indicates who in that group is chosen by using the "for three players" above.



  5. For 11 players; roll a d12, re roll 12's.



It's really fast and intuitive once you start doing this (it's easy to do the comparison in your head). It takes longer to describe it than to do it.



Notes:




  1. While I don't do this, @BenR suggests out that for 9 you could do it
    as you do for 7; roll a d10 and re roll 0's/10's. Since I prefer to
    roll just once; the 2d6 is one roll and the result is right
    there).



  2. While I have not done this, @Yakk suggests this for a d13 if you have
    13 players.




    • For 1d13, roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value.








share|improve this answer











$endgroup$









  • 3




    $begingroup$
    Why is the nine players one so complex? Just roll a d10 and reroll 10s.
    $endgroup$
    – Ben R.
    15 hours ago










  • $begingroup$
    @BenR. It isn't complex; you roll two dice once, and you know who it is by looking at the two dice. As I said, it is far easier to do than to explain. I've been doing stuff like that for a long time. it's the 'for three people" with two levels.
    $endgroup$
    – KorvinStarmast
    13 hours ago












  • $begingroup$
    What, no 13 players?
    $endgroup$
    – Yakk
    12 hours ago










  • $begingroup$
    @Yakk No, haven't run a game that big in over 30 years. On that one, use d14 and reroll 14's. ;) What's a d14? Those two six sided dice (red and green) and a d8. (There were some 1 to 14 tables in AD&D 1e DMG, page 26, gems, random chance). Roll the two d6 and the d8. Low green? Use the d8. High green? Add 8 to the d6. Easy to do, one roll.
    $endgroup$
    – KorvinStarmast
    12 hours ago












  • $begingroup$
    @KorvinStarmast That d14 isn't an even distribution; low numbers have 1/16 chance, high numbers 1/12 chance each. Try this: roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value. Or, just roll 1d8 reroll 8s; then use high-low d2 to maybe add 7. Then reroll 14s.
    $endgroup$
    – Yakk
    12 hours ago














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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









35












$begingroup$

Dice work just fine for the most common numbers of players



I work with probabilities a lot. Dice allow you to represent many common probabilities, but when you want to roll to pick from options, I've discovered the following "law": it stops making sense to people when you have to roll more than a single die at once. Luckily, the most common numbers of players can be chosen from using just a single die roll, or a single die roll that occasionally gets re-rolled when its result is invalid.



If you have a single player, no need to roll.



If you have three players, a d6 will do --- one player gets results 1-2, another 3-4 and the third one 5-6.



Five players, roll a d10. One player gets 1-2, the next 3-4, then 5-6, then 7-8, and the final one 9-10.



There is just a bit of math behind these two options: they work neatly because three and five are factors of six and ten, respectively. That means we can "partition" each dice's possible outcomes into that many equally probable sets, similar to how we can use "odd" and "even" to select from two options using a d6.



Seven players is the first tricky one because it's the least prime number that's not a factor of any of the common die sizes of 4, 6, 8, 10, 12 or 20. This means you cannot resolve a choice between seven players with any bounded number of these die rolls with complete fairness. You can, however, use a d8 with numbers 1-7 corresponding to each of the players and 8 indicating a reroll. The chances of having to roll the die impractically many times for a single decision are minimal.



Nine players or eleven players, similar to seven except use a d10 or d12, respectively. Since nine is divisible by three, you can also use two rolls of a d6: first divide the player into three groups of three, let the first roll decide which group you pick and the second roll which particular player from that group. The d10 method is simpler, though.



You can of course use arbitrarily large dice to cover any amount of players using the "reroll if too high" strategy, but it can get unwieldy for larger numbers.



Drawing from a deck of cards works for many sizes of groups



Using a deck of cards provides a simple solution for groups small and large, without having to remember any complex roll rules. Have each player draw a card from a standard deck of 52 cards. The player who gets the highest card (or lowest, if you prefer) is chosen. Remember to have a set convention on the ordering used (aces high or low and precedence of suits).



There's 52 cards in a normal French-style deck of cards, and each is unique, so you can choose a random player with everyone drawing a single card for groups up to 52 players. That's probably more than enough.



Draw from a bag



Yet another classic method: have each player write their name on a slip of paper. Put them in a box or bag, and resolve the decision by drawing one without looking. My experience is that this is quite awkward especially when table space is limited --- I appreciate the other methods' compactness. However, this method works quite well regardless of the size of the game, assuming you have a big enough box for the name slips and shuffle it well.



Random number list, if you want to keep it secret



If you know the amount of players you'll have in advance, you can create a list of random numbers and bring that to the game with you. When consulting the list, cross over the topmost number on it and select the player based on that number.



This solution is, for most games, over-engineered, but it has an advantage in that unlike die rolls and draws from the deck of cards, it is very subtle and allows invoking the randomness without signaling it to your players. My experience has been, though, that it's best to let the players know when you're actually choosing one of them at random in most cases --- otherwise, they can assume the choice is you being arbitrary.






share|improve this answer











$endgroup$













  • $begingroup$
    Re dice: If you have four or fewer players, I suppose you could use these. Just be very sure you don't accidentally mix them in with the regular d12s, or the DM will not be happy with you ("What do you mean, you rolled a 47 on 1d12?").
    $endgroup$
    – Kevin
    23 hours ago












  • $begingroup$
    If you go with drawing from a bag, I would recommend physical items. Differently-coloured tiddlywinks (or dice of the same size) if exact fairness is more important than immersion; cheap unpainted miniatures if the reverse.
    $endgroup$
    – TimLymington
    18 hours ago










  • $begingroup$
    Yeah, I think this is best. It's not perfect, but this works well, is fair, and rarely makes arguments. Maybe also mention apps or websites as an alternative, if "just getting a number" outweighs "being able to see how" for some people.
    $endgroup$
    – Jack V.
    14 hours ago
















35












$begingroup$

Dice work just fine for the most common numbers of players



I work with probabilities a lot. Dice allow you to represent many common probabilities, but when you want to roll to pick from options, I've discovered the following "law": it stops making sense to people when you have to roll more than a single die at once. Luckily, the most common numbers of players can be chosen from using just a single die roll, or a single die roll that occasionally gets re-rolled when its result is invalid.



If you have a single player, no need to roll.



If you have three players, a d6 will do --- one player gets results 1-2, another 3-4 and the third one 5-6.



Five players, roll a d10. One player gets 1-2, the next 3-4, then 5-6, then 7-8, and the final one 9-10.



There is just a bit of math behind these two options: they work neatly because three and five are factors of six and ten, respectively. That means we can "partition" each dice's possible outcomes into that many equally probable sets, similar to how we can use "odd" and "even" to select from two options using a d6.



Seven players is the first tricky one because it's the least prime number that's not a factor of any of the common die sizes of 4, 6, 8, 10, 12 or 20. This means you cannot resolve a choice between seven players with any bounded number of these die rolls with complete fairness. You can, however, use a d8 with numbers 1-7 corresponding to each of the players and 8 indicating a reroll. The chances of having to roll the die impractically many times for a single decision are minimal.



Nine players or eleven players, similar to seven except use a d10 or d12, respectively. Since nine is divisible by three, you can also use two rolls of a d6: first divide the player into three groups of three, let the first roll decide which group you pick and the second roll which particular player from that group. The d10 method is simpler, though.



You can of course use arbitrarily large dice to cover any amount of players using the "reroll if too high" strategy, but it can get unwieldy for larger numbers.



Drawing from a deck of cards works for many sizes of groups



Using a deck of cards provides a simple solution for groups small and large, without having to remember any complex roll rules. Have each player draw a card from a standard deck of 52 cards. The player who gets the highest card (or lowest, if you prefer) is chosen. Remember to have a set convention on the ordering used (aces high or low and precedence of suits).



There's 52 cards in a normal French-style deck of cards, and each is unique, so you can choose a random player with everyone drawing a single card for groups up to 52 players. That's probably more than enough.



Draw from a bag



Yet another classic method: have each player write their name on a slip of paper. Put them in a box or bag, and resolve the decision by drawing one without looking. My experience is that this is quite awkward especially when table space is limited --- I appreciate the other methods' compactness. However, this method works quite well regardless of the size of the game, assuming you have a big enough box for the name slips and shuffle it well.



Random number list, if you want to keep it secret



If you know the amount of players you'll have in advance, you can create a list of random numbers and bring that to the game with you. When consulting the list, cross over the topmost number on it and select the player based on that number.



This solution is, for most games, over-engineered, but it has an advantage in that unlike die rolls and draws from the deck of cards, it is very subtle and allows invoking the randomness without signaling it to your players. My experience has been, though, that it's best to let the players know when you're actually choosing one of them at random in most cases --- otherwise, they can assume the choice is you being arbitrary.






share|improve this answer











$endgroup$













  • $begingroup$
    Re dice: If you have four or fewer players, I suppose you could use these. Just be very sure you don't accidentally mix them in with the regular d12s, or the DM will not be happy with you ("What do you mean, you rolled a 47 on 1d12?").
    $endgroup$
    – Kevin
    23 hours ago












  • $begingroup$
    If you go with drawing from a bag, I would recommend physical items. Differently-coloured tiddlywinks (or dice of the same size) if exact fairness is more important than immersion; cheap unpainted miniatures if the reverse.
    $endgroup$
    – TimLymington
    18 hours ago










  • $begingroup$
    Yeah, I think this is best. It's not perfect, but this works well, is fair, and rarely makes arguments. Maybe also mention apps or websites as an alternative, if "just getting a number" outweighs "being able to see how" for some people.
    $endgroup$
    – Jack V.
    14 hours ago














35












35








35





$begingroup$

Dice work just fine for the most common numbers of players



I work with probabilities a lot. Dice allow you to represent many common probabilities, but when you want to roll to pick from options, I've discovered the following "law": it stops making sense to people when you have to roll more than a single die at once. Luckily, the most common numbers of players can be chosen from using just a single die roll, or a single die roll that occasionally gets re-rolled when its result is invalid.



If you have a single player, no need to roll.



If you have three players, a d6 will do --- one player gets results 1-2, another 3-4 and the third one 5-6.



Five players, roll a d10. One player gets 1-2, the next 3-4, then 5-6, then 7-8, and the final one 9-10.



There is just a bit of math behind these two options: they work neatly because three and five are factors of six and ten, respectively. That means we can "partition" each dice's possible outcomes into that many equally probable sets, similar to how we can use "odd" and "even" to select from two options using a d6.



Seven players is the first tricky one because it's the least prime number that's not a factor of any of the common die sizes of 4, 6, 8, 10, 12 or 20. This means you cannot resolve a choice between seven players with any bounded number of these die rolls with complete fairness. You can, however, use a d8 with numbers 1-7 corresponding to each of the players and 8 indicating a reroll. The chances of having to roll the die impractically many times for a single decision are minimal.



Nine players or eleven players, similar to seven except use a d10 or d12, respectively. Since nine is divisible by three, you can also use two rolls of a d6: first divide the player into three groups of three, let the first roll decide which group you pick and the second roll which particular player from that group. The d10 method is simpler, though.



You can of course use arbitrarily large dice to cover any amount of players using the "reroll if too high" strategy, but it can get unwieldy for larger numbers.



Drawing from a deck of cards works for many sizes of groups



Using a deck of cards provides a simple solution for groups small and large, without having to remember any complex roll rules. Have each player draw a card from a standard deck of 52 cards. The player who gets the highest card (or lowest, if you prefer) is chosen. Remember to have a set convention on the ordering used (aces high or low and precedence of suits).



There's 52 cards in a normal French-style deck of cards, and each is unique, so you can choose a random player with everyone drawing a single card for groups up to 52 players. That's probably more than enough.



Draw from a bag



Yet another classic method: have each player write their name on a slip of paper. Put them in a box or bag, and resolve the decision by drawing one without looking. My experience is that this is quite awkward especially when table space is limited --- I appreciate the other methods' compactness. However, this method works quite well regardless of the size of the game, assuming you have a big enough box for the name slips and shuffle it well.



Random number list, if you want to keep it secret



If you know the amount of players you'll have in advance, you can create a list of random numbers and bring that to the game with you. When consulting the list, cross over the topmost number on it and select the player based on that number.



This solution is, for most games, over-engineered, but it has an advantage in that unlike die rolls and draws from the deck of cards, it is very subtle and allows invoking the randomness without signaling it to your players. My experience has been, though, that it's best to let the players know when you're actually choosing one of them at random in most cases --- otherwise, they can assume the choice is you being arbitrary.






share|improve this answer











$endgroup$



Dice work just fine for the most common numbers of players



I work with probabilities a lot. Dice allow you to represent many common probabilities, but when you want to roll to pick from options, I've discovered the following "law": it stops making sense to people when you have to roll more than a single die at once. Luckily, the most common numbers of players can be chosen from using just a single die roll, or a single die roll that occasionally gets re-rolled when its result is invalid.



If you have a single player, no need to roll.



If you have three players, a d6 will do --- one player gets results 1-2, another 3-4 and the third one 5-6.



Five players, roll a d10. One player gets 1-2, the next 3-4, then 5-6, then 7-8, and the final one 9-10.



There is just a bit of math behind these two options: they work neatly because three and five are factors of six and ten, respectively. That means we can "partition" each dice's possible outcomes into that many equally probable sets, similar to how we can use "odd" and "even" to select from two options using a d6.



Seven players is the first tricky one because it's the least prime number that's not a factor of any of the common die sizes of 4, 6, 8, 10, 12 or 20. This means you cannot resolve a choice between seven players with any bounded number of these die rolls with complete fairness. You can, however, use a d8 with numbers 1-7 corresponding to each of the players and 8 indicating a reroll. The chances of having to roll the die impractically many times for a single decision are minimal.



Nine players or eleven players, similar to seven except use a d10 or d12, respectively. Since nine is divisible by three, you can also use two rolls of a d6: first divide the player into three groups of three, let the first roll decide which group you pick and the second roll which particular player from that group. The d10 method is simpler, though.



You can of course use arbitrarily large dice to cover any amount of players using the "reroll if too high" strategy, but it can get unwieldy for larger numbers.



Drawing from a deck of cards works for many sizes of groups



Using a deck of cards provides a simple solution for groups small and large, without having to remember any complex roll rules. Have each player draw a card from a standard deck of 52 cards. The player who gets the highest card (or lowest, if you prefer) is chosen. Remember to have a set convention on the ordering used (aces high or low and precedence of suits).



There's 52 cards in a normal French-style deck of cards, and each is unique, so you can choose a random player with everyone drawing a single card for groups up to 52 players. That's probably more than enough.



Draw from a bag



Yet another classic method: have each player write their name on a slip of paper. Put them in a box or bag, and resolve the decision by drawing one without looking. My experience is that this is quite awkward especially when table space is limited --- I appreciate the other methods' compactness. However, this method works quite well regardless of the size of the game, assuming you have a big enough box for the name slips and shuffle it well.



Random number list, if you want to keep it secret



If you know the amount of players you'll have in advance, you can create a list of random numbers and bring that to the game with you. When consulting the list, cross over the topmost number on it and select the player based on that number.



This solution is, for most games, over-engineered, but it has an advantage in that unlike die rolls and draws from the deck of cards, it is very subtle and allows invoking the randomness without signaling it to your players. My experience has been, though, that it's best to let the players know when you're actually choosing one of them at random in most cases --- otherwise, they can assume the choice is you being arbitrary.







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday

























answered yesterday









kviirikviiri

38.3k13142216




38.3k13142216












  • $begingroup$
    Re dice: If you have four or fewer players, I suppose you could use these. Just be very sure you don't accidentally mix them in with the regular d12s, or the DM will not be happy with you ("What do you mean, you rolled a 47 on 1d12?").
    $endgroup$
    – Kevin
    23 hours ago












  • $begingroup$
    If you go with drawing from a bag, I would recommend physical items. Differently-coloured tiddlywinks (or dice of the same size) if exact fairness is more important than immersion; cheap unpainted miniatures if the reverse.
    $endgroup$
    – TimLymington
    18 hours ago










  • $begingroup$
    Yeah, I think this is best. It's not perfect, but this works well, is fair, and rarely makes arguments. Maybe also mention apps or websites as an alternative, if "just getting a number" outweighs "being able to see how" for some people.
    $endgroup$
    – Jack V.
    14 hours ago


















  • $begingroup$
    Re dice: If you have four or fewer players, I suppose you could use these. Just be very sure you don't accidentally mix them in with the regular d12s, or the DM will not be happy with you ("What do you mean, you rolled a 47 on 1d12?").
    $endgroup$
    – Kevin
    23 hours ago












  • $begingroup$
    If you go with drawing from a bag, I would recommend physical items. Differently-coloured tiddlywinks (or dice of the same size) if exact fairness is more important than immersion; cheap unpainted miniatures if the reverse.
    $endgroup$
    – TimLymington
    18 hours ago










  • $begingroup$
    Yeah, I think this is best. It's not perfect, but this works well, is fair, and rarely makes arguments. Maybe also mention apps or websites as an alternative, if "just getting a number" outweighs "being able to see how" for some people.
    $endgroup$
    – Jack V.
    14 hours ago
















$begingroup$
Re dice: If you have four or fewer players, I suppose you could use these. Just be very sure you don't accidentally mix them in with the regular d12s, or the DM will not be happy with you ("What do you mean, you rolled a 47 on 1d12?").
$endgroup$
– Kevin
23 hours ago






$begingroup$
Re dice: If you have four or fewer players, I suppose you could use these. Just be very sure you don't accidentally mix them in with the regular d12s, or the DM will not be happy with you ("What do you mean, you rolled a 47 on 1d12?").
$endgroup$
– Kevin
23 hours ago














$begingroup$
If you go with drawing from a bag, I would recommend physical items. Differently-coloured tiddlywinks (or dice of the same size) if exact fairness is more important than immersion; cheap unpainted miniatures if the reverse.
$endgroup$
– TimLymington
18 hours ago




$begingroup$
If you go with drawing from a bag, I would recommend physical items. Differently-coloured tiddlywinks (or dice of the same size) if exact fairness is more important than immersion; cheap unpainted miniatures if the reverse.
$endgroup$
– TimLymington
18 hours ago












$begingroup$
Yeah, I think this is best. It's not perfect, but this works well, is fair, and rarely makes arguments. Maybe also mention apps or websites as an alternative, if "just getting a number" outweighs "being able to see how" for some people.
$endgroup$
– Jack V.
14 hours ago




$begingroup$
Yeah, I think this is best. It's not perfect, but this works well, is fair, and rarely makes arguments. Maybe also mention apps or websites as an alternative, if "just getting a number" outweighs "being able to see how" for some people.
$endgroup$
– Jack V.
14 hours ago













23












$begingroup$

If you have 5 players, roll a d6. If you get a 6, re-roll until you get a number other than 6. And so on, for other odd numbers. Yes, this is completely fair, if the die you're rolling is a fair die.






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    You can also form5 players roll a d10 assigning 1-2 to one player, 3-4 to another and so on until all five have a number range that's theirs.
    $endgroup$
    – Sarriesfan
    yesterday










  • $begingroup$
    This is actually much simpler than the accepted answer, and works with any number of player unlike that answer.
    $endgroup$
    – John
    1 hour ago


















23












$begingroup$

If you have 5 players, roll a d6. If you get a 6, re-roll until you get a number other than 6. And so on, for other odd numbers. Yes, this is completely fair, if the die you're rolling is a fair die.






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    You can also form5 players roll a d10 assigning 1-2 to one player, 3-4 to another and so on until all five have a number range that's theirs.
    $endgroup$
    – Sarriesfan
    yesterday










  • $begingroup$
    This is actually much simpler than the accepted answer, and works with any number of player unlike that answer.
    $endgroup$
    – John
    1 hour ago
















23












23








23





$begingroup$

If you have 5 players, roll a d6. If you get a 6, re-roll until you get a number other than 6. And so on, for other odd numbers. Yes, this is completely fair, if the die you're rolling is a fair die.






share|improve this answer









$endgroup$



If you have 5 players, roll a d6. If you get a 6, re-roll until you get a number other than 6. And so on, for other odd numbers. Yes, this is completely fair, if the die you're rolling is a fair die.







share|improve this answer












share|improve this answer



share|improve this answer










answered yesterday









John DallmanJohn Dallman

11.7k23263




11.7k23263








  • 1




    $begingroup$
    You can also form5 players roll a d10 assigning 1-2 to one player, 3-4 to another and so on until all five have a number range that's theirs.
    $endgroup$
    – Sarriesfan
    yesterday










  • $begingroup$
    This is actually much simpler than the accepted answer, and works with any number of player unlike that answer.
    $endgroup$
    – John
    1 hour ago
















  • 1




    $begingroup$
    You can also form5 players roll a d10 assigning 1-2 to one player, 3-4 to another and so on until all five have a number range that's theirs.
    $endgroup$
    – Sarriesfan
    yesterday










  • $begingroup$
    This is actually much simpler than the accepted answer, and works with any number of player unlike that answer.
    $endgroup$
    – John
    1 hour ago










1




1




$begingroup$
You can also form5 players roll a d10 assigning 1-2 to one player, 3-4 to another and so on until all five have a number range that's theirs.
$endgroup$
– Sarriesfan
yesterday




$begingroup$
You can also form5 players roll a d10 assigning 1-2 to one player, 3-4 to another and so on until all five have a number range that's theirs.
$endgroup$
– Sarriesfan
yesterday












$begingroup$
This is actually much simpler than the accepted answer, and works with any number of player unlike that answer.
$endgroup$
– John
1 hour ago






$begingroup$
This is actually much simpler than the accepted answer, and works with any number of player unlike that answer.
$endgroup$
– John
1 hour ago













8












$begingroup$

How I have done this over the years:



It makes for even probabilities.




  1. For three players: roll 1d6, divide result by 2, round up.


  2. For five players: roll 1d10, divide result by 2, round up


  3. For seven players: roll 1d8, re roll any 8.



  4. For nine players: Roll 2d6 of different colors. (Let's say red and
    green, since that is what my bag is filled with)




    • Green die indicates which range of three: 1 or 2 =1-3, 3 or 4 = 4-6, 5 or 6 = 7-9.

    • You now have a "three players" situation

    • Red die indicates who in that group is chosen by using the "for three players" above.



  5. For 11 players; roll a d12, re roll 12's.



It's really fast and intuitive once you start doing this (it's easy to do the comparison in your head). It takes longer to describe it than to do it.



Notes:




  1. While I don't do this, @BenR suggests out that for 9 you could do it
    as you do for 7; roll a d10 and re roll 0's/10's. Since I prefer to
    roll just once; the 2d6 is one roll and the result is right
    there).



  2. While I have not done this, @Yakk suggests this for a d13 if you have
    13 players.




    • For 1d13, roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value.








share|improve this answer











$endgroup$









  • 3




    $begingroup$
    Why is the nine players one so complex? Just roll a d10 and reroll 10s.
    $endgroup$
    – Ben R.
    15 hours ago










  • $begingroup$
    @BenR. It isn't complex; you roll two dice once, and you know who it is by looking at the two dice. As I said, it is far easier to do than to explain. I've been doing stuff like that for a long time. it's the 'for three people" with two levels.
    $endgroup$
    – KorvinStarmast
    13 hours ago












  • $begingroup$
    What, no 13 players?
    $endgroup$
    – Yakk
    12 hours ago










  • $begingroup$
    @Yakk No, haven't run a game that big in over 30 years. On that one, use d14 and reroll 14's. ;) What's a d14? Those two six sided dice (red and green) and a d8. (There were some 1 to 14 tables in AD&D 1e DMG, page 26, gems, random chance). Roll the two d6 and the d8. Low green? Use the d8. High green? Add 8 to the d6. Easy to do, one roll.
    $endgroup$
    – KorvinStarmast
    12 hours ago












  • $begingroup$
    @KorvinStarmast That d14 isn't an even distribution; low numbers have 1/16 chance, high numbers 1/12 chance each. Try this: roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value. Or, just roll 1d8 reroll 8s; then use high-low d2 to maybe add 7. Then reroll 14s.
    $endgroup$
    – Yakk
    12 hours ago


















8












$begingroup$

How I have done this over the years:



It makes for even probabilities.




  1. For three players: roll 1d6, divide result by 2, round up.


  2. For five players: roll 1d10, divide result by 2, round up


  3. For seven players: roll 1d8, re roll any 8.



  4. For nine players: Roll 2d6 of different colors. (Let's say red and
    green, since that is what my bag is filled with)




    • Green die indicates which range of three: 1 or 2 =1-3, 3 or 4 = 4-6, 5 or 6 = 7-9.

    • You now have a "three players" situation

    • Red die indicates who in that group is chosen by using the "for three players" above.



  5. For 11 players; roll a d12, re roll 12's.



It's really fast and intuitive once you start doing this (it's easy to do the comparison in your head). It takes longer to describe it than to do it.



Notes:




  1. While I don't do this, @BenR suggests out that for 9 you could do it
    as you do for 7; roll a d10 and re roll 0's/10's. Since I prefer to
    roll just once; the 2d6 is one roll and the result is right
    there).



  2. While I have not done this, @Yakk suggests this for a d13 if you have
    13 players.




    • For 1d13, roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value.








share|improve this answer











$endgroup$









  • 3




    $begingroup$
    Why is the nine players one so complex? Just roll a d10 and reroll 10s.
    $endgroup$
    – Ben R.
    15 hours ago










  • $begingroup$
    @BenR. It isn't complex; you roll two dice once, and you know who it is by looking at the two dice. As I said, it is far easier to do than to explain. I've been doing stuff like that for a long time. it's the 'for three people" with two levels.
    $endgroup$
    – KorvinStarmast
    13 hours ago












  • $begingroup$
    What, no 13 players?
    $endgroup$
    – Yakk
    12 hours ago










  • $begingroup$
    @Yakk No, haven't run a game that big in over 30 years. On that one, use d14 and reroll 14's. ;) What's a d14? Those two six sided dice (red and green) and a d8. (There were some 1 to 14 tables in AD&D 1e DMG, page 26, gems, random chance). Roll the two d6 and the d8. Low green? Use the d8. High green? Add 8 to the d6. Easy to do, one roll.
    $endgroup$
    – KorvinStarmast
    12 hours ago












  • $begingroup$
    @KorvinStarmast That d14 isn't an even distribution; low numbers have 1/16 chance, high numbers 1/12 chance each. Try this: roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value. Or, just roll 1d8 reroll 8s; then use high-low d2 to maybe add 7. Then reroll 14s.
    $endgroup$
    – Yakk
    12 hours ago
















8












8








8





$begingroup$

How I have done this over the years:



It makes for even probabilities.




  1. For three players: roll 1d6, divide result by 2, round up.


  2. For five players: roll 1d10, divide result by 2, round up


  3. For seven players: roll 1d8, re roll any 8.



  4. For nine players: Roll 2d6 of different colors. (Let's say red and
    green, since that is what my bag is filled with)




    • Green die indicates which range of three: 1 or 2 =1-3, 3 or 4 = 4-6, 5 or 6 = 7-9.

    • You now have a "three players" situation

    • Red die indicates who in that group is chosen by using the "for three players" above.



  5. For 11 players; roll a d12, re roll 12's.



It's really fast and intuitive once you start doing this (it's easy to do the comparison in your head). It takes longer to describe it than to do it.



Notes:




  1. While I don't do this, @BenR suggests out that for 9 you could do it
    as you do for 7; roll a d10 and re roll 0's/10's. Since I prefer to
    roll just once; the 2d6 is one roll and the result is right
    there).



  2. While I have not done this, @Yakk suggests this for a d13 if you have
    13 players.




    • For 1d13, roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value.








share|improve this answer











$endgroup$



How I have done this over the years:



It makes for even probabilities.




  1. For three players: roll 1d6, divide result by 2, round up.


  2. For five players: roll 1d10, divide result by 2, round up


  3. For seven players: roll 1d8, re roll any 8.



  4. For nine players: Roll 2d6 of different colors. (Let's say red and
    green, since that is what my bag is filled with)




    • Green die indicates which range of three: 1 or 2 =1-3, 3 or 4 = 4-6, 5 or 6 = 7-9.

    • You now have a "three players" situation

    • Red die indicates who in that group is chosen by using the "for three players" above.



  5. For 11 players; roll a d12, re roll 12's.



It's really fast and intuitive once you start doing this (it's easy to do the comparison in your head). It takes longer to describe it than to do it.



Notes:




  1. While I don't do this, @BenR suggests out that for 9 you could do it
    as you do for 7; roll a d10 and re roll 0's/10's. Since I prefer to
    roll just once; the 2d6 is one roll and the result is right
    there).



  2. While I have not done this, @Yakk suggests this for a d13 if you have
    13 players.




    • For 1d13, roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value.









share|improve this answer














share|improve this answer



share|improve this answer








edited 11 hours ago

























answered yesterday









KorvinStarmastKorvinStarmast

83.5k20258450




83.5k20258450








  • 3




    $begingroup$
    Why is the nine players one so complex? Just roll a d10 and reroll 10s.
    $endgroup$
    – Ben R.
    15 hours ago










  • $begingroup$
    @BenR. It isn't complex; you roll two dice once, and you know who it is by looking at the two dice. As I said, it is far easier to do than to explain. I've been doing stuff like that for a long time. it's the 'for three people" with two levels.
    $endgroup$
    – KorvinStarmast
    13 hours ago












  • $begingroup$
    What, no 13 players?
    $endgroup$
    – Yakk
    12 hours ago










  • $begingroup$
    @Yakk No, haven't run a game that big in over 30 years. On that one, use d14 and reroll 14's. ;) What's a d14? Those two six sided dice (red and green) and a d8. (There were some 1 to 14 tables in AD&D 1e DMG, page 26, gems, random chance). Roll the two d6 and the d8. Low green? Use the d8. High green? Add 8 to the d6. Easy to do, one roll.
    $endgroup$
    – KorvinStarmast
    12 hours ago












  • $begingroup$
    @KorvinStarmast That d14 isn't an even distribution; low numbers have 1/16 chance, high numbers 1/12 chance each. Try this: roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value. Or, just roll 1d8 reroll 8s; then use high-low d2 to maybe add 7. Then reroll 14s.
    $endgroup$
    – Yakk
    12 hours ago
















  • 3




    $begingroup$
    Why is the nine players one so complex? Just roll a d10 and reroll 10s.
    $endgroup$
    – Ben R.
    15 hours ago










  • $begingroup$
    @BenR. It isn't complex; you roll two dice once, and you know who it is by looking at the two dice. As I said, it is far easier to do than to explain. I've been doing stuff like that for a long time. it's the 'for three people" with two levels.
    $endgroup$
    – KorvinStarmast
    13 hours ago












  • $begingroup$
    What, no 13 players?
    $endgroup$
    – Yakk
    12 hours ago










  • $begingroup$
    @Yakk No, haven't run a game that big in over 30 years. On that one, use d14 and reroll 14's. ;) What's a d14? Those two six sided dice (red and green) and a d8. (There were some 1 to 14 tables in AD&D 1e DMG, page 26, gems, random chance). Roll the two d6 and the d8. Low green? Use the d8. High green? Add 8 to the d6. Easy to do, one roll.
    $endgroup$
    – KorvinStarmast
    12 hours ago












  • $begingroup$
    @KorvinStarmast That d14 isn't an even distribution; low numbers have 1/16 chance, high numbers 1/12 chance each. Try this: roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value. Or, just roll 1d8 reroll 8s; then use high-low d2 to maybe add 7. Then reroll 14s.
    $endgroup$
    – Yakk
    12 hours ago










3




3




$begingroup$
Why is the nine players one so complex? Just roll a d10 and reroll 10s.
$endgroup$
– Ben R.
15 hours ago




$begingroup$
Why is the nine players one so complex? Just roll a d10 and reroll 10s.
$endgroup$
– Ben R.
15 hours ago












$begingroup$
@BenR. It isn't complex; you roll two dice once, and you know who it is by looking at the two dice. As I said, it is far easier to do than to explain. I've been doing stuff like that for a long time. it's the 'for three people" with two levels.
$endgroup$
– KorvinStarmast
13 hours ago






$begingroup$
@BenR. It isn't complex; you roll two dice once, and you know who it is by looking at the two dice. As I said, it is far easier to do than to explain. I've been doing stuff like that for a long time. it's the 'for three people" with two levels.
$endgroup$
– KorvinStarmast
13 hours ago














$begingroup$
What, no 13 players?
$endgroup$
– Yakk
12 hours ago




$begingroup$
What, no 13 players?
$endgroup$
– Yakk
12 hours ago












$begingroup$
@Yakk No, haven't run a game that big in over 30 years. On that one, use d14 and reroll 14's. ;) What's a d14? Those two six sided dice (red and green) and a d8. (There were some 1 to 14 tables in AD&D 1e DMG, page 26, gems, random chance). Roll the two d6 and the d8. Low green? Use the d8. High green? Add 8 to the d6. Easy to do, one roll.
$endgroup$
– KorvinStarmast
12 hours ago






$begingroup$
@Yakk No, haven't run a game that big in over 30 years. On that one, use d14 and reroll 14's. ;) What's a d14? Those two six sided dice (red and green) and a d8. (There were some 1 to 14 tables in AD&D 1e DMG, page 26, gems, random chance). Roll the two d6 and the d8. Low green? Use the d8. High green? Add 8 to the d6. Easy to do, one roll.
$endgroup$
– KorvinStarmast
12 hours ago














$begingroup$
@KorvinStarmast That d14 isn't an even distribution; low numbers have 1/16 chance, high numbers 1/12 chance each. Try this: roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value. Or, just roll 1d8 reroll 8s; then use high-low d2 to maybe add 7. Then reroll 14s.
$endgroup$
– Yakk
12 hours ago






$begingroup$
@KorvinStarmast That d14 isn't an even distribution; low numbers have 1/16 chance, high numbers 1/12 chance each. Try this: roll a red, green and blue d12. If all 3 match, use the value. If red and green match, but not blue, use 13. Otherwise, use red. Accurate within a fraction of a percent to 1/13 for each value. Or, just roll 1d8 reroll 8s; then use high-low d2 to maybe add 7. Then reroll 14s.
$endgroup$
– Yakk
12 hours ago




















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