Skip to Content

Is this set-up too complicated?

10 replies [Last post]
farquhj
Offline
Joined: 12/31/1969

My initial game set-up: All players must NOT begin by holding a matching pair of cards where there are as many initial pairs possible as there are players.

For example, in a four-player game there are 4 pairs in a total of 8 cards such as 2 jacks, 2 queens, 2 kings, and 2 aces. Initially, all 8 cards are dealt to the four players. I need to begin the game where NO player has two-of-a-kind. In a four-player game, I believe the chances of this happening randomly are 1 in 4. That is... 3*2*1 or six valid combinations (where no pairs exist) in a total of 4*3*2*1 or 24 total combinations. For five players, the chances are one in five. For six players, one in six, etc.

The only mechanism that I can think of to appropriately distribute these eight cards unknowingly to all players and to avoid any pairs, is to deal the cards, then shuffle and re-deal them if a pair is formed. In a typical four-player game, this would often require re-dealing these cards 4 or more times before any play begins.

My questions are: Have I calculated the odds correctly? Is there a better mechanism for distributing these cards appropriately? And, if not, is this set-up mechanism overly complicated?

Thank you in advance for your thoughts.

jf

seo
seo's picture
Offline
Joined: 07/21/2008
Is this set-up too complicated?

Your math is right, AFAIK. In a deck made of pairs, you'll have 1/n chances of dealing a pair to at least one player, n being the number of pairs in the deck (which in your game is also the number of players).

I can't figure a simple method to blindly deal two non-paired cards to each player. But I think I have an improvement over your deal-again method.

If one or more players have a pair, each player discards just one card, and keeps the other. Discarded cards are shuffled and dealt again. This process is repeated until no player has any pairs.

Now, if my maths are correct, by doing this you're going to have less chances of dealing a pair than reshuffling all the cards. In the worst scenario, each player will keep a different card, hence the chances of a pair will remain the same as with your method. But some other times two players will each keep half a pair, thus reducing the probability of a pair being dealt.

Seo

farquhj
Offline
Joined: 12/31/1969
Thanks

seo -

Thanks for your reply.

There is an additional requirement that I did not mention in the first post, so let me elaborate. The game concept has one discrete set of cards dealt, then the second set. The first four cards are individual character cards. The second set of cards represent a relationship that the character has to another character. I need to avoid the situation where someone is assigned as having a relationship to oneself.

So, I believe I am stuck with the situation of having to re-deal the second set of four cards until no one has a relationship card matching their character card.

Thanks.

jf

seo
seo's picture
Offline
Joined: 07/21/2008
Is this set-up too complicated?

Yep, I'm affraid you're stuck with my worst case scenario. :-(

Maybe you can modify your relationship cards to include two names: if the first name is your character, you ignore it and use the second name. If the first name is not your character, you ignore the second name.

This will eliminate all the need for reshuffling, but will result in two characters being connected to a third. I don't know if this is posible in your game.

Or even more simple: add a rule "if the relationship card has your character name, ignore it and apply to the player on your right". That will only work if players characters are not kept secret.

Seo

Zomulgustar
Offline
Joined: 07/31/2008
Is this set-up too complicated?

If this isn't too complicated, one way to avoid reshuffling would be to construct a set of 'relationship' cards whose value was dependent on the character card held, i.e. a sample relationship card would say (presumably in a compact symbolic form) "If you are A, you have a relationship with B. If you are B, you have a relationship with C. If you are C, you have a relationship with A." You'd need to set them up just so, though, to avoid two relationships with the same character (if this is even possible). We might be able to help further with a better idea of the intended purpose behind this setup....

Hamumu
Offline
Joined: 12/31/1969
Is this set-up too complicated?

It seems a simpler and faster (but still a little awkward) way of doing this would be: Everybody gets their character. Then the relationships are scattered in the table middle face down. Then go around the table like so:

1 - Take a card and look at it.
2 - If it's yourself, take a different card, then put yours back in and rescatter the cards.
3 - Next player!

As I understand it, you'll only have as many cards as there are players, so shuffling would be a pain anyway with so few cards. Scattering is really the easier method. Rescattering is quick and easy, and while you may have to do it as much as once per person (well, once for everybody but the last person!), you are guaranteed to never have to do more than 1 per person.

EDIT: Oops, I see the flaw in my plan. It's okay for the first person, but each later person who draws themselves later on gets information about who has them. And kind of vice-versa too, when you see someone have to return a card, you know they drew themselves which is all kinds of information.

farquhj
Offline
Joined: 12/31/1969
Is this set-up too complicated?

Zomulgustar wrote:
If this isn't too complicated, one way to avoid reshuffling would be to construct a set of 'relationship' cards whose value was dependent on the character card held, i.e. a sample relationship card would say (presumably in a compact symbolic form) "If you are A, you have a relationship with B. If you are B, you have a relationship with C. If you are C, you have a relationship with A." You'd need to set them up just so, though, to avoid two relationships with the same character (if this is even possible). We might be able to help further with a better idea of the intended purpose behind this setup....

This is a very interesting idea that I will consider. Seems to add a different level of complexity -- that of appropriately interpreting your relationship card.

While I appreciate the suggestions for avoiding the regretable "re-dealing until conditions are met", no one has yet commented on whether they thought this limitation would be acceptable. Do I work hard to avoid it?

Thanks,

jf

sedjtroll
sedjtroll's picture
Offline
Joined: 07/21/2008
Is this set-up too complicated?

Can you pre-deal one of each card, then offset and deal one of each card again to create the pairs, then randomly distribute the pairs?

I.e.:

deal J, Q, K, A face down.
then on top of that deal K, A, J, Q face down.
Randomly distibute the piles to players 1, 2, 3, and 4.

Or if you like, set the pairs together, 'shuffle' the pairs so they're still pairs but in a random or unknown order, then deal them out and offset the top row by moving them over once, then distribute the new 'sets of 2 cards which are no longer pairs'.

- Seth

seo
seo's picture
Offline
Joined: 07/21/2008
Is this set-up too complicated?

I'll place it in the level of "if you can't find an elegant alternative, use the redealing method".

It might be a bit annoying, but it all depends on how the rest of the game is. If, after the dealing mess, you get to play an enjoyable game for an hour and a half, spending a couple minutes dealing the cards is not a huge problem.

OTOH, if the game only lasts a couple more minutes, then by all means you should try to find a better method.

Finally, if your game is long and boring, nobody will care about the dealing of the cards. Concentrate in making the game better. ;-)

Seo

seo
seo's picture
Offline
Joined: 07/21/2008
Is this set-up too complicated?

sedjtroll wrote:
Can you pre-deal one of each card, then offset and deal one of each card again to create the pairs, then randomly distribute the pairs?

I.e.:

deal J, Q, K, A face down.
then on top of that deal K, A, J, Q face down.
Randomly distibute the piles to players 1, 2, 3, and 4.

Or if you like, set the pairs together, 'shuffle' the pairs so they're still pairs but in a random or unknown order, then deal them out and offset the top row by moving them over once, then distribute the new 'sets of 2 cards which are no longer pairs'.

- Seth

But that would mean, unless I'm much mistaken, that everybody will be able to deduce what the other pairs are. If the character for each player is made public, you'll automatically have knowledge about their relationships too. If the players characters remain hidden, you'll still know what character is related to whom, though not what player has each pair. I'm not sure about how this will apply to your game in particular.

Seo

farquhj
Offline
Joined: 12/31/1969
What are the odds?

farquhj wrote:
My initial game set-up: All players must NOT begin by holding a matching pair of cards where there are as many initial pairs possible as there are players.

For example, in a four-player game there are 4 pairs in a total of 8 cards such as 2 jacks, 2 queens, 2 kings, and 2 aces. Initially, all 8 cards are dealt to the four players. I need to begin the game where NO player has two-of-a-kind. In a four-player game, I believe the chances of this happening randomly are 1 in 4. That is... 3*2*1 or six valid combinations (where no pairs exist) in a total of 4*3*2*1 or 24 total combinations. For five players, the chances are one in five. For six players, one in six, etc.

I am considering a modification to this arrangement, but don't know if I am calculating the odds correctly.

Let's say that in a four-player game, four character cards are shuffled and dealt out (ex: Ace, King, Queen, Jack). Then, eight more cards (two of each kind) are shuffled and dealt out (ex: 2 Aces, 2 Kings, 2 Queens, and 2 Jacks). What are the odds of any single player ending up with three-of-a-kind?

In my game, the three-of-a-kind set-up is an illegal formation and would require that the 8 cards be dealt again. So, I want this scenario to be infrequent. I believe the odds are one in sixteen. Am I correct?

Thanks,

jf

Syndicate content


forum | by Dr. Radut