The official 53 drinking game

http://www.fiftythree.org
Should every drinking game have its own website, or should every website have its own drinking game?

Authors:

One, Zero, PKH, and bluto.
HTML Markup by: Lloyd Worthnone, drunkard

The Rules:

You roll 4 dice.
You have to achieve 53 out of them or you must take a drink and roll again to make 53.
Each of the dice can only be used once for Valueness.
Each of the dice can only be used once for Pairness/Tripleness/Quadness.
You don't have to use all of the dice, but you have to take one drink for each of the remaining unused dice.
When you make 53, you do not have to drink, and dice are given to the next player to make 53.
If you roll a 5 and a 3, this is called 53 the easy way. It is a social. You still must drink for the remainder, if any.
If you roll 5 5 3 3, this is a waterfall.
There is a one drink penalty for using one pairness, however Tripleness and Quadness is free.
If a player makes 53 the hard way, everyone one else must drink

Valueness and pairness:

These 4 dice have 2 different properties. One which is the most familliar is to use the number on top of the dice as they are rolled. This is called the Valueness property. The 2nd way dice can be used is through multiple occurances of a particular number such as a pair or triple of a number. You can use this property called the pairness concurrently with valueness to reach 53. Pairness of 2 dice is '2'. Pairness (or tripleness) of 3 dice is 3. Beginning players often make the mistake of using "one pair" or '1' when the value of a pair is in fact '2'. Remember the penalty for using pairness, however Tripleness or Quadness is free.

How to reach 53:

To get 53 out of 4 dice, you have 2 options. One is to add, subtract, multiply, and divide the valueness and pairness of the dice to reach the number 53. This is called 53 the hard way. The other and more commonly used option is to add, subtract, multiply, and divide the valueness and pairness of the dice to achieve and individual 5 and an individual 3 while still only using each of the dice in accordance with rules #3 and #4.

Examples:

The authors of 53 realize this may not be the easiest game to grasp offhand, and so many examples are included.

First off, it's a social since there is a '5' and a '3' rolled. As it stands, the player must take 2 additional drinks because of the leftover dice. But the more intellegent player would notice the pairness of the 5's (which then gives us a '2') and divide it by the remaining '2' to give us a '1' that we can multiply against 53. So the player only has to take one additional drink because of the penalty for pairness. But the even more experienced player would add the 5's together to reach 10, and divide by 2 to get 5. Then with the '3' we have 53 without any penalty drink since all of the dice were used without pairness.

Solution: 2 + 3 is 5, and use the other '3' for 53. Take a penalty drink because of the unused '6'. Better solution: '6' minus a '3' gives us 3. '2' plus the other '3' is the 5.
No easy solution. Try your luck by taking a drink and rolling again. The authors note that the solution is: 2 / (pairness of 2) = 1. 1 + 4 =5. 2 / (pairness of 4) = 1. 4 - 1 = 3. You would have to do 2 drinks because of pairness penaltys.

Social. ((4 + 6) * 5) + 3 = 53 53 the hard way
1 + 4 = 5. 6 / 2 = 3.
6 / (Pairness of the fours) = 3. And we're given a 5. That would leave you with a one drink penalty. But why not do this: ((4 + 4) * 6) + 5 = 53. 53 the hard way

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	First off, it's a social since there is a '5' and a '3' rolled. As it stands, the player must take 2 additional drinks because of the leftover dice. But the more intellegent player would notice the pairness of the 5's (which then gives us a '2') and divide it by the remaining '2' to give us a '1' that we can multiply against 53. So the player only has to take one additional drink because of the penalty for pairness. But the even more experienced player would add the 5's together to reach 10, and divide by 2 to get 5. Then with the '3' we have 53 without any penalty drink since all of the dice were used without pairness.
	Solution: 2 + 3 is 5, and use the other '3' for 53. Take a penalty drink because of the unused '6'. Better solution: '6' minus a '3' gives us 3. '2' plus the other '3' is the 5.
	No easy solution. Try your luck by taking a drink and rolling again. The authors note that the solution is: 2 / (pairness of 2) = 1. 1 + 4 =5. 2 / (pairness of 4) = 1. 4 - 1 = 3. You would have to do 2 drinks because of pairness penaltys.
	Social. ((4 + 6) * 5) + 3 = 53 53 the hard way
	1 + 4 = 5. 6 / 2 = 3.
	6 / (Pairness of the fours) = 3. And we're given a 5. That would leave you with a one drink penalty. But why not do this: ((4 + 4) * 6) + 5 = 53. 53 the hard way