How to use D6s to select from 4 choices

[TheDS]

In various games I play, there are often 4 players, and some method must be used to choose a player (such as, for who goes first). With 6 players, it's a simple matter to roll a D6 to choose. Same for 3 or 2. For 5, it's just like for 6, except you reroll a 6; as this happens only one time in six, it's not a big deal. But when you have to choose between 4 things, rolling a die and rerolling on five or six has a 1 in 3 chance of needing a reroll, which can happen multiple times in a row. So I came up with a different way to do it: use 2 dice.

There are 36 possible combinations of 2D6, which is evenly divisible by 4, so that no rerolls are necessary. But how to divide them up in a way that you can easily remember, and doesn't require the hassle of remembering which die is the primary die (such as by doing two hi-low rolls)?

Once I figured it out, a memory aid suggested itself: 6, 5, 4, 3. You can assign those numbers to choices the same way you'd assign 1, 2, 3, 4.

The 6 group of results is "a 6 without a 5". If either die is a 6, and the other is NOT a 5, that's choice 6.

The 5 group of results is "a 5 without a 6". If either die is a 5, and the other die is NOT a 6, that's choice 5.

Pretty simple so far, but now we have to be a little more sophisticated.

The 4 group of results is "a 5 AND 6, OR a 4 without a 5 or 6". Above, we excluded the specific combination of 5, 6, but now we need to count it to fill in for the 4s. Otherwise it works just like our previous choice: either die can be the number we care about (4), and the other cannot be a higher number.

The 3 group of results is "nothing higher than 3". This is what's left over.

Now you too can amaze your friends by making them think you've memorized a complicated table of results!

 

[whartung]

Roll 2 different color dice.

if both are <=3, it's choice 1.
if both are > 3 it's choice 2.
if the black die is >=3 that's choice 3
if the red die is >=3 that's choice 4

It also works with 2 different coins.

Or get a d4, or a d8, or a d12, or a d20, or a percentile pair, or a d100....
Or an app for your phone.

 

[spank]

That was one reason I was looking at using D66, I wanted to select for 75% probability, and the best 2d6 will do is 72% or 83%. Similarly there is no 50% with 2d6, it's 41% or 58%.a D66 gives alot more probabilities.
For 25% chances you'd want: 1= 11+, 2= 24+ 3= 41+ and 4 = 54+
Or:

11-23	1
24-36	2
41-53	3
54-66	4

In various games I play, there are often 4 players, and some method must be used to choose a player (such as, for who goes first). With 6 players, it's a simple matter to roll a D6 to choose. Same for 3 or 2. For 5, it's just like for 6, except you reroll a 6; as this happens only one time in six, it's not a big deal. But when you have to choose between 4 things, rolling a die and rerolling on five or six has a 1 in 3 chance of needing a reroll, which can happen multiple times in a row. So I came up with a different way to do it: use 2 dice.

There are 36 possible combinations of 2D6, which is evenly divisible by 4, so that no rerolls are necessary. But how to divide them up in a way that you can easily remember, and doesn't require the hassle of remembering which die is the primary die (such as by doing two hi-low rolls)?

Once I figured it out, a memory aid suggested itself: 6, 5, 4, 3. You can assign those numbers to choices the same way you'd assign 1, 2, 3, 4.

The 6 group of results is "a 6 without a 5". If either die is a 6, and the other is NOT a 5, that's choice 6.

The 5 group of results is "a 5 without a 6". If either die is a 5, and the other die is NOT a 6, that's choice 5.

Pretty simple so far, but now we have to be a little more sophisticated.

The 4 group of results is "a 5 AND 6, OR a 4 without a 5 or 6". Above, we excluded the specific combination of 5, 6, but now we need to count it to fill in for the 4s. Otherwise it works just like our previous choice: either die can be the number we care about (4), and the other cannot be a higher number.

The 3 group of results is "nothing higher than 3". This is what's left over.

Now you too can amaze your friends by making them think you've memorized a complicated table of results!

Roll 2 different color dice.

if both are <=3, it's choice 1.
if both are > 3 it's choice 2.
if the black die is >=3 that's choice 3
if the red die is >=3 that's choice 4

It also works with 2 different coins.

Or get a d4, or a d8, or a d12, or a d20, or a percentile pair, or a d100....
Or an app for your phone.

 

[mike wightman]

2 players on your left, 2 on your right.

Roll 1d - 1-3 left, 4-6 right

You now have 2 players to select from

Roll 1d - 1-3 left, 4-6 right

Use different coloured dice and you can do both at the same time.

 

[coliver988]

or just pick a number between 1 and 20 or something and have them guess and closest goes first. Not all random decisions need to be dice-driven. Though I do like rolling dice.

 

[Dragoner]

Op is interesting; I think I usually have everyone roll 2d6 and go in order.

 

[atpollard]

Op is interesting; I think I usually have everyone roll 2d6 and go in order.

Or roll 1d6 and go in order (reroll to break a tie or make simultaneous).
I am a simple man with simple solutions.

 

[Badenov]

These are all very interesting alternatives, though one person clearly sees the simple solution as best. Because "1d6; reroll 5's and 6's" means 66% of the time you only need one die roll, with the side benefit of no obscure systems to work out. Alternately, 2d6,with 36 potential rolls, does break into 4 pieces neatly, if not in an easily memorable way:

Choice 1: 2, 3, 7
Choice 2: 5, 6
Choice 3: 8, 9
Choice 4: 4, 10, 11,12

Obviously you can switch out equivalent numbers like 2 and 12 or 3 and 11.

 

[kilemall]

Reminds me of divination by oracles, casting lots and studying entrails/thrown chicken bones- wonder what exotic arguments they had to get their throw right.

 

[Spinward Scout]

There's an easier way than d66.

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[Starglim]

2 players on your left, 2 on your right.

Roll 1d - 1-3 left, 4-6 right

You now have 2 players to select from

Roll 1d - 1-3 left, 4-6 right

Use different coloured dice and you can do both at the same time.

Sure, anything that lets you use a d6 quickly as a d2:

• Both odd: choice 1
• Both even: choice 2 (or rearrange these to make it choice 4)
• Die 1 odd, die 2 even: choice 3
• Die 1 even, die 2 odd: choice 4 (or 2)

Or draw a matrix and read off results immediately, which also works with dice that have unusual symbols or something.