A question of dice -- answer only if you're old enough.

[Zander]

(d4 - 1) * 11 + d11 + 5

So my d11 has a use after all.

 

[hagor]

For those of you who live in a state that offers some sort of lottery ticket system, have you ever picked your numbers by using your gaming dice? ... And rolled them in front of strangers as you filled out the card?

Be honest.

Apart from the rolling in front of strangers part, yes.
Didn't work though (and that was a big surprise ).

I've used some modified percentile dice roll, with a re-roll when the number was too high.

Hagor

 

[Cor Azer]

(If you're not old enough to legally buy into a lottery, you can't answer this one.)

For those of you who live in a state that offers some sort of lottery ticket system, have you ever picked your numbers by using your gaming dice? ... And rolled them in front of strangers as you filled out the card?

Be honest.

I don't play the lottery very often, and those few times I do, I either use the auto-pick or just grab a scratch ticket. No dice.

Still, I do have friends that always have their dice with them, and they have been known to pull them out on strange occasions (Philly cheesesteak versus the fish'n'chips questions are raised every now and then...)

 

[Nadaka]

(d4 - 1) * 11 + d11 + 5

nice try, but the range is 1to49, not 6 to 49.

to get a number in the range of 1 to 49 we take the log7(49)=2. so we can express this with 2 dice (same applies to log10(100) = 2, hence how 2d10 can be used as percentile).
so the actual expression would be (1d7-1)*7+1d7, roll that 6 times to get 6 numbers. of course if each of the 6 has to be different, you would get a different expression for each one.

And yes, you can theoretically do something similar for all non-prime numbers.
for instance a d26 can be expressed as (1d2-1)*13+1d13

 

[GoodKingJayIII]

nice try, but the range is 1to49, not 6 to 49.

to get a number in the range of 1 to 49 we take the log7(49)=2. so we can express this with 2 dice (same applies to log10(100) = 2, hence how 2d10 can be used as percentile).
so the actual expression would be (1d7-1)*7+1d7, roll that 6 times to get 6 numbers. of course if each of the 6 has to be different, you would get a different expression for each one.

And yes, you can theoretically do something similar for all non-prime numbers.
for instance a d26 can be expressed as (1d2-1)*13+1d13

 

[Zander]

nice try, but the range is 1to49, not 6 to 49.

to get a number in the range of 1 to 49 we take the log7(49)=2. so we can express this with 2 dice (same applies to log10(100) = 2, hence how 2d10 can be used as percentile).
so the actual expression would be (1d7-1)*7+1d7, roll that 6 times to get 6 numbers. of course if each of the 6 has to be different, you would get a different expression for each one.

And yes, you can theoretically do something similar for all non-prime numbers.
for instance a d26 can be expressed as (1d2-1)*13+1d13

I think it's right to say that if x/y is a whole number, then Dx=(D[x/y]-1)*y+Dy where "D" indicates a die.

 

[Henry]

For those of you who live in a state that offers some sort of lottery ticket system, have you ever picked your numbers by using your gaming dice? ... And rolled them in front of strangers as you filled out the card?

I don't play games of any sort. 'Tis a foul habit, and the Devil's work.

 

[Driddle]

The biggest problem with most of the equations/methods listed here as examples is that any time you add two dice together, you end up with a bell curve of some sort in which certain outcomes are more likely than others.

I prefer something like 1d4 or 1d6 (and reroll 6s) to determine the first digit (depending on your lottery range, of course), and 1d10 for the second digit. Might still have to discard some results beyond the appropriate range. But at least it's a flat-line outcomes chart.

I don't play games of any sort. 'Tis a foul habit, and the Devil's work.

But you'll still put a gambler NPC in your campaign to tempt the innocent who might stumble into his shadowy lair at the back of the tavern...

 

[Terwox]

Much simpler way to get a random number from 1-56 with your own set of dice:
roll percentile dice. If you get 57-00, reroll. done.

 

[Daijin]

Wow..rolling the die for numbers..that never occured to me... hmm interesting. I should check into it and instead of using one pick of random numbers, use the die.... Thanks!