Simple math puzzle

[Muse]

I have a pair of dice.

To pick an integer between 1 and 6, I can simply throw one of them.

Can I determine a random integer between 1 and 12 with them?

 

[Muse]

I thought this over a few minutes and decided

Spoiler: This may be the answer

AFAIK, the answer's no. But you can pick randomly a number between 1 and 11 using dice. You just need one. You throw that one and mark its result. You throw it a 2nd time and if the integer that comes up is even, you throw it a 3rd time and add the result to that from the first throw and subtract 1 and that's your end result, otherwise the first throw is the end result.

 

[JujuFish]

Not by adding the pips together, which would be at least 2 and be a bell curve of probability, with 7 being the most probably and 2/12 being the least.

I suppose you could treat one die as a multiplier, odds = x1, evens = x2?

 

[Muse]

- - -

 

[iCyborg]

Throwing the first dice gives you N.
From the second dice - if result<=3, final number is N. If result>=4, final result is N+6.

Does this do it? I assume I'm restricted to one throw each, otherwise what's the point of the 2nd dice...

 

[Fenixgoon]

Throwing the first dice gives you N.
From the second dice - if result<=3, final number is N. If result>=4, final result is N+6.

Does this do it? I assume I'm restricted to one throw each, otherwise what's the point of the 2nd dice...

Seems reasonable... At least when i mathed it in my head

 

[Ajay]

Not by adding the pips together, which would be at least 2 and be a bell curve of probability, with 7 being the most probably and 2/12 being the least.

I suppose you could treat one die as a multiplier, odds = x1, evens = x2?

The probability would be a flat line, not a curve. Unless I've lost my marbles again. Each die has an equal probability of landing on any of it's 6 sides.

 

[BarkingGhostar]

What if they are quantum-entangled dice? I just watched a PBS show in which they used dice to layman explain QE.

 

[JujuFish]

The probability would be a flat line, not a curve. Unless I've lost my marbles again. Each die has an equal probability of landing on any of it's 6 sides.

The only way to get 2 is snake eyes (1 pip on each die). 1/6 * 1/6 = 1/36. Ditto with 12. 3 can be gotten with either 1+2 or 2+1. 4 is 1+3, 3+1, 2+2. And so on.

 

[JujuFish]

Not by adding the pips together, which would be at least 2 and be a bell curve of probability, with 7 being the most probably and 2/12 being the least.

I suppose you could treat one die as a multiplier, odds = x1, evens = x2?

If the second die is simply +0 for odds or +6 for evens, that should do it. Multiplication doesn't work.

 

[Ajay]

The only way to get 2 is snake eyes (1 pip on each die). 1/6 * 1/6 = 1/36. Ditto with 12. 3 can be gotten with either 1+2 or 2+1. 4 is 1+3, 3+1, 2+2. And so on.

Ah, my bad. Geez, I need to go re-read my prob&stat book from university, because, well, that was a long time ago!

 

[pete6032]

delete

 

[[DHT]Osiris]

Can I determine a random integer between 1 and 12 with them?

No 1. 2 and 12.

 

[Muse]

Using the basic method explained by @iCyborg here you can roll a couple of dice and come up with a random member of these sets (1-6, 1-12, 1-18, 1-24, 1-30, 1-36):

1-6: First cube
1-12: First cube + (0 if 2nd cube shows 1-3, first +6 otherwise)
1-18: First cube + (0 if 1 or 2, 6 if 3 or 4, 12 if 5 or 6 on second cube)
1-24: First cube + (0 if 1, 6 if 2, 12 if 3, 18 if 4... having rolled 2nd cube until 1,2,3, or 4 returns)
1-30: First cube + (0 if 1, 6 if 2, 12 if 3, 18 if 4, 24 if 5... having rolled 2nd cube until 1,2,3,4 or 5 returns)
1-36: First cube + (0 if 1, 6 if 2, 12 if 3, 18 if 4, 24 if 5, 30 if 6 on 2nd cube)

 

[Paperdoc]

Muse, you got that entire scene correct. The last one, using 2 dice to generate in the full range 1 to 36, is similar to "base 6" numbers, when zero is not part of your number system. Die #1 is the right-most (least significant) digit, and Die #2 is the left-most (most significant) digit. So when the sequence (Die2 / Die1) is, say, 34, you read that as 3*6 + 4 = 22 in base-10 numbers. OR, if you do what Muse did in his last system and treat Die #2 as having the numbers 0 through 5, 34 is (3-1)*6+4 = 16. But if we do it the proper way with BOTH dies having the digits 0 through 5 for a base-6 number system, then 34 is read as (3-1)*6 + (4-1) = 15, and Snake Eyes (11) is zero, and "66" is 35.

If you wanted to generate any random number between 0 and 99 using two dice, you'd need two with TEN faces each, numbered 0 though 9 on each die.

 

[shortylickens]

This is irrelevant, but in Shadowrun you roll only six sided dice, and you are trying to beat a target number, which counts as a Success.
For most tasks a single Success is usually enough to accomplish the goal. For opposed checks you and the other guy roll and total up your successes and whoever has the most wins the opposition. Such as trying to hold closed a door when the other guy is trying to force it open.

If you wanna beat a target number above 6, you have to roll again any dice that hit a 6 on the first roll. This is why it helps to have lots of dice for the first roll. If your strength is a 7 then you get to roll 7 dice with the attempt to hit a high target number, say 10. Odds are pretty decent at least one of them will be a 6, then you get to re-roll that one and hopefully get at least a 4, beating the target number. Since 4, 5, and 6 all will work, you got a 50 percent chance.

This also means if you are weak, or unskilled, theres always a tiny possibility you may succeed, even on a super high target number. It makes role-playing fun.

As for OP's problem, I wonder if something like that would work.
You roll one die. If you are lucky enough to get a 6, you can roll a second die for a max of 12.

But that means not all values have an equal likelihood of being achieved.....
Its still technically random though.

 

[SKORPI0]