Hi, Veeky Forums. First-time poster here. I have a question regarding an upcoming d&d puzzle I have planned and I thought that you might be able to help me.
Unfortunanly I'm a bit of a dumbass when it comes to math problems so I figured I'd ask you.
My PCs will be tasked with mixing together 5 different liquids in the right order to concoct an antidote curing them of a poison. All liquids must be used in the mixture.
My questions are: how many different possibilities for a cure would there be? Is this question too hard for the average person, or am I just a dumbass? Are there any other math related puzzles you know of that I could use?
tl;dr (5 potions need to be mixed in the right order, how many possibilities are there?)
Could you explain the reasoning being this equation please (as simply as you can)?
David Ortiz
Not user who replied. But you can pick from five potions to begin with. There are five ways to do this, one for each potion. Then you have four potions left to pick from. There are four ways to do this for each previous potion. So there are now 5*4 ways to pick a potion.
You now continue till you run out of potions, which makes for 5! Ways to combine the potions.
Parker Ortiz
Awesome. Thank You! I imagine this is a slow board so I'm going to leave this thread open in case anyone has fun answers for my last question (about related puzzles), but I'll try to delete it if that'd be preferable.
Charles Jones
Two complications apply to .
First, 5*4*3*2*1 is the number of possible orders in which you can select the 5 liquids. But this is not quite the number of possible mixtures. If you choose order [red, green, purple, blue, yellow], start by mixing red and green, and then add in purple, blue, and yellow in order -- does that give you a different potion than if you choose [green, red, purple, blue, yellow]? In that case, you ALSO start by mixing green and red. Is that a different potion? (If the first ingredient has to simmer for five minutes in a cauldron before adding the second, the answer is probably "yes". If it's just a matter of mixing, the answer is probably "no".) If not, divide 's number by two.
Second, can you for example do the following: - mix red and green. - mix yellow into the red/green. - mix blue and purple. - mix the red/green/yellow with the blue/purple. If all the mixing needs to happen in a single cauldron, probably not. If you can just mix vials together, probably so. If so, that yields a different number than said. This one is a bit harder to compute in detail.
Josiah Miller
And if you want to be even more completionist you can have 3 or more colors mixed simultaneously. And have 3 or more mixed batches mixed simultaneously.
Colton Phillips
>This one is a bit harder to compute in detail. Lazy way: Compute the first few numbers, then look up in OEIS. Here it is: oeis.org/A001147
Justin Green
Ah, yeah, I assumed OP was just having them mix in one potion at a time and that the order they were mixed in mattered, so that there would be only 120 different potions.
Cooper Ramirez
Now try this one
Isaac Miller
OP here; it seems like this question is a lot more complicated than intended.
I'm going to try to simplify the problem for the sake of simplifying it for myself.
1) A single person drinks a poison.
2) They have 5 liquids in vials laid out before them.
3) They must mix the 5 liquids in the correct order (a,b,c,d,e) in order to concoct an antidote.
I'm note sure how to word this in "mathematical terms", but I'm looking for a simple 1+2+3+4+5=antidote equation. Would I be better off (under the assumption that my PCs are of average intelligence) going with a 3 liquid equation?
Bentley Roberts
No, the answer is simply 4! = 24 in that case. But trivial problems are no fun.
Evan Jones
Typo? Shouldn't that be 5!, or 5!/2 if the order of the first two doesn't matter?
Tyler Gray
This I'm not finding in the OEIS. Could be not there or I'm miscomputing the terms.
Thanks, I now (think I) understand the correct equation to answer this type of problem.
I understand that this type of problem might not be fun, but they're sometimes neccesary in a d&d campaign. Do you guys have any "fun" math problems that you could turn into a D&D mission?
oeis.org/A000311 looks like the answer, which means my 201 is missing something.
Ian Butler
OP here, once again (sorry).
Would a (3) liquid equation be 3*2*1 resulting in (18) different results?
Also, just to make sure this thread isn't a complete waste, do you guys have any other equations that you would consider "fun enough to incorporate into a D&D quest"?
Nolan Cox
screw all the complications, just do it based on order alone, which for 5 liquids gives 120 combos, or 4 liquids gives 24 combos
throw a monty hall problem or three utility problem for luls? ask them to find non trivial solutions to the riemann zeta function?
ask them to do a finite quantum walk with multiple absorbing boundaries? I dunno
Luke Evans
was missing ((ABC)D)E 20 (AB)(CD)E 15
3*2*1 = 6 Do you need your players to compute the number of possibilities or try them all? In either case, you'd need to be very clear about what the rules are for how you mix stuff. If it works out to just counting the number of orders, that's something normally taught in high school, although many people will have forgotten.
Christopher Long
You're going to have at least one player and probably 3 or 4 who will immediately know this. Might want to go for something harder.
Joshua Gutierrez
The "solution" of the problem would simpy be mixing the liquids in the right order; either through an equation or random mixing.
They will lose a percentage of HP for every "wrong" mixture.
