"Anyone here a pro at using Anydice?" Topic
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Extra Crispy  05 Feb 2020 11:18 a.m. PST 
I am trying to figure some odds. A player is rolling D6s. He needs, say, a 5 or 6. BUT he only needs one 5 or 6. Excess are ignored. So what are the chances of getting a 6 on X number of dice? On anydice.com I used the following program – did I do it wrong? (Example is for 3 dice, obviously) output [count 6 in 3d6] 
kevin smoot  05 Feb 2020 11:45 a.m. PST 
You have a 1 in 6 chance of rolling just the six per D6. Without dong the math, your odds are about 4045% on 3D6 (gambling tip – play odds, not chance) 
tabletopwargamer  05 Feb 2020 11:45 a.m. PST 
Isn't the chance of getting one six always going to be one in six? 
colgar6  05 Feb 2020 12:04 p.m. PST 
You need to do it the other way around. If you need just a single 5 or 6 on a number of d6 then think about the probabilities of *not* getting a 5 or 6. For one die, the chance of *not* getting the desired result is 4 in 6. The other 2 in 6 results are either a 5 or a 6. For two dice, the chance of not getting a 5 or 6 on *either* die is 4/6 * 4/6, or 16/36. All the other results [i.e. 20/36] will have at least one 5 or 6, possibly 2. For 3 dice, the chance of not getting a 5 or 6 is 4/6 * 4/6 * 4/6, or 64/216. Therefore there is a 152/216 chance of at least one 5 or 6. …and so on. 
Extra Crispy  05 Feb 2020 12:14 p.m. PST 
@TableTop: Yes, on one die. I need to know, what is the chance of at least one six if I roll X number of dice? @Colgar6 The program above turns out a chart showing the chance of 0 sixes, one six, two sixes, etc. up to the number of dice thrown. I just need to be sure it means what I think it means! 
Don Perrin  05 Feb 2020 2:04 p.m. PST 

Stryderg  05 Feb 2020 6:06 p.m. PST 
colgar6 has it. I used Excel and laid out the possible die rolls for 3d6. Came up with the same numbers. 
Sgt Slag  06 Feb 2020 7:47 a.m. PST 
I was working on a dice game, last night, for a promotional event at an upcoming game convention. We were planning on selling rolls, for prizes. Foam dice must be rolled through a 4foot tall, acrylic Dice Tower, to ensure randomness; it will be supervised by our staff, to ensure fairness, with prizes being awarded immediately. You are guaranteed to win a prize each roll… We needed to know the percentage chances for each number rolled, as we have Tiers of prizes to be handed out. At the lowest Tier, you get a prize of four, random polyhedra dice. Each Tier above that, increases the value of the prize: a complete polyhedra set of matching dice, including d24 and d30; an RPG set of metal polyhedra dice; a set Hickory wood coasters, engraved with custom fantasy images; and a Hickory Dice Tower, engraved with a custom fantasy image. We will be at AustiCon, Austin, MN USA, the end of May. If you are in the area of southeastern Minnesota, join us, and roll some dice for prizes! We rolled 200 times, last night, to get an idea of how frequently the numbers would reach into the higher prize Tiers, so we could make sure we didn't lose our shirts! The AnyDice site nailed it for us! Thank you! Cheers! 
Russ Lockwood  06 Feb 2020 7:35 p.m. PST 
Interesting site. >For 3 dice, the chance of not getting a 5 or 6 is 4/6 * 4/6 * 4/6, or 64/216. Therefore there is a 152/216 chance of at least one 5 or 6. Just over 20 years ago, in a Panzerschiffes WWII game of Savo Island or one of the naval battles around Guadalcanal, the torpedo attack mechanic generates x number of d6s based on number of torps fired, distance from firing point, speed of target, and angle of attack. Usually, you get a couple or three or four or so to toss, with a 1 in 6 chance of getting a hit. Well, I plotted well…really well…extraordinarily well…and generated 18 d6s against one target, 18d6s against a second target, and 6d6 against a third target. You guessed it. NOT ONE SINGLE HIT in 42 dice. Yes, it's gaming group lore. And yes, one of my oh so clever buddies gave me a die with stickers and inkedin pips to represent a hit on all six sides as a birthday present (but they never let me use it!). Anyway, if I read your odds calculation right, that means 5/6 * 5/6 * 5/6 …. and another 39 multipliers of 5/6s to NOT get a hit? Then you flip it around to get the 1 in [some huge number] oddsdefying roll that I got? 
