New answers tagged

0

Here is a good way to intuitively understand your odds at this... With one d10 there is a 90% chance of failure, that is, NOT rolling a 1. With two dice there is a 90% chance of 90% chance of failure. That is .9 x .9 = .81 = 81% you will not roll a 1. With three dice there is a 90% chance of 90% chance of 90% chance of failure. That is .9 x .9 x .9 = ....


4

There's another way of looking at this, which works particularly well with a ten-sided die. Think of the single-digit numbers from 0-9. How many of them have a 1? One, out of ten, or 10%. Now, out of all of the combinations of two digits, from 00-99, how many have any 1s? Well, there's 01, plus all ten of the numbers from 10-19, and then 21, 31, 41, 51, 61,...


2

Whenever you have an probability experiment where there are exactly two outcomes and you can assign a probably to the one of the outcomes, it's called a binomial experiment: True, false, yes, no, black, white, etc. In this case, you get a one or you don't. The number of dice would constitute the number of trials, since it wouldn't matter if you rolled a ...


10

Think of that way: if you roll a die, there's a 10% chance of it being a 1. If you roll two dice, it's the same as if you first rolled one of them (since both rolls are independent of each other), then the other. Say you make the experiment 100 times (and equate occurrences with percents). With a single die, 10 times you lose, 90 times you win. With two ...


5

Putting together a layman's answer/example. To further simplify, I will use the example of a coin flip to emphasize the results. Given a single coin, you have a 50% change of flipping a "heads". Given a second coin (or flipping the same coin a second time), you now have a separate and independent 50% chance for each coin flip. With two coin flips you now ...


68

Yes it does. Your instinct is right. The more dice, the more likely you are to roll some 1s. If I'm reading you right you're just interested in whether any 1s appear in your rolled pool. It may not seem obvious but the easiest way to think of this is to model the probability of rolling all 2s-through-10s. That is $$P(\text{not }1)=\frac 9 {10} = 0.9$$ ...


11

Yes, the odds of rolling at least one of anything increase as the number of dice goes up. I don't know the actual function to determine how those odds increase, but as an example, here's an Anydice graph showing a 65.13% chance of getting at least one 1 on 10d10: http://anydice.com/program/8b09 and only 27.10% chance of getting at least one 1 on 3d10: http:/...


0

Another option is a nice Dice-Box, with a separate labelled pocket for every dice. You can put it one the table, it can look really nice (with leather and nice fabric) and you can put each die in its own place. After rolling you can keep the die you rolled on the table and put it back if you need another die.


4

You had the right idea by specific colors. But make sure they all have the same color scheme. It can be confusing if one D20 is green but for another player it is red. If that does not work have them draw each die (just the outline shape) and write the number in it. The act of drawing and labeling it will cause the brain to produce a stronger memory of ...


0

At the start of your session, encourage those players that don't yet have the dice memorized to take a few moments to lay one of each die on the table, turn each to its highest value (the d10 should be turned to the 0 but still treated as a ten for the next step), then place the dice in ascending order in a spot designated for dice not being actively used. ...


11

Give them a playmat with pictures and labels of each die, plus other important info for the game system The most effective way I have found is to give them a playmat which has both a picture of each die and a space for it. Savage Worlds has a popular fan made playmat that has a picture of each die and a space for where it goes (along with other important ...


0

I know in DND, throwing dice around is a ritual but I generally don't let my players to roll their dices. I roll for them, thus no need for them to hunt for the die they need. Saves time and a lot of trouble if there are players tempted to cheat around. Another plus side: you might use your mobile phone to roll the dice without killing the immersion.


3

Keep them in one of those compartmentalised trays (the kind you get dips and snacks in). Use labelled cups, beakers etc. If you're artistic, you could even draw pictures on them.


3

These techniques have work for me both with younger kids and casual gamers: Use a sample set Matching shapes is easy. Remembering stuff is hard. Set up a set of dice in front of you. Point at the die when you refer to the type of die/dice they need to roll. Positive thinking Encourage the players to arrange their dice with the highest numbers up. Then ...


10

There really should not be any reason for anybody to feel ashamed of picking up the wrong die. We all had to learn to tell which was which, and it didn't happen immediately. You wouldn't feel ashamed of, say, not being able to distinguish orthoclase from plagioclase, or a reed warbler from a nightingale, would you? At least not if you weren't already an ...


6

When I introduce new players to the game, I place all the dice on the table onto a piece of paper with labelled areas for each dice size, so that when they need a die they can take it from the appropriate area. I ask them to replace the dice to the correct place in the dice pool, so they get used to recognising the dice as they use them.


2

Positive reinforcement Nothing to add to the answers advising you to take your time besides: Reward them when they pick the correct dice on the first try (with chocolates, small xp or whatever). Ok, this might actually sound pretty dumb, but positive reinforcement has been proven to increase efficiency in learning. Might be worth giving a try. (As long as ...


4

use a cheat sheet while learning Take a picture of each polyhedron, create a document that shows the value count next to them and maybe their most frequent uses, then bring a print for each of your players. Combine this with colored dice and your players will have an easier time learning the different dice.


4

Though you have already tried it in a way, I still would like to contribute what I believe is the easiest way to get started: If you just mention the color that they need to roll, it is very easy for players to pick the right one. EXAMPLE Please roll the red D20 Though I have not actually used this in a D&D setting, I used it in various ...


19

This answer is absolutely right in saying that the best solution is to get practice and experience with dice in play. So let's power-level your players' dice XP! Take a session or two to play a game that forces frequent interaction with all the polyhedrons at once rather than one a time. Games which put emphasis on picking up multiple different-shaped dice ...


8

Partial answer: The only technique I've used that hasn't been mentioned in the question or in other answers is placing the largest number pointing up after I've finished using a die. It's slightly less tedious than keeping all dice in order (although I've done that for some games), but it does still take player overhead. (It also feeds into the common ...


72

Short answer: really, practice is the only answer It's like any kind of memorization task, eventually you're going to get it, and you'll have trouble until then. But there are ways to make the memorization easier. You are going to have to correct them sometimes. Don't think of that as a failure. Just make the correction and move on. Don't default to ...


13

Most of the time players will need only two dice: A d20 and the damage die for their weapon. Instead of handing the new players a full set, give them just these two and introduce them to the other dice when they need to roll them.


0

Probability The answers provided effectively cover the probability for every result, 1 through 20, for advantage/disadvantage with 2d20. For completeness, the probabilities follow: Expected Value When rolling 2d20, and keeping the Maximum value from each of the 400 permutations, the expected value is 13.825. By contrast, the expected value when you ...



Top 50 recent answers are included