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Probability and statistics in games
ElderCat
Registered User, ClubPA regular
Registered User, ClubPA regular
What is the percent chance of rolling 6d6 and getting at least one 6. If you answered 100% you would be wrong. According to probability, its 67%.
This was something I heard on the old portent boards and I never really quite believed it, but there is an article that goes over this in detail.
http://www.gamasutra.com/features/20061018/sigman_01.shtml
For the 6d6 explaination above, I will go ahead and quote.
"This mistake can lead you down the dark path of believing that the chance of rolling a “6†on a six throws of the dice is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1.0 = 100% = SureThingTM. Unfortunately, there is not a 100% chance of rolling a “6†on six throws of a die.
The chance of rolling at least one “6†in six throws is found by looking at the chance of rolling no “6â€s. On each throw, the chance of no “6â€s is 5/6. Cumulatively, in order for no “6â€s to come up in six throws, the chance is just the product of all six throws: 5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = .33 = 33%. So, the chance of at least one “6†coming up is 1.0 – .33 = .67 = 67%. Thus, you should see at least one “6†on six throws about 2/3 of the time. Now go use this to make money off somebody."
This was something I heard on the old portent boards and I never really quite believed it, but there is an article that goes over this in detail.
http://www.gamasutra.com/features/20061018/sigman_01.shtml
For the 6d6 explaination above, I will go ahead and quote.
"This mistake can lead you down the dark path of believing that the chance of rolling a “6†on a six throws of the dice is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1.0 = 100% = SureThingTM. Unfortunately, there is not a 100% chance of rolling a “6†on six throws of a die.
The chance of rolling at least one “6†in six throws is found by looking at the chance of rolling no “6â€s. On each throw, the chance of no “6â€s is 5/6. Cumulatively, in order for no “6â€s to come up in six throws, the chance is just the product of all six throws: 5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = .33 = 33%. So, the chance of at least one “6†coming up is 1.0 – .33 = .67 = 67%. Thus, you should see at least one “6†on six throws about 2/3 of the time. Now go use this to make money off somebody."
ElderCat on
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of course, saying that, all the dice will stick on their edges
Be part of something big, even if you are small.
One of the things that took me a while to get my head around in Statistics, but that made a shitload of sense when I finally grasped it, was that you cannot combine probabilities of doing something once with a set number of outcomes with doing the task a large amount of times and ever expect the probability to reach 100%.
Throwing a d6 three times does not mean that the chance of a six coming up is 3:6. They are not cumulative. Each throw is a separate entity, and the chance of each throw is 1:6. However, the chance of getting a single outcome each time is cumulative, if I recall correctly. Throwing a d6 six times and getting a six each time is 6x1:6, or 1/6 x 1/6 x 1/6 x 1/6 x 1/6 x 1/6, or 1:46,656.
Oh god, looking back at that now makes me realize it has been twelve years since my statistics class, and that I am very likely talking out of my ass. Can anyone confirm if the method above is true?
I had a rough argument with a friend over probability in Magic: the Gathering. In simplified form, the situation was "Given that a player has 4 of a certain card in his deck, given that he has one card in hand and no copies of that card in play/graveyard, given that the odds that he is holding that card are X%, and given no other information, what are the odds that he has at least one copy of that card after drawing one card?" My friend was adamant that it would be 2X -- in other words, the odds doubled since he now has 2 cards in hand. Obviously, there is no way you can even remotely calculate these probabilties without knowing a lot of other information (the number of cards left in their library, at the very least), but I just could not convince him of the fact.
People hate and fear numbers.
Doesn't work that way. Statistically, it is incredibly unlikely that you won't, but you never, ever have a 100% sure thing when it comes to probability like that.
I remember a DM who used loaded dice to make sure parties would get fucked over, 99.9% of the time.
Haven't used them yet, but I'm still trying to come up with a reason to pull them out for the party I'm GMing for.
It was interesting because it implied that the root of all that stuff might be the misguided common-sense approach to probability.
Misunderstanding probability is what gives rise to notions of alien intelligences, here, now, abusing livestock and kidnapping farmers. :roll:
The distinction is quite important...
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1: 0.01%
2: 2%
5: 23%
6: 3%
I don't really feel like figuring out 3 or 4...
hmm.... I might give that a shot over the weekend, just out of sheer perversity
If you've got 6d6, and you're counting the chance of any amount of numbers coming up, shouldn't the chance for 1 (all numbers come up the same) and 6 (all numbers come up different) be the same?
Edit: although I'm coming up with closer to 1.5% for 6
I guess I see what you mean, though. I hate statistics. >>
He's handed a pair of dice and manages to swap one without anybody noticing, but not both.
1) Assuming he doesn't get caught, has he improved his odds?
2) Does either die benefit him more than the other?
3) Which of the two die is more likely to get him busted by the guy running the table, assuming all of the dice look identical?
(assume for the purposes of this discussion that you "win" on 7 or 11, "lose" on 2 or 12, and any other value is more or less a push)
swapping the "5's only" die for one of the fair one gives 40% chance of a "win" and no chance at all of a "lose" (if one of your numbers is always a 5, there's no way you can get either a 2 or a 12).
swapping the "2 or 6" die for one of the fair one gives approx 27% chance of a "win" and 9% chance of a "lose"
Assuming the croupier is at all intelligent, he's going to notice the "5's only" swap first, because the odds of getting at least one 5 are 2/36 or not quite 6%. Getting one every time is pretty unlikely. Also, it's human nature to notice streaks of a single number before streaks of "one of two", especially when mixed in with the results of the fair die.
Whether swapping the "2 or 6" die in for a fair one is worth it or not depends on your level of risk aversion. The chance of winning and the chance of losing go up by about the same amount.
Or just the all time classic: odds involved in rolling a 3d6. If you are rolling hard core (3d6, no re rolls) it is extremely unlikely that an entire party of D&D characters would have a single score of 18. Or that 10+d8 gives very different results than 3d6 reroll 10 or less.
I agree on that the 5s only will be caught first, but for a different reason- even if he's not paying too much attention to the individual dice results (a 2 and a 5, a 5 and a 3, a 4 and a 5, hey, what gives?) he'll probably catch that there has been no total less than six before too long.
I like this question
I'll use it if I ever teach a probability class.
1: 0.01%
2: 1.99%
3: 23.15%
4: 50.15%
5: 23.15%
6: 1.54%
I'm pretty sure my formulas are working right, as they do add up to 100% (+/- .01% round off error).
Basically, the way I worked it was to figure out the odds of each for 1d, then used that to figure out the odds for 2d, then used the results for 2d to figure out the 3d, etc. and used Excel to do the actual math for me. The table looks something like this:
I can post the formulas I used if anyone wants to be sure what I did was legit from a statistical point of view -- it has been more years than I want to admit since I studied stats.
I once worked it out from a theoretical point of view, but I'm too lazy to do it again right now, and my old work is apparently at the office (I just checked my stack of math at home and didn't find it). I'll post the correct solution once I find it, but from what I remember, the percentages you posted above look just about right
They're worse. MUCH worse. D20 chances = 1/20 = 5%.
2D6 + 1 D8 chances = 1/6 * 1/6 * 1/8 (you have to nail all 3 of them individually to get the 20) ~ 0.00347222222 ~ 0.34%
Chance of rolling a 20: 5% (1 in 20)
Chance of rolling two 6's on a d6 and an 8 on a d8? Approximately 0.35% (1 in 288)
edit: JINX
And while you may not get a 20, you have a better chance of getting massive numbers
Like a 7% chance
Yeah, thats a good one.
With the hardcore 3d6, the probability that one of your stats will be an 18 is less than 3%.
But if you're rolling d10+8, the probability that one of your stats will be an 18 is just about 55%.
(The easiest way to calculate this is actually figuring out what the probability of NOT rolling an 18 is, compound that 6 times and then subtract from 1. For my own amusement I also wanted to figure out the probability of an 18 on 4d6 with a throwaway, but it's a friday afternoon and my brain hurts.)
No, rolling dice and adding them always creates a bell curve distribution. Rolling a single die is always a linear distribution (eg: on 1dX the odds of rolling any given number is exactly the same. You are just as likely to roll a 20 on a d20 as a 10 or a 1 or any other number). The more dice you roll and add, the more likely it is to roll near the average value and the less likely you are to roll extreme values.
So with 2d6+d8 you will get a lot of rolls in the 10-12 range and very few low or high rolls.
Consistent 10 or 12 rolls are better than taking risks with low rolls.
But what if you were flipping 10 D2s? You are likely to get a higher number. And if you were rolling 20 D1s, you would always get 20. So basically, increase the number of dice to attain the desired "range" of rolls. And therefore by induction, it stands to reason that 1 D20 would be the worse, but that apparently is not the case, therefore, what is the optimal number and combination of dice to roll?
10 D2's puts the minimum roll at 10, and the max at 20, with a very high percentage of the rolls being near the mean (15). 1 D20 drops the minimum roll down to 1.
I don't quite understand the question, either - what is the optimal number and combination of dice to roll for what?
It totally depends what you mean by "optimal". If we are talking about a D&D game here, then at lower levels you want a 1d20 because the die roll is much more important. A level 1 character will frequently need a 15 or higher to hit. Low level characters would be seriously screwed by 2d6+d8 (for example) compared to a single d20.
At higher levels the roll becomes irrelevant compared to all the other factors (base attack, ability bonus, magic) and getting consistant rolls in the 10 range will still be able to hit and thus be better than risking a low roll.
If we assume the designer of a game to be at all competant (eg: not the guy who made FATAL) then there will be a damn good reason when they choose to add multiple dice vs. rolling a single die.
Edit: Even in the extremely degenerate case of 10d2, if a player needed a 19 or 20 to hit they would still be better off rolling a single d20 despite there being no chance of rolling less than 10. The odds of rolling a 20 on 10d2 is only 1/1024 as opposed to 1/20 on a d20.
As you increase the number of dice for a range of values, you make it more and more probable that you roll the median. If what you want is for the throw to revolve around the median and rarely stray to the highest and lowest values, that's probably the way to do it.
Rolling one die for each digit lets you generate a range that's impractical for a single die. Have you ever tried to read an actual d100? It's a pain in the ass.
Some games also use multiple dice in a single roll to indicate different things. This tends to be awkward.
Have you ever tried to roll an actual d100?
We play it don't count if it's on the floor, and there's no way that d100 isn't rolling right off the table. :P
Totally. It's like trying to pick out a divot on a golf ball.
golf balls suck 2d10 4 lyfe