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Dice math.
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Bagpuss
Legend
It isn't really that much use however, since dice don't have memory of what they last rolled or any knowledge of statistics and so it's entirely possible (however unlikely) for your D20 to roll 1 all its life (which no doubt would be very short as you are likely to microwave it or slam it in a vice as an example to your other dice to buck their ideas up).
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MerakSpielman
First Post
It's easy for gamers to get caught up in the "Gambler's Fallicy," which is, in a nutshell, that if a bad result has been repeatedly coming up (dice, roulette, cards, whatever) that it is increasinly likely for a good result to come up. That is, if I've been rolling 1's all night, surely a few 20s are coming up. This is, however, untrue. Each roll of a d20 has a 5% chance of being any particular number, no matter how many or few times any particular number has come up previously (assuming a fair die). As Bagpuss said, the die has no memory of its previous activity.
This is a common problem for people who really want a child of a particular sex. After five boys, they often start thinking, "surely the next one has to be a girl. What are the odds of having 6 boys?" This is incorrect thinking. The odds of any particular child being a boy are 50/50. The odds of six 50/50 situations ending in the same result are vanishingly slim, but each individual situation remains unimpacted by those that have gone before.
As far as dice are concerned, the average roll on a d4 is 2.5. This makes sense because, on a d4, one-quarter of all rolls will result in a "1". One quarter of all rolls will result in a "2". One quarter of all rolls will result in a "3". One quarter of all rolls will result in a "4". So you can figure out the average value rolled by taking the average of these possiblities:
(1+2+3+4)/4 = 2.5
The equation you posted is a mathmatical shortcut. It assumes that each die is numbered sequentially beginning at 1 and ending with a number equal to the number of sides. This is, of course, true for all dice we actually use. A theoretical six-sided die that is numbered 2, 4, 7, 9, 14, and 17 would produce an average result of 8.83.
Your equation is good for all situations that will occur, and PC's and mine are good for all situations that can occur (but really never will).
This is a common problem for people who really want a child of a particular sex. After five boys, they often start thinking, "surely the next one has to be a girl. What are the odds of having 6 boys?" This is incorrect thinking. The odds of any particular child being a boy are 50/50. The odds of six 50/50 situations ending in the same result are vanishingly slim, but each individual situation remains unimpacted by those that have gone before.
As far as dice are concerned, the average roll on a d4 is 2.5. This makes sense because, on a d4, one-quarter of all rolls will result in a "1". One quarter of all rolls will result in a "2". One quarter of all rolls will result in a "3". One quarter of all rolls will result in a "4". So you can figure out the average value rolled by taking the average of these possiblities:
(1+2+3+4)/4 = 2.5
The equation you posted is a mathmatical shortcut. It assumes that each die is numbered sequentially beginning at 1 and ending with a number equal to the number of sides. This is, of course, true for all dice we actually use. A theoretical six-sided die that is numbered 2, 4, 7, 9, 14, and 17 would produce an average result of 8.83.
Your equation is good for all situations that will occur, and PC's and mine are good for all situations that can occur (but really never will).
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Zappo
Explorer
Technically, you perform the sum of all the values multiplied by their individual probability. In most cases, this is a bit complex, but with dice you can simplify a lot.
Since the probability for any given value of a die is 1/N where N is the number of faces, the formula for the average ends up being the sum of all values divided by N.
However, the values on a normal die generally just go from 1 to N. This makes it very easy to compute the sum as N*(N+1)/2.
Which leaves the formula as (N+1)/2, or N/2 + 0.5. That's about all.
Edit: thx Merak
Since the probability for any given value of a die is 1/N where N is the number of faces, the formula for the average ends up being the sum of all values divided by N.
However, the values on a normal die generally just go from 1 to N. This makes it very easy to compute the sum as N*(N+1)/2.
Which leaves the formula as (N+1)/2, or N/2 + 0.5. That's about all.
Edit: thx Merak
Even when you don't have memory, FEAR always works.Bagpuss said:
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wocky
Masterwork Jabberwock
Well... of course that dice have no memory and that knowing the average of 1d20 is no good...
Still, we have the fabulous Central Limit Theorem, thanks to which the most dice we roll, the closest the expected result (the average) matches what we actually roll... that is: Roll 10d6 and see how closely you match the average (35) each time. Roll 20d6 repeatedly and you should see that most times you're even closer to the average (70) than before.
Also, there's another interesting result from probability theory. Each time you roll 1d20 you have exactly the same probability of rolling a natural 20: 1:20 (or .050). So, even if you rolled your dice 19 times without getting a 20, your chances are still the same: .050 . However, the odds of rolling 1d20 20 times, and never getting a 20 are .358 . which is much lower than our initial .095 expectation.
So, probabilities are no good for individual cases, but the real deal when you're dealing with a series of rolls. It's possible for a single gaming night to be below or over average, but balanced die should represent averages in the long run.
Still, we have the fabulous Central Limit Theorem, thanks to which the most dice we roll, the closest the expected result (the average) matches what we actually roll... that is: Roll 10d6 and see how closely you match the average (35) each time. Roll 20d6 repeatedly and you should see that most times you're even closer to the average (70) than before.
Also, there's another interesting result from probability theory. Each time you roll 1d20 you have exactly the same probability of rolling a natural 20: 1:20 (or .050). So, even if you rolled your dice 19 times without getting a 20, your chances are still the same: .050 . However, the odds of rolling 1d20 20 times, and never getting a 20 are .358 . which is much lower than our initial .095 expectation.
So, probabilities are no good for individual cases, but the real deal when you're dealing with a series of rolls. It's possible for a single gaming night to be below or over average, but balanced die should represent averages in the long run.
MerakSpielman
First Post
OK, since we're having fun with probablilty, who (besides me) can answer these:
Question 1:
You have a fair coin that is flipped 4 times, resulting in heads or tails.
Possible results:
All heads
All Tails
Three heads and 1 tail
Three tails and 1 head
2 heads and 2 tails
Which combination(s) are most likely to come up?
Question 2:
What famous mathmatical diagram can be used to predict probabilities for problems such as this? (don't know if I'm phrasing this question properly, but if you know the answer, you should recognize it)
Question 1:
You have a fair coin that is flipped 4 times, resulting in heads or tails.
Possible results:
All heads
All Tails
Three heads and 1 tail
Three tails and 1 head
2 heads and 2 tails
Which combination(s) are most likely to come up?
Question 2:
What famous mathmatical diagram can be used to predict probabilities for problems such as this? (don't know if I'm phrasing this question properly, but if you know the answer, you should recognize it)
