basic probability confusion
however if i roll 3 dice a million times i can only expect to see a 6 a little over 42%. is counter intuitive as i would expect 50%
here is the challenge. explain this to me verbally without resorting to 5/6 * 5/6 * 5/6. i want an explanation that show understanding of the nature of the odds NOT a regurgitation of statistics 101.
talk to a philosopher not a mathematician.
😊 thank you
the question is clear and contains all of the data necessary so please do not answer with a question, thx.
Is that what you had in mind?
I do not see it as counter-intuitive.Quote: dpcjsrif i roll 1 dice a million times i can expect a 6 to be rolled a little over 16%. makes sense. 1/6
however if i roll 3 dice a million times i can only expect to see a 6 a little over 42%. is counter intuitive as i would expect 50%
Then the same logic would follow for 4 dice at 4/6
5 dice = 5/6
6 dice = 6/6
7 dice = 7/6
Hey, is not the probability of an event between 0 and 1 inclusive?
So this method can not be correct for a probability
but it is correct for an average
Well, (5/6)^3 still is not the proper answer.Quote: dpcjsrhere is the challenge.
explain this to me verbally without resorting to 5/6 * 5/6 * 5/6.
i want an explanation that show understanding of the nature of the odds NOT a regurgitation of statistics 101.
The explanation has to do with what are or are not mutually exclusive events. (can't happen at the same time)
It is really that simple.
math is fun
http://www.mathsisfun.com/data/probability-events-mutually-exclusive.html
Good Luck
added: this also shows the difference between a probability and an average
1/6 + 1/6 = 2/6 or 33.3%
But this over-counts the successful outcomes when both dice are 6s.
1,6
2,6
3,6
4,6
5,6
6,6 -->2 6s, but this counts as only one success
6,5
6,4
6,3
6,2
6,1
So out of the 36 possible 2 dice permutations
only 11 of them have at least 1 6 in it.
The correct probability is then
11/36 and Not 12/36
n/N
n=number of possible successes
N = total number of outcomes
added:
what the 1/6 + 1/6 does represent is the total expected number of 6s (average) when tossing 2 dice.
Not the probability of rolling at least one 6.
2/6 = 1/3 or .33...
3 dice = 3/6 (not the probability) but the total expected number of 6s when tossing 3 dice
so for 7 dice we have 7/6 expected number of 6s
for averages we do count ALL the possible successes, every 6 that can show.
for 2 dice that is now 12 total 6s out of 36 possible sequences
still n/N
But when we count ALL the 6s that could roll, we end up with an average.
Averages can also be in the form of N/n. That answers a different question.
and sometimes the average does equal the probability, 1/6 and 1/6
but one value (average) can be number from 0 to infinity and the other (probability) is a value between 0 and 1
Quote: EdCollinsThis article MIGHT help. I wrote it 15 years ago:
http://www.edcollins.com/backgammon/diceprob.htm
So you are HE!
Two thumbs up!
Ed's Bikini Babes!
Of course I did read your site over a few years ago
Thanks!
Quote: dpcjsrif i roll 1 dice a million times i can expect a 6 to be rolled a little over 16%. makes sense. 1/6
however if i roll 3 dice a million times i can only expect to see a 6 a little over 42%. is counter intuitive as i would expect 50%
here is the challenge. explain this to me verbally without resorting to 5/6 * 5/6 * 5/6. i want an explanation that show understanding of the nature of the odds NOT a regurgitation of statistics 101.
Imagine the dice game with a little trick: Each time you roll a 6 you get paid a buck (or something else you like).
You would rather like to throw the dices individually, instead of the groups of 3 - right ? Because you intuitively understand that if you hit more than a single 6 in a group, you only get paid on the first 6 (and get screwed on the other remaining 6s).
The money you get after 1 million rounds reflect the probability of the events you described. So you now know why you get a lesser probability on the 3-dice game than you would expect for the single dice game.
Quote: dpcjsrtalk to a philosopher not a mathematician.
The easiest way to do this calculations is to first calculate the probability of it not happening.
Calculate the probability that each dice is NOT a six.
So if you are throwing 1 die, the probability that it is NOT a six is 5/6. The probability that it IS A six, is 100%-5/6.
If you are throwing 2 die, the probability that neither is a six is Probability the first is not a six TIMES the Probability that the second is not a six or 5/6*5/6.
The probability that it IS a six is 100% - 5/6*5/6
In a similar way for 3 die the calculation is:100% - 5/6*5/6*5/6
The following table results:
P(get a six) - Number of Dice thrown
16.67% 1
30.56% 2
42.13% 3
51.77% 4
59.81% 5
66.51% 6
72.09% 7
76.74% 8
80.62% 9
83.85% 10
...
97.39% 20
...
99.93% 40
As we would expect, if we throw a huge number of dice the probability that one of them is a six approaches certainty.
Nice example.Quote: MangoJImagine the dice game with a little trick:
Each time you roll a 6 you get paid a buck (or something else you like).
You would rather like to throw the dices individually, instead of the groups of 3 - right ?
Because you intuitively understand that if you hit more than a single 6 in a group,
you only get paid on the first 6 (and get screwed on the other remaining 6s).
The money you get after 1 million rounds reflect the probability of the events you described.
So you now know why you get a lesser probability on the 3-dice game than you would expect for the single dice game.
And if he was paid $1 for EVERY 6 that did show rolling the 3 dice, {6,2,6} = $2 paid instead if $1
he would show close to $500,000 (on average) after 1 million rounds.
That would be the average number of 6s to show per round.
1/6 + 1/6 + 1/6 = 3/6 = .5
averages vs probabilities
12 dice = 1/6 *12 = 12/6 = 2 6s on average each round
