Rolling Dice statistics question (For someone very maths minded)

[Deleted]

Hey there

I have a bit of an odd statistics question for you. It involves dice and I am searching for the lowest number possible.

Say I roll a dice once, the chance that the lowest number is a 1 is 1/6 (as i am only rolling one dice), the chance that the lowest number is a 2 is also 1/6 (again as I only have one dice) etc. So to calculate my average value if I roll one die its just 1/6*1 + 1/6*2 etc = 3.5.

Now say I roll 2 dice. REMEMBER, I am looking for the lowest number possible.

If I roll the 2 dice the chances that the lowest number showing is a 1 are 11/36:

1.1
1.2
1.3
1.4
1.5
1.6
2.1
3.1
4.1
5.1
6.1

The chances that the lowest number is a 2 are 9/36:

2.2
2.3
2.4
2.5
2.6
3.2
4.2
5.2
6.2

Now, my average lowest value if I roll 2 die is : (11/36 * 1) + (9/36 * 2) etc giving me 2.53.

For 3 dice, (I wont show the iterations as there are so many) you get 2.04.

For the first 3 times I have managed to come up with a general formula that works - but I am struggling to extend it past 3 rolls, it looks like this.

Say we are looking at the probability that the lowest number is a 1:

When we roll once: 6^0 = 1 so its 1/6
When we roll twice: (6^1) + (6-1) =11 so its 11/36
When we roll thrice: (6^2) +(6-1)*(12-1) = 91 so its 91/216

If I try extending this for the 4th roll (assuming that the second term follows the pattern 6, 12, 18 etc: (6^3) + (6-1)*(12-1)*(18-1) = 1151 so 1151/1296.

This is where I know I have gone wrong as when we look at the same formula but for finding the probability of a 2.

We have:

When we roll once: 5^0 = 1 so 1/6
When we roll twice: 5^1 + (6-2) = 9 so 9/36
When we roll thrice: 5^2 + (6-2)(12-3) = 61 so 61/216

Extending this for 4 rolls: 5^3 + (6-2)(12-3)(18-3) = 665 so 665/1296

Clearly I have the probability of the lowest number being a 1 plus the probability of it being a 2 being more than 1. So my formula is wrong.

However I dont know if I have gone wrong by assuming that it follows the 6, 12, 18 pattern. OR If its because I did (6-1)*(12-1)*(18-1) in the case where I am looking for 1. And then (6-2)*(12-3)*(18-3) in the case where I am looking for 2. Something just doesnt add up

I know that this is quite lengthy and hard to digest but any thoughts would be appreciated.

Cheers

[Deleted]

the chance you roll a predicted number is always 1/x where x is amount of sides. unless you want to sound like a complete retard and for example you flip a coin 2 times, two times head as result and you say that 100% heads roll on that coin.

you have to take limit of tries->infinity and you'll always have 1/x where x is amount of sides. op you're missing out classes, go back in there.

[Darsithis]

Moved to General Off Topic

[Deleted]

I don't know, I suck at maths :/

[Vexxilus]

For 3 dice, (I wont show the iterations as there are so many) you get 2.04.

This is how you know that the method you are attempting is wrong, or at least being applied wrong, since the lowest result cannot be below 3.

Now, my average lowest value if I roll 2 die is : (11/36 * 1) + (9/36 * 2) etc giving me 2.53.

This shows how you went wrong. You are taking 11/36 * 1 + 9/36 * 2 + 7/36 * 3 + 5/36 * 4 + 3/36 * 5 + 1/16 * 6 = 2.527... which is wrong because the values with a 11/36 chance of occurring range from 2 to 7 with an average of 4.72... If you apply the correct averages throughout for 2 dice you get a result of 7 which is just a long and drawn out way of calculating the average expected roll.

If all you want to do is determine the average expected roll it is very simple since the results follow a normal distribution. You just add the lowest and highest possible results and divide by two, or n(s + 1)/2, where n is the number of dice and s is the number of sides on the dice. For 6-sided dice that simplifies to 3.5n (with n still representing the number of dice).

[Blueobelisk]

The way I do it is:
Die One rolls a 1. Die Two can roll any number from 1-6. Die Three can roll any number from 1-6. = 1 * 6 * 6 = 36 possibilities.
Die Two rolls a 1. Die One can roll any number from 2-6 (we already used the 1 before). Die Three can roll any number from 1-6. = 5 * 1 * 6 = 30 possibilities.
Die Three rolls a 1. Die One can roll any number from 2-6 (we used the 1 before). Die Two can roll any number from 2-6 (we used the 1 before). = 5 * 1 * 5 = 25 possibilities.

So the total possibilities with a 1 in it is 36 + 30 + 25 = 91 for three dice.

I think your formulas are wrong. Maybe later when my mind is off this other math homework I'm doing I'll try to help you with the formula, but in the mean time you can look at the way I think about it and maybe rework your formulas.

[Deleted]

When we roll once: 6^0 = 1 so its 1/6
When we roll twice: (6^1) + (6-1) =11 so its 11/36
When we roll thrice: (6^2) +(6-1)*(12-1) = 91 so its 91/216

This here is wrong, you have the correct probabilities in the end but the formula is more like this:

P(to have at least a 1) = 1 - P(to never have a one)

P(to never have a 1) = (5/6)^n for n iterations
P(to have at least a 1) = 1 - (5/6)^n for n iterations

So for one dice, the probability that the lowest number is a 1 is 1-(5/6)^1 = 1-(5/6) = 1/6
for two dices, 1-(5/6)^2= 11/36
three dices, 1-(5/6)^3 = 91/216
four dices, 1-(5/6)^4 = 671/1296

The probability that the lowest number is a 2 is calculated like that:

P(the lowest number is a 2) = P( to never have a 1) * P(to have at least a 2 knowing that there is no 1) It's conditional probability.
P(to have at least a 2 knowing that there is no 1) = 1 - (4/5)^n
Then P(the lowest number is a 2)= (5/6)^n * (1 - (4/5)^n) = (5^n-4^n)/(6^n)

for one dice: P(the lowest number is a 2) = (5-4)/6=1/6
for two dices: (5² - 4² )/6² = 9/36
three dices: (5^3 - 4^3)/(6^3) = 61/216
four dices: 369/1296

[Blueobelisk]

If all you want to do is determine the average expected roll it is very simple since the results follow a normal distribution. You just add the lowest and highest possible results and divide by two, or n(s + 1)/2, where n is the number of dice and s is the number of sides on the dice. For 6-sided dice that simplifies to 3.5n (with n still representing the number of dice).

He wants the average value if you roll a specific number on the die (1-6), not just the overall average for any given number of dice, which makes it a little trickier.

[Vexxilus]

He wants the average value if you roll a specific number on the die (1-6), not just the overall average for any given number of dice, which makes it a little trickier.

Yes, but he is summing those results together as well and since the sum of the probabilities is 1 he is just finding the average.

[darkwarrior42]

So we're looking at finding the average value of the lowest number. In general, the formula that will express this is of the form:

P(1) + 2P(2) +3 P(3) + 4P(4) + 5P(5) + 6P(6), where P(X) is the probability that X is the lowest number.

Expected behavior: the more dice we roll, the closer the average lowest number will come to 1; if we were to roll an infinite number of dice, the average lowest value would be 1.

Definitions:
P(X): probability that X is the lowest number rolled
P(X|~Y): Probability that X is the lowest number given that Y is not the lowest number
Q(X): probability that X is not the lowest number rolled given that all positive integers less than X are also not the lowest number rolled
N: Number of dice rolled

P(1): P(1) is most easily calculated by finding the probability that 1 is *not* the lowest number, which we'll call Q(1), since P(1) = 1 - Q(1).
---Q(1) is simply (5/6)^N, where N is the number of dice we're rolling.
Thus, P(2) given that 1 isn't the lowest number is P(2|~1)
---With 2 dice, Q(1) = (5/6)^2 = 25/36. With 3, Q(1) = (5/6)^3 = 125/216, etc.
So, P(1) = 1 - (5/6)^N
-----Verification on 1, 2, and 3 dice cases:
-----1 die: P(1) = 1 - (5/6)^1 = 1 - (5/6) = 1/6
-----2 dice: P(1) = 1 - (5/6)^2 = 1 - (25/36) = 11/36
-----3 dice: P(1) = 1 - (5/6)^3 = 1 - (125/216) = 91/216

P(2): Probability that 2 is the lowest number; we can use the same methodology as above, keeping in mind that 1 isn't the lowest number.
---Q(2): Given that 1 isn't the lowest number, the probability that 2 isn't the lowest number is Q(2) = (4/5)^N
Thus, P(2|~1)= 1 - (4/5)^N.
And P(2) = Q(1) * P(2|~1) = (5/6)^N * (1 - (4/5)^N) = ((5^N-4^N)/6^N)
-----Verification on 1, 2, and 3 dice cases:
-----1 die: P(2) = ((5^1-4^1)/6^1) = 1/6
-----2 dice: P(2) =((5^2-4^2)/6^2) = (25-16)/36 = 9/36
-----3 dice: P(2) =((5^3-4^3)/6^3) = (125-64)/216 = 61/216


P(3): We'll do this the same way; P(3) = 1 - Q(3)
---Q(3): Given that 1 and 2 are not the lowest numbers, Q(3) = (3/4)^N
P(3|~1,~2) = 1 - (3/4)^N
P(3) = Q(1) * Q(2) * (1-(3/4)^N) = (5/6)^N * (4/5)^N * (1- (3/4)^N) = (4^N-3^N)/6^N
-----Verification on 1, 2, and 3 dice cases:
-----1 die: P(3) = (4-3)/6 = 1/6
-----2 dice: P(3) = (16-9)/36 = 7/36
-----3 dice: P(3) = (64-27)/216 = 37/216

P(4): Same deal, P(4) = 1 - Q(4)
---Q(4): Given that 1, 2, and 3 are not the lowest numbers, Q(4) = (2/3)^N
P(4|~1,~2,~3 = 1 - (2/3)^N
P(4) = Q(1) * Q(2) * Q(3) * (1 - (2/3)^N) = (3^N-2^N)/6^N

The pattern continues, so that P(5) = (2^N-1^N)/6^n = (2^N-1)/6^N, and P(6) = 1/6^N.

Taken altogether, the average value of the lowest number rolled on N d6 is:

A = P(1) + 2P(2) + 3P(3) + 4P(4) + 5P(5) + 6P(6)
A = (1 - 5^N/6^N) + 2[(5^N-4^N)/6^N] + 3[(4^N-3^N)/6^N] + 4[(3^N-2^N)/6^N] + 5[(2^N-1)/6^N] + 6(1/6^N)
A = (6^N + 5^N + 4^N + 3^N + 2^N + 1) / 6^N <-- Simplification courtesy of Vexxilus

Edit: Thanks Vexxilus for catching a copy/paste error, and providing the final simplification

Examples:
N =1, we get (1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 1) = (3.5)
N=2, we get (1-25/36) + 2((25-16)/36) + 3((16-9)/36) + 4((9-4)/36) + 5((4-1)/36) + 6(1/36) = 11/36 + 18/36 + 21/36 + 20/36 + 15/36 + 6/36 = 2.52777... or about 2.53

No guarantees I didn't make a mistake somewhere, also talking about prestige classes with my wife while doing this, but that looks rightish to me.

- - - Updated - - -

Yes, but he is summing those results together as well and since the sum of the probabilities is 1 he is just finding the average.

Not exactly. He's looking for the average value of the lowest number. If you look at the case of 2 dice, the average roll is in fact 7, consistent with your formula of 3.5n (n=2). However, the average value of the lowest number rolled will certainly be less than that; if you look at the n=2 case, there are 36 possible results. What follows is the enumerated list of possibilities, with the lowest number of a given pair in bold, then listed after the row and summed:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) Lowest digits: 1,1,1,1,1,1 Sum: 6
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6) Lowest digits: 1,2,2,2,2,2 Sum: 11
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6) Lowest digits: 1,2,3,3,3,3 Sum: 15
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6) Lowest digits: 1,2,3,4,4,4 Sum: 18
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6) Lowest digits: 1,2,3,4,5,5 Sum: 20
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) Lowest digits: 1,2,3,4,5,6 Sum: 21

If you take out the bold numbers and average them, you find the average lowest number rolled. Total sum of the lowest digits is 91, divide by 36 to find the average: 91/36=2.53 (rounded to 2 decimals).

[Vexxilus]

Not exactly. He's looking for the average value of the lowest number. If you look at the case of 2 dice, the average roll is in fact 7, consistent with your formula of 3.5n (n=2). However, the average value of the lowest number rolled will certainly be less than that; if you look at the n=2 case, there are 36 possible results. What follows is the enumerated list of possibilities, with the lowest number of a given pair in bold, then listed after the row and summed:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) Lowest digits: 1,1,1,1,1,1 Sum: 6
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6) Lowest digits: 1,2,2,2,2,2 Sum: 11
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6) Lowest digits: 1,2,3,3,3,3 Sum: 15
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6) Lowest digits: 1,2,3,4,4,4 Sum: 18
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6) Lowest digits: 1,2,3,4,5,5 Sum: 20
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) Lowest digits: 1,2,3,4,5,6 Sum: 21

If you take out the bold numbers and average them, you find the average lowest number rolled. Total sum of the lowest digits is 91, divide by 36 to find the average: 91/36=2.53 (rounded to 2 decimals).

Ok, I completely misunderstood what he wanted, I was thinking of the set and not the lowest individual die in the set.

No guarantees I didn't make a mistake somewhere

Math looks right but there a couple copy-paste errors if someone is just skimming to the bold.

P(5) = (2^N-1^N)/6^n = 2^N/6^N

That should be: P(5) = (2^N-1)/6^N

A = (1 - 5^N/6^N) + 2[(5^N-4^N)/6^N] + 3[(4^N-3^N)/6^N] + 4[(3^N-2^N)/6^N] + 5[(3^N-2^N)/6^N] + 6(1/6^N)

Should be: A = (1 - 5^N/6^N) + 2[(5^N-4^N)/6^N] + 3[(4^N-3^N)/6^N] + 4[(3^N-2^N)/6^N] + 5[(2^N-1)/6^N] + 6(1/6^N)

which I believe simplifies down to: A = (6^N + 5^N + 4^N + 3^N + 2^N + 1) / 6^N

[Mongo42]

Here's a Penny Arcade comic about bad dice rolls. http://www.penny-arcade.com/comic/20...5/counterspell

[Deleted]

Thank you so much for all the help here. Algar62 and Darkwarrior42 have summed this up pretty perfectly for me. It is for an economics essay regarding searching costs. Basically showing that if you can go to many different stores with different prices, that the more stores you go to, the lower the expected price. As you said, if you could go to an infinite amount of stores (or do an infinite amount of rolls of the dice) then you would pay the lowest price possible: 1.

Thanks

[iengdixuecun]

I think about it and maybe rework your formulas.

[darkwarrior42]

Math looks right but there a couple copy-paste errors if someone is just skimming to the bold.

Fixed in my post, thanks