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Haku1234567890

1,720 posts

Nomad

Posted August 18, '11 10:40pm UTC

I really need help, I looked all over internet, and tried calculating but nothing helped
can anyone fill this chart up for me

dx1* [upper numbers] (means how many six sided dices are thrown, dx3 would mean 3)
1-5 [numbers on right] (means how many of certain number you got, for example 2 means two numbers, like two sixes)
so "dx3 ... 2" would mean Rolling 3 six sided dices, and getting two wanted numbers (what is percentage of that happening)

Empty squares need to be filled with percentage, that's what I'm asking for...

6 Replies

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EmperorPalpatine

9,438 posts

Jester

Posted August 18, '11 11:55pm UTC

OMG this is long:
well, on a 6-sided dx1, for getting the desired number in 1 throw, you have a 1/6 chance. the first space should be 1/6 (16.67%)

I'll refer to the spaces on the grid as
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

Spaces 6, 11, 12, 16, 17, 18, 21, 22, 23, and 24 are not possible and therefore are shaded.

Space 2: To roll dx2 and get 1 desired number is an addition of each die's percentage seperately because each has an independent chance of hitting that number, so it's 1/3 (33.33%)

Space 3: add another 16.67% to the last one, so it's 1/2 (50%)

Space 4: add another 16.67% to the last one, so it's 2/3 (66.67%)

Space 5: add another 16.67% to the last one, so it's 5/6 (83.33%)

Space 7: all the possible outcomes of dx2 total 36(1,1;1,2;1,3;1,4;1,5;1,6;2,123456;3,123456;4,123456;5,123456;6,123456). To have that desired number show up twice only occurs once per number. Therefore it's a 1/36 chance.

I'll skip around for now because I found a pattern somewhere else.

Space 13: since 1@dx1 = 1/6 chance, 2@dx2 = 1/36 chance, then 3@dx3 = 1/216 (multiply denominator by 6)

Space 19: 1/1296

Space 25: 1/7776

Jeez this is a lot of work. I'm sure there's a very simple formula to figure this out.

OMG this is long:\r\nwell, on a 6-sided dx1, for getting the desired number in 1 throw, you have a 1/6 chance. the first space should be 1/6 (16.67%)\r\n\r\nI'll refer to the spaces on the grid as\r\n1  2  3  4  5\r\n6  7  8  9  10\r\n11 12 13 14 15\r\n16 17 18 19 20\r\n21 22 23 24 25\r\n\r\nSpaces 6, 11, 12, 16, 17, 18, 21, 22, 23, and 24 are not possible and therefore are shaded.\r\n\r\nSpace 2: To roll dx2 and get 1 desired number is an addition of each die's percentage seperately because each has an independent chance of hitting that number, so it's 1/3 (33.33%)\r\n\r\nSpace 3: add another 16.67% to the last one, so it's 1/2 (50%)\r\n\r\nSpace 4: add another 16.67% to the last one, so it's 2/3 (66.67%)\r\n\r\nSpace 5: add another 16.67% to the last one, so it's 5/6 (83.33%)\r\n\r\nSpace 7: all the possible outcomes of dx2 total 36(1,1;1,2;1,3;1,4;1,5;1,6;2,123456;3,123456;4,123456;5,123456;6,123456). To have that desired number show up twice only occurs once per number. Therefore it's a 1/36 chance.\r\n\r\nI'll skip around for now because I found a pattern somewhere else.\r\n\r\nSpace 13: since 1@dx1 = 1/6 chance, 2@dx2 = 1/36 chance, then 3@dx3 = 1/216 (multiply denominator by 6)\r\n\r\nSpace 19: 1/1296\r\n\r\nSpace 25: 1/7776\r\n\r\nJeez this is a lot of work. I'm sure there's a very simple formula to figure this out.

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Gantic

11,891 posts

King

Posted August 19, '11 1:52am UTC

Are you sure you got the instructions down right? The probability of getting unique numbers is different from repeated numbers.

For 2 dx2
There is a 1/36 chance of getting a 6 and a 6 in any particular order.
There is a 2/36 chance of getting a 6 and a 1 in any particular order.

For 2 dx4
There is a 162/1296 chance of getting a 6 and a 6 in any particular order.
There is a 302/1296 chance of getting a 6 and a 1 in any particular order

Also, EmperorPalpatine did 2-5 incorrectly.

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EmperorPalpatine

9,438 posts

Jester

Posted August 19, '11 3:48am UTC

EmperorPalpatine did 2-5 incorrectly

Yeah I haven't done these in a long time. I'm generally good at math, just not probability. I think I see what went wrong. Should space 2 be 10/36?

Like if the desired number was 1,
(1,1;1,2;1,3;1,4;1,5;1,6;2,123456;3,123456;4,123456;5,123456;6,123456).)
but the first 2 1s don't count because they both showed up in the same roll. Is that right?

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Haku1234567890

1,720 posts

Nomad

Posted August 19, '11 8:39am UTC

OMG this is long:
well, on a 6-sided dx1, for getting the desired number in 1 throw, you have a 1/6 chance. the first space should be 1/6 (16.67%)........

I see, but you didn't do it well,
I found out percentage for getting one wanted number
1 17%
2 31%
3 42%
4 52%
5 60%

but I still need percentage for getting more of same numbers
It needs to be somewhere..hmm

Are you sure you got the instructions down right? The probability of getting unique numbers is different from repeated numbers.

For 2 dx2
There is a 1/36 chance of getting a 6 and a 6 in any particular order.
There is a 2/36 chance of getting a 6 and a 1 in any particular order.

For 2 dx4
There is a 162/1296 chance of getting a 6 and a 6 in any particular order.
There is a 302/1296 chance of getting a 6 and a 1 in any particular order

Also, EmperorPalpatine did 2-5 incorrectly.

Yes, now I see, I was thinking about getting two of the same numbers

[quote]OMG this is long:\r\nwell, on a 6-sided dx1, for getting the desired number in 1 throw, you have a 1/6 chance. the first space should be 1/6 (16.67%)........[/quote]\r\n\r\nI see, but you didn't do it well,\r\nI found out percentage for getting one wanted number\r\n1 17%\r\n2 31%\r\n3 42%\r\n4 52%\r\n5 60%\r\n\r\nbut I still need percentage for getting more of same numbers\r\nIt needs to be somewhere..hmm\r\n\r\n[quote]Are you sure you got the instructions down right? The probability of getting unique numbers is different from repeated numbers.\r\n\r\nFor 2 dx2\r\nThere is a 1/36 chance of getting a 6 and a 6 in any particular order.\r\nThere is a 2/36 chance of getting a 6 and a 1 in any particular order.\r\n\r\nFor 2 dx4\r\nThere is a 162/1296 chance of getting a 6 and a 6 in any particular order.\r\nThere is a 302/1296 chance of getting a 6 and a 1 in any particular order\r\n\r\nAlso, EmperorPalpatine did 2-5 incorrectly.\r\n[/quote]\r\n\r\nYes, now I see, I was thinking about getting two of the same numbers

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Joe96

2,226 posts

Peasant

Posted August 20, '11 2:42am UTC

I'm too tired to actually think about what you said (sorry :P), but a general rule of probability: for the likeliness of specific outcomes more than once, multiply.
Ex. probability of rolling a "2" 3x in a row on a ten-(I'm going with that, because it's an easy fraction-percent multiple)sided die...
1/10 x 1/10 x 1/10 = 1/1,000 probability

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Gantic

11,891 posts

King

Posted August 21, '11 12:03am UTC

There is an easy way to do this by subtracting the probability of undesired outcomes from the probability of all outcomes possible.

I'll use 2 dx2 as an example.

(1) first desired number
(2) second desired number

The probability of all possible outcomes is obviously 36/36. Each dice has a 6/6 probability for any number. Two dice is

6/6 * 6/6 = 36/36.

Now, the undesired outcomes are ones that do not include (1) or (2).
The probability for not including (1) is 5/6 for each dice. There are two dice so

5/6 * 5/6 = 25/36

The probability for not including (2) is also 5/6 for each dice, so

5/6 * 5/6 = 25/36.

Now adding those to, you get

25/36 + 25/36 = 50/36

probability of not including 1 or 2, which doesn't make sense since that is greater than the probability of all outcomes. That is because there are overlaps. For outcomes that do not include (1) there are also some that do not include (2) and vice versa. We need to get rid of the overlap to get the probability of undesired outcomes. This overlap includes all outcomes that do not include (1) and (2). The probability of neither (1) and (2) is 4/6. There are 2 dice, so

4/6 * 4/6 = 16/36

Now, it's possible to calculate the probability of undesired outcomes.

50/36 - 19/36 = 34/36

So

36/36 - 34/36 = 2/36

That means there are 2 out of 36 possibilities that will give you the 2 desired numbers.

Notice that when you get to 3 dice. You remove not (1), not (2), not (3). This includes overlap of not (1)(2), not (1)(3), and not (2)(3), so subtract those. However, that overlap includes the overlap of not (1)(2)(3) that you removed, so you need to add that back in.

Fiddle around with that enough and you come upon this lovely summation, that can also work for more than just dice.

P = the probability of getting the desired numbers
n = number of desired numbers
d = number of dice

There might be s simpler way to do this but I can't remember it because Prob & Stat wasn't my best math class.

There is an easy way to do this by subtracting the probability of undesired outcomes from the probability of all outcomes possible.\r\n\r\nI'll  use 2 dx2 as an example.\r\n\r\n(1) first desired number\r\n(2) second desired number\r\n\r\nThe probability of all possible outcomes is obviously 36/36. Each dice has a 6/6 probability for any number. Two dice is \r\n\r\n6/6 * 6/6 = 36/36.\r\n\r\nNow, the undesired outcomes are ones that do not include (1) or (2).\r\nThe probability for [i]not[/i] including (1) is 5/6 for each dice. There are two dice so\r\n\r\n5/6 * 5/6 = 25/36\r\n\r\nThe probability for [i]not[/i] including (2) is also 5/6 for each dice, so\r\n\r\n5/6 * 5/6 = 25/36.\r\n\r\nNow adding those to, you get\r\n\r\n25/36 + 25/36 = 50/36\r\n\r\nprobability of not including 1 or 2, which doesn't make sense since that is greater than the probability of all outcomes. That is because there are overlaps. For outcomes that do not include (1) there are also some that do not include (2) and vice versa. We need to get rid of the overlap to get the probability of undesired outcomes. This overlap includes all outcomes that do not include (1) and (2). The probability of neither (1) and (2) is 4/6. There are 2 dice, so\r\n\r\n4/6 * 4/6 = 16/36\r\n\r\nNow, it's possible to calculate the probability of undesired outcomes.\r\n\r\n50/36 - 19/36 = 34/36\r\n\r\nSo\r\n\r\n36/36 - 34/36 = 2/36\r\n\r\nThat means there are 2 out of 36 possibilities that will give you the 2 desired numbers.\r\n\r\nNotice that when you get to 3 dice. You remove not (1), not (2), not (3). This includes overlap of not (1)(2), not (1)(3), and not (2)(3), so subtract those. However, that overlap includes the overlap of not (1)(2)(3) that you removed, so you need to add that back in.\r\n\r\nFiddle around with that enough and you come upon this lovely summation, that can also work for more than just dice.\r\n\r\n[img]http://img.photobucket.com/albums/v226/blackcairn/DiceNumberProbability.png[/img]\r\nP = the probability of getting the desired numbers\r\nn = number of desired numbers\r\nd = number of dice\r\n\r\nThere might be s simpler way to do this but I can't remember it because Prob & Stat wasn't my best math class.

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