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Math Question - Taem - 04-04-2018
I need to create a table showing the probability of ten possibilities. I need to know the probability of rolling two 10-sided dice, however my criteria aren't like the formulas easily to found on Google;
In every case, the first die MUST be the roll of a 10 on a 10-sided die.
The second die must roll a specific number, OR higher.
In the first entry, the second die must roll 10 also, so the chance is 10 x 10, or 1 in 100, easy enough.
In the last entry, the second die must roll any number 1 through 10, so the second die can be discounted and the odds are simply 1 in 10.
The rest, I'm not entirely certain of. The second die in the second entry must roll either 9 OR 10. The third entry must roll either a 8 OR 9 OR 10. Etc.
Table
First Die - Second Die - (Outcome)
10 - 10 - (1 in 100)
10 - 9 or 10
10 - 8 or 9 or 10
10 - 7 or 8 or 9 or 10
10 - 6 or 7 or 8 or 9 or 10
10 - 5 or 6 or 7 or 8 or 9 or 10
10 - 4 or 5 or 6 or 7 or 8 or 9 or 10
10 - 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10
10 - 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10
10 - 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10 - (1 in 10)
It seems logical that as the second die requires less numbers to reach its potential out of the possibilities, that the odds would decrease incrementally by 10%, so the first entry is 1 in 100, the second would be 1 in 90, then 1 in 80, 1 in 70, etc. all the way down to 1 in 10, but I'm merely hypothesizing based on the first and last entries and what I know of odds of rolling the same number of two or more different dice. This formula tweaks my brain because its not rolling the same number on both die, but a specific number on the first die, and a variable rate OR higher on the second die. It seems odd to me that using this method shows the odds of rolling a 10 on one die AND a 5-10 on the second die has a 1 in 50 chance of success.
RE: Math Question - Taem - 04-05-2018
I think I may have figured it out; in this formula, variable starts at 1 being the number you want to roll and increases by a factor of 1 with the maximum on both dice still being 10
(10 * 10) / variable
Results:
10 + 1 = 1 in 100
10 + 1 or 2 = 1 in 50
10 + 1 or 2 or 3 = 1 in 33
10 + 1 or 2 or 3 or 4 = 1 in 25
10 + 1 or 2 or 3 or 4 or 5 = 1 in 20
10 + 1 or 2 or 3 or 4 or 5 or 6 = 1 in 17
10 + 1 or 2 or 3 or 4 or 5 or 6 or 7 = 1 in 14
10 + 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 = 1 in 13
10 + 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 = 1 in 11
10 + 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10 = 1 in 10
RE: Math Question - kandrathe - 04-05-2018
I think you got there.
RE: Math Question - Lissa - 04-05-2018
Overall, you're looking at the Combination (and a lesser extent Permutation) forumla.
C(n,r) = n! / [r! x (n - r)!]
Where n is the set size (10 in this case) and r is the set that makes up the valid choices (so in the case where 9 or 10 is what you're looking for, it would be 2).
For Permutation, the formula removes the multiple of r!
P(n,r) = n! / (n - r)!
RE: Math Question - Taem - 04-06-2018
(04-05-2018, 11:42 PM)Lissa Wrote: Overall, you're looking at the Combination (and a lesser extent Permutation) forumla.
C(n,r) = n! / [r! x (n - r)!]
Where n is the set size (10 in this case) and r is the set that makes up the valid choices (so in the case where 9 or 10 is what you're looking for, it would be 2).
For Permutation, the formula removes the multiple of r!
P(n,r) = n! / (n - r)!
Hi Lissa, and thank you for your response. I have a few questions;
Using the Combination formula, I seem to be getting a bell curve when I expected instead to see increasing returns. How might the output be interpreted? I may be applying the formula incorrectly. FACT in Excel applies a Factorial (!) to whatever is in "()".
RE: Math Question - Taem - 04-06-2018
(04-05-2018, 07:17 PM)kandrathe Wrote: I think you got there.
I'm pretty sure that's the answer, but I'm curious to Lissa's post. Perhaps its a more accurate number than mine, however from what I've seen of the formula in use, the outcome was not as I expected.
RE: Math Question - kandrathe - 04-09-2018
(04-06-2018, 09:59 PM)Taem Wrote:If I understand the problem...(04-05-2018, 07:17 PM)kandrathe Wrote: I think you got there.
I'm pretty sure that's the answer, but I'm curious to Lissa's post. Perhaps its a more accurate number than mine, however from what I've seen of the formula in use, the outcome was not as I expected.
In your example of two dice, one die must = 10 in order to consider the second die, so it is always 1 in 10 or 10%, and you would multiply that times the probability of the second die.
.1 x .1 = ten & ten = .01 or 1 in 100 or 1%
.1 x .2 = ten & (10 or 9) 2 ways of ten = .02 or 2 in 100 or 2%
...
.1 x .9 = ten & (10, 9, 8, 7, 6, 5, 4, 3, or 2) = .09 or 9%
.1 x 1 = ten & (any) 1 in 10 or 10%