Hello all,
I have a SIMPLE issue with some "dice odds". I need the following to be computed:
3d6s, odds of a PAIR being rolled.
I want to compare this to "dice odds" of 2d6, roll 8 or higher (42%). And 2d6 roll 7 or higher (58%).
Just as a comparison ... to see if the odds are in the SAME or SIMILAR range.
Many thanks.
96/216
Or
44,4%
I think there are 216 different combinations of 3d6 results, yeah?
So out of all those, how many result in at least a pair being in the results? X3M suggests 96 of them will allow you to do that.
Personally, I'm too stinkin' lazy at the moment to do the math for you, but I'm certain a lookup table is posted somewhere in the Internet to help you out.
...Maybe Troll will help?
TBH, I counted.
I am sure there is logic to it.
Lets see...
The first die rolls X.
I give X a 1. The other 2 dice give 36 options.
Meaning that the number of 2 dice being equal in the set of 1 will simply be multiplied by 6. But so would the 36x6=216
The second and third die can both roll 1 to 6.
6 of them can be equal in any number. Then each of the die could have rolled a 1. Thus adding 2 times 5 as well.
We get to 16 now. 16x6=96
It gets more complicated with more dice. But I suggest making an excel template for it...
16 possible PAIRs per dice value. Which means 16 x 6 = 96 PAIRs.
Each die has 6 values and there are 3 dice. Which means 6 x 6 x 6 = 216 outcomes.
96/216 = 44.4%
Thank you for providing me with the insight and solving the Math problem... Thanks for BOTH @X3M and @Let-off for providing me with the information that I needed. I did not use Troll ... But I worked backwards from the 96/216.
Wasn't too difficult once I understood the nature of the problem.
Many thanks again for the help solving this Math conundrum.
Cheers!
Thanks for the original answer, both you and let-off helped in solving the Math behind the scenes. Yeah, I saw you explained it too... Same thinking as my own. So good stuff and many thanks!
That includes triples, I believe.. which may or may not be correct depending on what you mean :)
I think there are 90 (3 * 5 * 6) pairs that aren't also triples.
Yeah so there are exactly SIX (6) Triples... I was originally going to go with rolling HIGHER than a seven (7+). But the Triples with only 6/216 = 2.8% ... I could have a special BONUS of "Earn TWO (2) Loot, if you roll a TRIPLE!"
So in fact you are correct that there are 90 pairs and 6 triples... And this gives both a 44.4% chance of earning "AT LEAST ONE (1) LOOT". Plus the odd and low occurring triple which earns you DOUBLE the amount of "Loot"!
In problems like this, you usually work out the chance of NOT getting what you're after. In this case, the chance of not getting a pair.
The second die has a 5/6 chance of not matching the first die.
The third die has a 4/6 = 2/3 chance of not matching either of the first two dice.
So your chance of not getting a pair is 5/6 x 2/3, which works out to 10/18 or 5/9.
So your chance of getting a pair (including 3 of a kind) is 4/9.
So now comes the ADVANCED question, something that I am planning on ... but need to understand the implications before deciding.
Okay so here I go:
Instead of triple (which is very low probability), what IF I had two (2) WHITE dice and one (1) BLACK die?
The results would look something like this:
A> Two (2) WHITE dice pair = 2 LOOT
B> White (1) + Black (1) pair = 1 LOOT
So I need probabilities for these results. I'm trying to determine if it's BETTER ODDS than a "triple" (which is much too low = 2.86%)
Thank you so much for helping. Math can be a challenge for me (at times). And believe me, it can be challenging for OTHER people too! (LOL)
Note #1: So please correct me if I am wrong!
A> The two (2) WHITE dice pair = 1/6 odds or 16.67% (not bad). The Black die is inconsequential (so no special treatment for that die).
B> White (1) + Black (1) pair = 2/6 odds or 33.33%
Total = 50% odds(!!!) Is this CORRECT?? Would be AMAZING if this was TRUE.
It depends.
Do you want to give preference to the 2 white dice pair, over the white/black one?
Also, the chance of the 2 white being NOT a pair is important to add before continuing with testing on the white/black pair.
The 2 white NOT being a pair is 5/6th. This is test zero (0).
Testing on each white/black pair is 1/6th again.
However, 2 tests doesn't mean doubling the results.
If test 1 fails, 5/6th, test 2 is 1/6th.
If test 1 succeeds, 1/6th, test 2 is not needed.
So in total, you have
Test 0: 1/6th on succeeding.
Test 1: 5/6th (test 0 failed) * 1/6th. Is 5/36th on succeeding.
Test 2: 31/36th (test 0 AND 1 failed) * 1/6th. Is 31/216th on succeeding.
Now, these 3 can be added up.
If every test would yield the same reward in loot. The total would be: (36+30+31)/216=97/216th
Or 44,91%
However, you suggest having different loot.
2 for test 0, 1 for test 1 or 2.
The average loot can be calculated now.
2*36+30+31=133
133/216=0.616 loot per roll.
Another way to look at it... out of what you had before (4/9 chance of a pair or triple), for any given pair (e.g. 224 or 656) there are two ways to have mismatched colors and one way to have matched colors. Triples I assume you'd still maybe want separate, which means
60/216 for white and black
30/216 for two whites
6/216 for a triple.
You had 44% chance of doubles/triples before. Coloring the dice did not increase that to 50% :)
You had 44% chance of doubles/triples before. Coloring the dice did not increase that to 50% :)
Hmm... A guy can dream, no?! I'm going to review the explanations in order to understand. I'm not 100% sure I understood both of your explanations.
6 x 6 x 6 = 216 outcomes.
Triples: 6/216. I understood.
2x White: 30/216. 5 x 6 pairs. I understood.
White + Black: 60/216. 5 x 6 pairs x 2 White dice. (I think?)
Many thanks @X3M and @Jay103!
There are 90 combinations of pure doubles, as follows:
Assume the first die is a 1.
If the second die is a 1, then the third can be any value 2..6 and give you pure double. 5 combinations.
If the second die is NOT a 1, then the third must be a 1. The second die (not being a 1), must be 2..6. 5 combinations.
Also, you get pure doubles if the second and third die themselves are a pair (but not a pair of 1's, which would give a triple). 5 combinations.
15 total pairs.
Six distinct values for the first die, not just the 1 we were assuming for the example. So 6 x 15 = 90.
Now, imagine you have a pure double. Two matching dice and one not. The dice that make up the pair are either WW WB or BW. 30 combinations each. (which is also "the pair is any value 1..6, and the leftover die is any different value, so 6 x 5 = 30")
BW and WB are indistinguishable, so 60 for that result, and 30 for WW.
Okay so here I go:
> Instead of triple (which is very low probability), what IF I had two (2) WHITE dice and one (1) BLACK die?
The results would look something like this:
A> Two (2) WHITE dice pair = 2 LOOT
B> White (1) + Black (1) pair = 1 LOOT
So I need probabilities for these results. I'm trying to determine if it's BETTER ODDS than a "triple" (which is much too low = 2.86%)
Thank you so much for helping. Math can be a challenge for me (at times). And believe me, it can be challenging for OTHER people too! (LOL)
Note #1: So please correct me if I am wrong!
A> The two (2) WHITE dice pair = 1/6 odds or 16.67% (not bad). The Black die is inconsequential (so no special treatment for that die).
B> White (1) + Black (1) pair = 2/6 odds or 33.33%
Total = 50% odds(!!!) Is this CORRECT?? Would be AMAZING if this was TRUE.
The chance of the two white dice matching, and the black dice not matching, is 1/6 x 5/6 = 5/36.
The chance of the two white dice not matching, and the black dice matching one of them, is 5/6 x 2/6 = 10/36 = 5/18.
The two situations are mutually exclusive (that is, they can't both happen in the same roll). So to get the chance of one or the other happening, you just add them: 5/36 + 10/36 = 15/36 = 5/12.
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Or
44,4%
Can you "explain" the results/math??? I'd be curious on how to compute this result! Thanks @X3M.