Chance of getting 33 out of 3d100?

[AstralFire]

I have no idea what kind of roleplaying system requires this calculation, but my friend's apparently in a game where he needs to know this, and I don't remember covering this in Psych Statistics. Googling wasn't any help; I don't even recognize some of the symbols I encountered. Help?

[Tyndmyr]

.65%, give or take about .02% for rounding errors. At least, that's if you're going for a total of 33 or less on 3d100.

Anydice.com is really your best bet for this.

[AstralFire]

I meant just 33, not 33 or less. Brainfart. Thanks!

[Quimquay]

Well, with three dice, there are 10^6 possible combinations, 496 of which have the result 33, so the probability is 0.000496 (or, 0.0496%).

[Hylas]

The odds of rolling a 33 on 3d100 is 2.97%
The odds of rolling exactly 1 33 on 3d100 is 2.94%
The odds of rolling exactly 2 33s on a 3d100 is 0.03%
The odds of rolling exactly 3 33s on a 3d100 is less than 0.01%

I use this calculator for figuring out strange dice rolling systems and for game testing. http://www.fnordistan.com/smallroller.html

Oh you mean total. Nevermind, the guy above me answers.

[nedz]

I meant just 33, not 33 or less. Brainfart. Thanks!

Why are you calling Tyndmyr Brainfart ?

BTW What exactly do you mean by 3d100 ?
For example you could mean: if I roll d100 3 times, then what is the chance of getting a 33 ? The answer to this would be 3% ?

[Tyndmyr]

Why are you calling Tyndmyr Brainfart ?

BTW What exactly do you mean by 3d100 ?
For example you could mean: if I roll d100 3 times, then what is the chance of getting a 33 ? The answer to this would be 3% ?

I presume he's saying that he had a brainfart(common slang in some areas) in not specifying.

That said, anydice.com does also answer that quite efficiently.

[Binks]

I love die roll calculations .

Total of 33: 0.326%
See at least 1 33 on a die: 2.97%
See exactly 1 33 on a die: 2.94%
See exactly 2 33s: 0.0297%
See exactly 3 33s: 0.0003%

Calculations:

Spoiler

Show

Total of 33 - We have basically Sum(X+Y+Z) = 33 and we're trying to find the possible values of X/Y/Z. To do this simply we can set one of them and derive the number of possibilities of the others, say Z=1 then there are 32 (31+1,30+2 etc) integer combinations of X and Y that make this equation true. Set Z=2 and there are 31 combinations. Z=3 means 30 combinations, etc. So we can say that whatever we set Z to we get 33-Z combinations, so it's a simple sum(33-Z) over Z from 1 to 31 (as 32 and above aren't valid).
sum(33-Z) over 1<=Z<=31 = 528 possible combinations where we have a total of 33.

There are 161700 possible combinations of 3d100 (100 choose 3) so the odds of getting a total of exactly 33 are 528/161700 = 0.00326530612, or 3 tenths of a percent.

See 33 on a die - With each die there is a 1/100 chance of a 33 coming up. Thus the odds of seeing at least of of those dice come up is the inverse of the odds that all 3 come up non-33s, or 1-(99/100*99/100*99/100) = 0.029701 so about a 3% chance of seeing a 33 come up.

The odds of seeing exactly 1 33 are the odds that 1 die has a 33 and 2 others do not, or 1/100*99/100*99/100 = 0.009801, times 3 possible combinations = 0.029403

The odds of seeing exactly 2 33s are the odds that 2 dice have 33 and 1 does not, or 1/100*1/100*99/100 = .000099, times 3 possible combinations = 0.000297

The odds of seeing exactly 3 33s are the odds that 3 dice have 33, or 1/100*1/100*1/100 = .000001, times 3 possible combinations = 0.000003

Summing those three values gives us 0.029703, the same odds of seeing at least 1 33 (within rounding error bounds).