What are the odds? - Stat geeks help!

[reveal]

Ok... I've created an Epic level monk who gets 4 attacks per round normally and 6 attacks per round in a flurry of blows. (not important how or why; that's just the number of attacks)

Anyway, I'm thinking of taking the Epic Feat Vorpal Strike. It allows the unarmed strike to be considered a Vorpal weapon. So on a natural 20, it kills.

What are the odds of me rolling a single natural 20 out of the four dice I roll for a normal round?

What are the odds of rolling a single natural 20 out of the six dice I roll for a full-round action?

What are the odds of rolling 4 natural 20's out of the 4 dice I roll for a normal round?

What are the odds of rolling 6 natural 20's out of the 6 dice I roll for a full-round action?

Mainly I'm just concerned about the first two questions but I thought the last 2 would be fun. I really would like to know this because it would help me determine if I want to take the feat or not.

 

[kayn99]

If I remember

If I remember from my old stats class, each die roll is independent of each other. That means you 5% out of 100 on each die. There is a 160,000 different combos of numbers on those 4 dice and only 1 will produce the effect you want so maybe 1 out of 160,000.

 

[DanMcS]

Ok... I've created an Epic level monk who gets 4 attacks per round normally and 6 attacks per round in a flurry of blows. (not important how or why; that's just the number of attacks)

Anyway, I'm thinking of taking the Epic Feat Vorpal Strike. It allows the unarmed strike to be considered a Vorpal weapon. So on a natural 20, it kills.

Actually, on a natural 20, it threatens a kill; you still have to roll to confirm the crit.

What are the odds of me rolling a single natural 20 out of the four dice I roll for a normal round?

The odds of rolling at LEAST 1 natural 20 are 18.55%.

What are the odds of rolling a single natural 20 out of the six dice I roll for a full-round action?

The odds of rolling at LEAST 1 natural 20 are 26.49%.

What are the odds of rolling 4 natural 20's out of the 4 dice I roll for a normal round?

1 in 20^4, or 1 in 160,000.

What are the odds of rolling 6 natural 20's out of the 6 dice I roll for a full-round action?

1 in 20^6, or 1 in 64,000,000.

 

[reveal]

Thanks!

Here's another question: My crit range is 18-20. Using the 4 questions above, and assuming that me rolling an 18 or a 19 does actually hit, what are the odds of me rolling a critical threat in each scenario?

 

[Creamsteak]

Depends on the AC of your opponent.

 

[nhl_1997]

Threat range 18-20 (assuming an 18 hits):

Four Dice
Exactly one threat: 37%
At least one threat: 48%
Four threats: 5 in 10000

Six Dice
Exactly one threat: 40%
At least one threat: 62%
Six threats: 1 in 100000

Note, these are just threats.... confirming the threat has not been done.

 

[Zoatebix]

Maybe I'm using the binomial distribution equation wrong, but I didn't get what you got, DanMcS...

Edit: Wait, I think there's an error in how I entered it into my calculator...

Edit number 2: In fact, I'm using the incorrect equation all together!

 

[nhl_1997]

What are the odds of me rolling a single natural 20 out of the four dice I roll for a normal round?

What are the odds of rolling a single natural 20 out of the six dice I roll for a full-round action?

Unless the core rules change for epic characters, remember that you still need to use a full-attack four the four attacks in a round.


Anyway, I also am getting some slightly different results for your first question than what was posted previously. (edit) Upon further reflection, these are in fact the same results as previously posted.


Four Dice
Exactly one 20: 17%
At least one 20: 18.5%
Four 20s: 6 in a million

Six Dice
Exactly one 20: 23%
At least one 20: 26.5%
Six 20s: 1.5 in a hundred million

 

[reveal]

Depends on the AC of your opponent.

That's why I changed it from critical hit to critical threat. Assuming an 18 hits, that's a critical threat. I know it would be much harder to figure out a critical hit since, as you said, you have to know the AC.

 

[Zoatebix]

Yup I was using the equation for exactly one natural 20 in four tries.

What's the equation for at least "r" number of successes, given the probability "p" for success and "N" number of tries? I don't have a good text book on hand: I was using my memory (from 6 years ago in 11th grade) and Google for reference.
-George