What are the dice odds for a 'take highest' roll?

Thanks.

Eh?  Not much to work off of here. I assume you are asking for rolls like 'Everyone roll X dice and the guy who rolled highest wins!"

In which case the odds are essentially dependent upon the number of players. Between two I think that the average to be the 'high roller' works out to... wait for it... one in two.  50/50.

Obviously, three rollers becomes 'one in three' or roughly 33% and so on...

Provided everything else was equal.

Er, whoops.

I meant something like, roll NdX and take the highest dice - I'm looking for a general formula that'll take any NdX.

Hrm... well, you also have to take into account open ended dice (a la earlier editions of Shadowrun) and so on..

Napkin math says that NdX start with the odds where N equals 1 (one die) and computing the odds for X to a given value (1d6, means a 1 in six chance to get a six, 15% more or less).  As N increases the chance to get X out of any given roll (where X is the highest value on the die, again...) increases fairly evenly. Thus 2d6 would produce a 2 in six value, or 1/3 ratio for getting one six (33%, or roughly double the chance of 1d6) and so forth.

Make sense?

Calculator says that comes out to 1.83~/6, or 30.5~%.

Thanks, that should work into an equation.

I'm skeptical that it's that simple.

I feel that this is a special topic which was covered in a class I took, but at the moment the only approaches I can think of are essentially brute force (list every combination) and monte carlo (run a large number of simulations with a good random number generator).

I've got an inkling of another approach but I can't look into it right now.

For your best result to be n, you have to get:

a) no results higher than n, and
b) at least 1 result of n.

Let Y be the number of rolls.
Let d be the size of the dice.

Looking first at a)

The chance of getting a result higher than n on a single roll is (d-n)/d.

The chance of not getting a result higher than n on a single roll is n/d.

The chance of not getting a result higher than n on Y rolls is (n/d)^Y.

So a) is (n/d)^Y

Looking at b):

Given that a) is true, what are the odds of getting at least 1 result of n.

Given that you haven't gotten any results higher than n, ie all your rolls have been 1 to n, your chance of getting n on a given roll is 1/n.

Your chance of not getting n on  a single roll is (n-1)/n.

Your chance of not getting n for Y rolls is ((n-1)/n)^Y.

Therefore your chance of getting n at least once, for Y rolls, given that no roll has been over n, is 1-(((n-1)/n)^Y)


Therefore, your chance of both a) and b) being true should be

((n/d)^Y) * (1-(((n-1)/n)^Y))


[I've tried it with n=3, Y=2, d=6  and n=1, Y=6, d=6. It gives the right answer both times]

...I shall take a look at that when my brain is not fried from programming a roguelike.

AoF's method is what I was feeling my way toward, so I agree, it's probably right.

Thank you. That should help nicely.

Sorry if this is a silly question, but what does "^" mean? If I was taught that in school, I can't remember. And by the way, what if you roll several dice of different sides? I ask because in Usagi Yojimbo you roll several different dice and take the highest, so it would be useful to know.

Thank you in advance.

'To the power of'. It denotes an exponent.

Quote from: Claudius;283588Sorry if this is a silly question, but what does "^" mean? If I was taught that in school, I can't remember. And by the way, what if you roll several dice of different sides? I ask because in Usagi Yojimbo you roll several different dice and take the highest, so it would be usefult to know.

Thank you in advance.

That's right, '^' means 'to the power of'.

X to the power 2 is X times X.

X to the power 3 is X times X times X.

X to the power 4 is X times X times X times X , and so on.


As for rolling several different dice: I'll try to work it out as soon as I can.

Great! Thank you, both of you!

Age of Fable: I'll be waiting eagerly. One of the PC rolls 1d12+2d6 for attack, and the other one rolls 1d10+1d8+1d6, and I've got no idea which one is best.

Quote from: Claudius;283619Great! Thank you, both of you!

Age of Fable: I'll be waiting eagerly. One of the PC rolls 1d12+2d6 for attack, and the other one rolls 1d10+1d8+1d6, and I've got no idea which one is best.

Do you count the 2d6 as two seperate totals, or as a single total?