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Thread title is pretty self explanatory.

Would anyone who is particularly math-inclined care to share the exact probability of rolling Fortune between two Farseers (and lay out the math for me?). Thanks a bunch!

It depends what your process actually is.

If you're going all-out for Fortune - that is, you're taking 3 rolls with each Farseer on Fate and you're not going to take the primaris until the third roll (if ever), then a Farseer has a 1/6 chance of rolling Fortune on his first roll. 5/6 of the time, he'll fail to get Fortune and will accept another power. Then on the second roll he has a 1/5 chance of rolling Fortune (because he rerolls if he gets the same power twice). 4/5 of the time he fails and takes another power. Then on the third roll he has a 1/4 chance of rolling Fortune.

So for one Farseer the best chance you can ever have of getting Fortune is (1/6 + 5/6*(1/5 + 4/5*1/4)) = 50%.

Then the chance of getting it on at least one Farseer out of two is just 1 - .5^2 = 75%.

Edit: The really easy way of seeing that this is right is to note that if you're taking 3 random powers from Fate, excluding the primaris, then you're just getting one of the possible sets of 3 powers chosen at random. There are six total powers, so half of the sets of powers will have any particular power in them. 50%.

I don't actually know what odds there are, but I assume you are rolling a bunch of d6 and hoping for one specific number. Here's how to calcualte the odds of getting something with re-rolls, this will help you with twin linked weapons too if you're curious about that.

Here's how you do it: first, I assume you have a 1/6th chance of getting the power you want (this starting number can be anything you want, depending on what the odds actually are).

I'll just assume a warlock gets to roll 2 dice for the example, but it's easy to get whatever you want.

so, the odds of success are 1/6 which means the odds of failure are (1-1/6=) 5/6. If you get to roll twice then the odds of failure = (previous odds of failure)^2, which comes out to (5/6)^2 = 25/36 or 0.69. Thus, the odds of success (i.e. that one warlock getting the power you want) is (1-0.69)= .31 or just under a third.

If you're rolling more d6 just replace the "^2" with "^x" where x is the number of dice you're rolling.

NOTE: this doesn't take into account re-rolling if you get the same power on both dice, but I doubt that will skew the effects too far wither way, and quite honestly I don't know how to calculate that in short of Monte Carlo (i.e. generating it randomly a bunch of times and graphing the probability curve).


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Ninjad, and by someone with much better grasp of the rules. I hope you get what you need between the two of us (and yea I realize it's not a warlock now).

Heh yeah dementedwombat the idea here is a farseer gets Three rolls on a table to get one power fortune. Three rolls, and say you dont get it, you can take that power, and now that power is not an option. So for example if you roll a "1" and take that power, any further rolls of 1 are rerolled to get other powers. In this case we have 2 farseers with 3 rolls on the same table attempting to get fortune. Anything else to add now that you know more about it?

THANKS PANTSONHEAD! and hopefully this will be useful to someone other than myself.

Yep, just figure out the odds for one "instance", i.e. for one model to get the roll you want (that's not official probability terms or anything, just something I made up) then use my method with the number of instances and the probability you already found.

The first post got it right. I just wasn't clear on how the system worked (playing Tau you can hopefully understand why) although the method I gave tweaked for 3 rolls is only off by about 6%...

Resurrecting this thread from the grave.

I am calculating a 100% chance for two or more level 4 psykers to roll a specific power at least once.

That can't be correct, even with re-rolls it still cannot be a 100%.
Can anyone tell me the actual chance?

Very close to around 88%

chance of one ML4 psyker getting a specific power is 4/6. Chance if either psyker getting a specific power is 1-(1-4/6)^2 = 8/9. If they both had re-rolls the chance is 1-(1-4/6)^4 ~ 98.8%

It doesn't matter if you had 100 psykers. The chance will never hit 100%.