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How many 6s will you get when rolling 6d6?

Well, what we would expect, that being the result being perfectly divided up between all the possible outcomes, would be one 6. However, and most of you know this, sometimes you will roll 6d6 and get no 6s or 2 sixes.

This is because getting a 6 on 6d6 is not the PROBABLE outcome, that being the most likely outcome on any single event.

So what are the chances of getting a 6 on 6d6 well it comes out to 66%

What you do is take 1and subtract the number of possible failures per dice which in this case is 5, and divide number of possible outcomes 6 (leaving you with only the success), then factor the number of dice which you make the exponent, you will end up something like this.

1-(5÷6)^6 = 66%

So we have a 66% chance of getting a 6 on 6d6.

Why is this useful? Well, if you for example are casting Tzeentch Firestorm and want to know the probablity of getting 1 wound when rolling the 9d6, all you have to do is change the exponent which is currently a 6 to a 9.

1-(5÷6)^9 = 80.6% for a wound.

So you can reasonably expect to get one wound 80% of the time but there is the possibility that there are so many dice that getting only one wound is unlikely. It may be the case that you are more probable of getting more than one wound. This equation only checks to see if you will get at least one wound. So that is not a concern here.

I'm not sure why people insist on using Ork math but stop it it's annoying.

And all math and probability really doesn't matter in the end. There will be the game where nothing works, and there will be the game where everything is on fire.

Yeah. In my opinion the difference between a Good and a Great player is that a Good player can do statistics and predict the most likely outcome, but a Great player can compensate for when nothing goes as planned.

I played against some great players at NOVA (the three superheavy tank list, even with the Index, vaulted me way high for a bit after my first couple games before it got creamed lol), and Alex Fennel (of recent played-Tony-during-the-infamous-game-at-LVO fame) fought me. It was a brutal game, and my dice were absolutely on fire, but he crushed me fairly handily, because he had Plan A, Plan B, and Plan C. While me, being merely a fairly good player with a strong list, saw Plan A coming, and when he shifted to plan B after some of my hot dice, I gradually understood and countered it, but Plan C got off without a hitch as I struggled to compensate for my own good dice rolls (finding new targets for the tanks main guns was sloppy, and lots of really powerful firepower went into Brimstone Horrors and whatnot just for lack of visible targets. Had I moved better, this would not have been the case.).

 Thousand-Son-Sorcerer wrote:


Why is this useful? Well, if you for example are casting Tzeentch Firestorm and want to know the probablity of getting 1 wound when rolling the 9d6, all you have to do is change the exponent which is currently a 6 to a 9.

1-(5÷6)^9 = 80.6% for a wound.

So you can reasonably expect to get one wound 80% of the time but there is the possibility that there are so many dice that getting only one wound is unlikely. It may be the case that you are more probable of getting more than one wound. This equation only checks to see if you will get at least one wound. So that is not a concern here.

Except this is not what you claimed in the other thread. You were trying to work out the number of wounds per casting of a spell and you made a couple of mistakes, not least by suggesting that you would get 'average 1 6 per 9 d6 rolled not 1.5'. Sure, you have an 80% chance of at least 1 wound, but that doesn't mean you'll get exactly 1 wound all of the time (from memory I think it's around 35% for one wound, and around 45% for 2+ wounds).

 Thousand-Son-Sorcerer wrote:
How many 6s will you get when rolling 6d6?

Well, what we would expect, that being the result being perfectly divided up between all the possible outcomes, would be one 6. However, and most of you know this, sometimes you will roll 6d6 and get no 6s or 2 sixes.

This is because getting a 6 on 6d6 is not the PROBABLE outcome, that being the most likely outcome on any single event.

So what are the chances of getting a 6 on 6d6 well it comes out to 66%

What you do is take 1and subtract the number of possible failures per dice which in this case is 5, and divide number of possible outcomes 6 (leaving you with only the success), then factor the number of dice which you make the exponent, you will end up something like this.

1-(5÷6)^6 = 66%

So we have a 66% chance of getting a 6 on 6d6.

Why is this useful? Well, if you for example are casting Tzeentch Firestorm and want to know the probablity of getting 1 wound when rolling the 9d6, all you have to do is change the exponent which is currently a 6 to a 9.

1-(5÷6)^9 = 80.6% for a wound.

So you can reasonably expect to get one wound 80% of the time but there is the possibility that there are so many dice that getting only one wound is unlikely. It may be the case that you are more probable of getting more than one wound. This equation only checks to see if you will get at least one wound. So that is not a concern here.

I'm not sure why people insist on using Ork math but stop it it's annoying.

But that was not your claim. You made two claims in the other thread: That the average number of sixes rolled with 9D6 is 1 (it is 1.5). And that the chance of rolling a six on a D6 is 11% (it is 16.7%). Both of those claims are wrong.

Was this thread really needed? I'm not saying this to be mean, I mean this is entire thread is literally responding to the discussion in another thread.

 MechaEmperor7000 wrote:
Was this thread really needed?

No.