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MattDakka



Joined: Oct 09, 2007

Post   Posted: May 18, 2024 - 19:31 Reply with quote Back to top

To be accurate, Blitz! chance is 8.33 %, not 9%.
Rolling a 10 with 2D6 is 3/36 = 1/12 = 8.33%.
Primarch



Joined: Dec 14, 2003

Post   Posted: May 18, 2024 - 20:35 Reply with quote Back to top

I never said I lose because of the dice, I just asked what the odds were for 3 blitzes in a row, lol. I also get that I deployed poorly, that is why I said I was trying to learn how to play them, obviously the combo of trying to learn, and rolling 3 blitzes in a row is a recipe for disaster.
RDaneel



Joined: Feb 24, 2023

Post   Posted: May 18, 2024 - 22:18 Reply with quote Back to top

-

_________________
To judge a man, one must at least know the secret of his thoughts, his misfortunes, his emotions, Balzac


Last edited by RDaneel on %b %18, %2024 - %23:%May; edited 1 time in total
RDaneel



Joined: Feb 24, 2023

Post   Posted: May 18, 2024 - 22:21 Reply with quote Back to top

the number of events of the kick off table is 11: 2 : get the ref, 3, time out et... 12 pitch invasion
so the probability that 1 of this event on a space of event = 11

probability of single occurrence is 1/11

(edit: this is wrong)

_________________
To judge a man, one must at least know the secret of his thoughts, his misfortunes, his emotions, Balzac


Last edited by RDaneel on %b %18, %2024 - %23:%May; edited 1 time in total
MattDakka



Joined: Oct 09, 2007

Post   Posted: May 18, 2024 - 22:35 Reply with quote Back to top

The probabilities of these 11 results are not all the same, because 2D6 are rolled:
the probability of rolling a 10 (Blitz! event) is 8.33%;
the probability of rolling a 2 (Get the Ref) is 2.77%;
the probability of rolling a 12 (Pitch Invasion) is 2.77%;
the probability of rolling a 7 (Brilliant Coaching) is 16.66%.

The probability of any event would be 1/11 if a D11 were rolled, but 2D6 are rolled, not a D11.


Last edited by MattDakka on %b %18, %2024 - %22:%May; edited 1 time in total
RDaneel



Joined: Feb 24, 2023

Post   Posted: May 18, 2024 - 23:13 Reply with quote Back to top

mmmmmmmm too many vodka
with 2d6 the number of possibilities (space of events ) is 36
blitz can happen only if these rolls happen
5+5
4+6
6+4

so the possibility of a blitz! is 3(success) / 36 (number of possibilities)
8,3%
you were right

_________________
To judge a man, one must at least know the secret of his thoughts, his misfortunes, his emotions, Balzac
JackassRampant



Joined: Feb 26, 2011

Post   Posted: May 19, 2024 - 06:59 Reply with quote Back to top

Even better: on 2dX, the curve is a 45ยบ spike, with N-1 permutations (out of X^2) for every positive integer N up to X+1, then 1+2X-N permutations for integers above X. So on 2d6, it's 6 chances to roll a 7, 5 chances to roll a 6 or 8, 4 chances to roll a 5 or 9, etc. Blitz comes on a 10, thus 3 chances, but you can use this math for any kickoff result or any other 2d6 table, including AV and Injury.

Also useful in this game: if you're rolling 2 dice and angling for the right one, like on a 2d block, your chance of getting any given result N as your highest roll is (2N-1)/X^2, so 11/36 to get 6+ with a reroll or an open pow (if that's what you want), 9 chances that your highest roll will be a 5 (or a pow/push or whatever), 7 for the next, then 5, 3, and 1 for doubleskulls or snakes. Similarly, your chance of succeeding on two consecutive equal 1d6 rolls is 1/1, minus that chance, which is equal to N^2/X^2 (because the difference between two consecutive perfect squares is always the sum of the roots).

Also also useful is the 3d formula, which is (1+6T(N-1))/X^3. So for a 6, that's 1, plus six times the fifth triangle number, or 1 plus 6(1+2+3+4+5), or 91. So if you're throwing a 3d block and there are two good results, your success chance is 91+61 out of 216, or 152/216. If you don't like factoring out powers of 2 to reduce, you could do this on the individual dice: 2/6 becomes 1/3, so your chance of hitting that score is (1+6T(2))/3^3, or 19/27. The converse, the failure chance or the odds on 3 consecutive equal dice, are N^3/X^3 (and X^3 is 216 on 3d6, but of course it'd be 512 on a scatter).

This game is good for your arithmetic muscles, if you let it be.

_________________
Lude enixe, obliviscatur timor.
MattDakka



Joined: Oct 09, 2007

Post   Posted: May 19, 2024 - 11:41 Reply with quote Back to top

Or, if you don't want to bother with probability calculations, use this:
http://www.elyoukey.com/sac/
It's better to be able to calculate the odds with your own mind, but sometimes, in the heat of a match, it could not be easy.
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