38 coaches online • Server time: 15:28
Forum Chat
Log in
Recent Forum Topics goto Post Finishing the 60 Gam...goto Post ramchop takes on the...goto Post Borg Invasion
SearchSearch 
Post new topic   Reply to topic
View previous topic Log in to check your private messages View next topic
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.
Display posts from previous:     
 Jump to:   
All times are GMT + 1 Hour
Post new topic   Reply to topic
View previous topic Log in to check your private messages View next topic