Dice Math

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Dice Math

Postby VampireM on Sat Dec 14, 2013 8:27 am

I was just curious if someone could shed some light on the math of how our dice stats percentages are worked out

3 dice v 2 dice should win 65.595% of the time. Is it incorrect to assume that my overall 3v2 battle outcomes have the same results?

Assault 35295 29751 = 0%
Defend 17088 20508 = -1%

I win my assaults only 54.2%
My opponent wins their assault 54.5%
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Re: Dice Math

Postby Swifte on Sat Dec 14, 2013 11:11 am

It's true that a in a 4 v 2 battle, where you start with 3 dice and roll against 2 until one side or the other wins, the attacker should ultimately win the war 65.59% of the time.

However, what your dice stats are showing is only the result of the first roll (the one roll where you actually have 3 dice against 2). On that roll, you could win 2, win 1 lose 1, or lose 2. On that roll, your dice stats are showing that you kill an army 54.2% of the time and lose one 45.8% of the time.

That sounds roughly correct, because on a 3 v 2 battle, you have roughly 37.2% chance of killing 2, 33.6% chance of killing 1, 29.2% chance of losing 2. If you run the math on that, you get an expected kill rate of 53.8%, so your average assault rate is very close to the expected outcome for a single 3 v 2 roll.

Assault odds are accounting for the fact that after that first roll, there is a chance you could still win the war with a 3 v 1 or even a slim chance you could come back and win a 1 v 2. So you start from your 37.2% chance of winning 2 outright, and add to that your chance of wining 1 on the first roll and ultimately winning the fight, plus the slim chance of losing 2 on the first roll but still winning the fight, and you get to 65.59%, which is what assault odds is showing.

That's the difference... hope that helped.
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Re: Dice Math

Postby VampireM on Sat Dec 14, 2013 5:22 pm

yep thank you, should of just pm'ed u
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