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1775: Rebellion» Forums » General

Subject: Funny battle rss

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Bill H
United States
Rensselaer
New York
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Just a quick story I wanted to share. Last evening, I attacked a lone Loyalist with 2 Continental and 3 Militia armies. The Loyalist whiffed, but had nowhere to retreat to.

I then rolled my 5 dice for the attack... All runners.
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Barry Miller
United States
Saint Charles
Missouri
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The stats nerd in me, if I'm doing my math correctly, informs you that your probability of rolling five "Runs" in that situation was .001 !

I.e., One out of a thousand!

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Lewis Karl
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Vienna
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Oh, no no no. 1.286 in a thousand. laugh
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Bill H
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The result just cracked me up. I just pictured the scenario afterward. A small army being attacked and vastly outnumbered 5-1 with nowhere to run mounts a feeble defense by saying "Boo!".

Then the attacking army gets spooked and runs for the hills.

The British ended up winning the game too, which seems appropriate after that fiasco.
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Alex Drazen
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bgm1961 wrote:

The stats nerd in me, if I'm doing my math correctly, informs you that your probability of rolling five "Runs" in that situation was .001 !

I.e., One out of a thousand!



I get 1.3% chance of five runners for 3 militia/loyalist and 2 regular army units:

(0.5) * (0.5) * (0.5) * (0.333333) * (0.33333)

It's not 1775, but this reminds me, I swear my 1812 dice are loaded or cursed. Several times I have sent an American force to recapture Detroit only to roll 4-5 runners each and every freakin' time!
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Gary Selkirk
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Truro
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This is a really neat thing to see. The 1775 & 1812 games work perfectly this way. I've had battles in both games where the defender was badly outnumbered and stayed in battle to win.
 
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Barry Miller
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alexdrazen wrote:
I get 1.3% chance of five runners for 3 militia/loyalist and 2 regular army units:

(0.5) * (0.5) * (0.5) * (0.333333) * (0.33333)

Hmmm... are we using the same dice?

- A Continental "Runs" on one out of six sides (.1666)
- A Militia "Runs" on two out of six sides (.3333)

So the math would be (I believe):

(.1666)*(.1666)*(.3333)*(.3333)*(.3333)= .00102 = Probability of rolling 5x"Run" results out of 2xContinental and 3xMilitia dice is 1/1000[/b]

(I think I got that right. Please correct me if I'm wrong. I'm not a smart stats nerd!)



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Alex Drazen
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bgm1961 wrote:

alexdrazen wrote:
I get 1.3% chance of five runners for 3 militia/loyalist and 2 regular army units:

(0.5) * (0.5) * (0.5) * (0.333333) * (0.33333)

Hmmm... are we using the same dice?

- A Continental "Runs" on one out of six sides (.1666)
- A Militia "Runs" on two out of six sides (.3333)

So the math would be (I believe):

(.1666)*(.1666)*(.3333)*(.3333)*(.3333)= .00102 = Probability of rolling 5x"Run" results out of 2xContinental and 3xMilitia dice is 1/1000[/b]

(I think I got that right. Please correct me if I'm wrong. I'm not a smart stats nerd!)



You're right. I really haven't played 1775 or 1812 in a while and I inadvertently calculated the hit probability for five dice, not the flee probability. And not even that, as I assigned the Regular Hit as the Irregular Flee, and vice-versa.

It sure seems like half my "irregular" units flee, though

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