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What are the odds? DG? Little help?

Started by Berkut, January 12, 2010, 01:08:36 PM

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Berkut

So I am playing the new Battles for Normandy game. part of the landing process is to drop your naval support units into place to provide NGF support.

I just put down ten naval units. Each unit rolls 1 1d10 to see if it hits a mine and is removed from the game.

Rolled 5 10s out of the ten dice.

What are the odds of rolling x 10s on ten d10 dice?

I know is is not as simple as one would think, but is there a formula to figure out just how unlikely this result would be?

An "average" roll is obviously 1, but how do you figure how much more unlikely 2 is than 1, 3 than 1, 4, than 1, etc?
"If you think this has a happy ending, then you haven't been paying attention."

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Habbaku

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Tamas

Err, we learned this in college it can't be that hard... altough I dont remember much of it : (

frunk

If X is the number of 10s:

((9/10)^(10-X)) * ((1/10)^X) * (10!/(X!*(10-X)!))

0: .349
1: .387
2: .194
3: .057
4: .011
5: .001

Tamas


frunk

Generalizing, if X is the number of desired results, P is the decimal probability of getting the result and N is the number of attempts:

(1-P)^(N-X) * P^X * N! / (X! * (N-X)!)

Berkut

"If you think this has a happy ending, then you haven't been paying attention."

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The Brain

It is unlikely that very unlikely events occur.
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Grey Fox

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frunk

Factorial.  3! is 3*2*1.  10! is 10*9*8*7*6*5*4*3*2*1

Grey Fox

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ulmont

Quote from: frunk on January 12, 2010, 02:04:35 PM
If X is the number of 10s:

((9/10)^(10-X)) * ((1/10)^X) * (10!/(X!*(10-X)!))

0: .349
1: .387
2: .194
3: .057
4: .011
5: .001

You don't need all of that effort since the probabilities aren't unconnected.  To get 5 out of 5, he had to roll every die a 10, meaning it's just (10%) * (10%) * (10%) * (10%) * (10%).  Still .001% though.

Berkut

But we are not talking about 5/5, but 5/10.
"If you think this has a happy ending, then you haven't been paying attention."

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frunk

Quote from: ulmont on January 12, 2010, 03:05:58 PM
You don't need all of that effort since the probabilities aren't unconnected.  To get 5 out of 5, he had to roll every die a 10, meaning it's just (10%) * (10%) * (10%) * (10%) * (10%).  Still .001% though.

He's not talking about 5 dice rolling 10s, he's talking about 10 dice, with 5 of them coming up 10s.  If it was 5 dice needing all 10s you'd be right.  Note that my numbers are decimal, not percentage, so 0.1% for rolling 10 dice and 5 of them coming up 10s.

ulmont

Quote from: Berkut on January 12, 2010, 03:08:18 PM
But we are not talking about 5/5, but 5/10.

Ah, I'm sorry.  I thought you were whining about getting diced, not just having a weird streak.