this for odds masters
17 posts
• Page 2 of 2 • 1, 2
The only thing you need to answer these types of questions are:
1) how combinatorials work
2) common sense
The formula Adam gave is simply the hypergeometric distribution, which is the appropiate distribution. (This is a distribution that's tougher to memorize than derive.)
Let's say you had 7 lands in a deck of 60 and you wanted to know the probability of drawing exactly 4 lands in 7 picks without replacement.
There are 60c7 different ways to pick 7 cards from 60 (this that 'how combinatorials work' thing) of cards you can pick.
There are 7c4 different ways to pick 4 lands (without reguard to the other cards).
If you pick 4 lands there are there will be 3 non-lands. There will be 53c3 ways to pick those non-lands.
There are a total of 7c4*53c3 ways to pick your 4 lands (with reguard to non-lands).
So the probability of picking exactly 4 lands is 7c4*53c3/60c7 (note that this is the same thing Adam got).
n! =
n*[(n-1)*(n-2)*...2*1] =
n*[(n-1)!]
so...
n! = n*[(n-1)!]
plugging 1 in for n shows:
1!=1*0!
1 = 0!
1) how combinatorials work
2) common sense
The formula Adam gave is simply the hypergeometric distribution, which is the appropiate distribution. (This is a distribution that's tougher to memorize than derive.)
Let's say you had 7 lands in a deck of 60 and you wanted to know the probability of drawing exactly 4 lands in 7 picks without replacement.
There are 60c7 different ways to pick 7 cards from 60 (this that 'how combinatorials work' thing) of cards you can pick.
There are 7c4 different ways to pick 4 lands (without reguard to the other cards).
If you pick 4 lands there are there will be 3 non-lands. There will be 53c3 ways to pick those non-lands.
There are a total of 7c4*53c3 ways to pick your 4 lands (with reguard to non-lands).
So the probability of picking exactly 4 lands is 7c4*53c3/60c7 (note that this is the same thing Adam got).
red_pen0 wrote:0! is by definition equal to 1 so the equation holds.
n! =
n*[(n-1)*(n-2)*...2*1] =
n*[(n-1)!]
so...
n! = n*[(n-1)!]
plugging 1 in for n shows:
1!=1*0!
1 = 0!
-

suitedaces84 - Posts: 2398
- Joined: Sun Dec 12, 2004 8:13 pm GMT
- Location: A van down by the river
17 posts
• Page 2 of 2 • 1, 2
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