# how do you figure out these odds?

i have 52 different cards in a standard deck.

i take out 20 cards from that deck.

now how do i figure that out? plz leave ur calc’s to help me understand how you came up with your answer.

Are the twenty cards random, or do you KNOW that they contain the cards that you want?

The number of ways of choosing 20 cards from 52 is ( by definition) 52C20.
How many of these contain the 5 required cards? Put those cards in and we can then choose the other 15 from the remaining 47 cards. This can be done in 47C15 ways.
The probability is thus 47C15/52C20, roughly 0.006, for odds of about 167-1.

it’s a well-shuffled deck so you don’t know what you’re getting in the 20 you pick from the full 52.

This is an odd question, and I’m not entirely sure that I understand it, but I’ll have a shot.

Let’s split it out into 2 parts,

1, what are the odds that when you remove your 20 cards the flush is in the 20 you picked out are, well the odds that the first card is in there is 20/52, which is the same for each card. So the odds that all of your cards are in the 20 are (20/52)^5

2, Now let’s assume that your cards are in the 20, so you have a chance of picking your flush. We start picking cards randomly. The odds of getting a good card on the first go is 5/20, on the second go it’s 4/19, so we see the odds of getting them all are

5/20 * 4/19 * 3/18 * 2/17 * 1/16

Which is (15! * 5!)/20!

So the odds of doing both step 1 followed by step 2 is

((20/52)^5)*(15! * 5!)/20!)

Here’s my approach:

Imagine 52 slots, with a divider separating twenty spots on the right from 32 spots on the left. If we put one card at random in each slot, what is the probability that the ace of spades is to the right of the divider?

It’s 20/52.

Now, given that there is an ace of spades to the right of the divider, what is the probability that the king of spades is to the right of the divider too? It’s 19/51, since one spot is already taken up.

So IMHO, the probability is:

(20/52) * (19/51) * (18/50) * (17/49) * (16*48)

This works out to about 0.006

So it’s probably equivalent to Jabba’s approach.

I take it you just want those five cards in the group of twenty.

20/52 * 19/51 * 18/50 *17/49 * 16/48

This appears to be the same answer as Jabba.

And pretty damn similar to lucwarms, lol.

I’ve got to learn to save my answer and hit refresh before posting.