Probability of poker card draw?

Question

you are dealt 5 cards in a game of draw poker, and 3 of them are 5's what is the probability you will get another 5 if you draw 2 cards?

2/47
2/49
1/6
2/13


Answers please!

Answer 1

Exactly 2 / 47, so the previous answerer was wrong.

You are drawing from 47 unknown cards.
Chance of (5, not 5) = (1 / 47) * (46 / 46)
Chance of (not 5, 5) = (46 / 47) * (1 / 46) = same.

Answer 2

It depends if there is a possibility of drawing the cards you discarded (which we know weren't 5s). (I'm not sure about the rules here.) Assuming there isn't, then:

the probability of NOT getting a 5 with the first card is 48/49

and of NOT getting a 5 with the 2nd card is 47/48 (assuming the first one wasn't a 5)

so the probability of NOT getting a 5 at all is (47/48)*(48/49)

and therefore the probability of getting a 5 is 1 - (47/48)*(48/49)=2/49


And now assuming that you can't draw a card you have already discarded:

P(not 1st one)= 46/47
P(not 2nd one)= 45/46
P(neither)= (46*45)/(47*46)
P(either card is a 5)=1- P(neither)= 2/47

Answer 3

There are two ways to solve this problem, both of them should yield the correct answer. In this situation they are both easy to understand, but for some calculations one approach is easier than the other. The first is to treat the problem as a series of dependent events. In this form the question your are asking is "What is the chance of drawing a card, and then drawing four more of the same suit from the same deck?" The chance for the first card is 52 in 52, since you don't care what card it is. After that there are 12 remaining cards of the same suit. And the deck has only 51 cards left. A 12 in 51 chance. Each further draw has one less possible card, and a one smaller deck. For two events to both happen you simply multiply their probability together. Thus your final answer is 52/52 * 12/51 * 11/50 * 10/49 * 9/48 ~= 0.2%. The second way is the combinatorial approach. Here the question we are asking is "How many flushes can be drawn from a deck compared to the total number of hands?" How many possible hands there are is simple. We use something called the binomial coefficient, otherwise know as the choose operator. For example (52c5) = 2598960 says how many ways are there to select 5 cards from 52, and the answer is about 2.5mil. The ordering of the cards doesn't matter here (as they don't in poker). To count the flushes we then consider one suit, which has 13 cards. So (13c5) = 1287 says there are 1287 ways to choose 5 cards from 13 -- all the flushes for that suit. This is possible for each suit, so we have 1287 x 4 = 5148. Now to get the chance any flush happening we simply divide the number of flushes by the total number of hands to get 5148 / 2598960 ~= 0.2%.

Answer 4

P = 1/47 + (46/47)(1/46) = 2/47

Answer 5

1/47 + 1/47 = 2/47

Answer 6

2/47.
Heuristically, as long as you don't the outcome of each of the 2 cards. Each card has equal change.

Mathematically, you need conditional probability.
P(of getting a 5) = P(card 1 is 5 | given card 2 is not) + P(card2 is 5 | card1 is not).