Math-Probability?

Question

Math-Probability?

A poker hand (5 cards) is drawn from an ordinary deck of 52 cards. Find the probability for the following events. (Round your answers to 6 decimal places.)
(a)
The first four cards are the four aces.
(b)
Any four cards are the four aces.

Answer 1

a) (4/52*3/52*2/50*1/49)*(4!)= 0.000089
b) not sure could be 0.000089 x2= 0.000177

the person above me is wrong for sure (says someone that took calculus already and have a lot of math competition experience)

Answer 2

The probability of drawing a card is: 1/52

The probability of drawing an ace is: 4/52 or 1/13


a) The first four cards are the four aces.

Probability of one ace: 1/13
Probability of second ace: Now there are only 3 aces left, and 51 cards, so: 3/51
Prob of third ace: 2/50
Fourth ace: 1/49

1/13 x 3/51 x 2/50 x 1/49 = .00000369 = .000369%

b) Any four cards are the four aces.

The various ways this can happen is:

N = Non-ace
A = Ace

AAAAN
AAANA
AANAA
ANAAA
NAAAA

There are 5 possible combinations that place four aces in the hand. There are 311,875,200 combinations of 5 cards in the deck (found using factorial 52! / (52-5)! )

This gives a probability of:

5/311,875,200 = .0000000160 = .00000160 %

Answer 3

a) 1/52 x 1/51 x 1/50 x 1/49 x 1/48 = 0.00000000032 or 0.000000032 %

b) 0.00000000032 x 1/47 = 0.0000000000064 or 0.00000000064%

Answer 4

Why r u asking me ? I am neither a poker player nor a maths expert!