probability.. who can answer this??

Question

in a card game, if a hand is holding 5 cards, find the probability that there will be 3 aces.

Answer 1

use the key nCr, its easier to use than !..

its just a probability of an event so

A=the event of having 3 aces and 2 of any kind other than aces.

4C3 or 4!/3!(4-3)!= number of ways of having 3 aces

48C2 or 48!/2!(48-2)!=number of ways of having 2 of any kind

nA= 4C3 (48C2)= 4512

N= 52C5 or 52!/5!(52-5)!=2,598,960

req'd: P(A)

solution:

P(A)= nA/N

= 4512/2,598,960

= 94/54,145 or 0.001736

Answer 2

(4C3 x 48C2 + 4C4 x 48C1)/(52C5) = 0.00175454797...

Oh dear...

First you need some background in probability to make these easier to explain. When we choose x from a set of y we can calculate the number of combinations using the formula
y!/((y-x)!x!

This is typically written yCx ... this gives us from a set size y the number of combinations of size x. The proof that this formula is correct I'll not get into.

So, you want 3 cards out of the 4 aces...
that's 4C3 possible combos => 4!/(1!3!) = 4

The other 2 cards don't matter...
That's 48C2 possible combos => 48!/(46!2!) = 1128

In total, that's 4x1176=4512 possible ways to have this hand with 3 aces.

All possible hands? 52C5 = 52!/(47!5!) = 2598960

But wait... are we missing something? Hell yes! A 4-Ace hand is not included in this calc!

How many possible hands is this? 4C4 x 48C1 = 48

Final probability? 4560/2598960 = 0.00175...

I've included a reference to a chart which shows this calculation; the probability I've calculated is naturally (slightly) higher than a strict 3-ace (no more or less) hand; in fact, it's exactly what you get adding the two (3 & 4-ace) probabilities together!

Sure, the 0.000018 probability some people will miss is not huge, but it's there.

Answer 3

No of Aces in dec of card = 4
Total number of cards =52

Probability of selection 5 cards out of 52 is (5/52)
Probability of 3 ace out of 5 is (3/4)
so it is 5/52 * 3/4

Answer 4

the total number of possibilities, if you have 5 cards is:

52! / ( 5! x 47! ) = 2598960 ( '52 choose 5' in mathematical terms)

the number of possibilities that have 3 aces, is

4! / ( 3! x 1! ) x 48! / ( 2! x 46! ) = 4512
('4 choose 3' times '48 choose 2', because there are 3 aces (from 4) and 2 other cards (from 52-4=48))


So, the chance is 4512 / 2598960 = 0,001736079 (approximately)

Answer 5

Here is my take:
- First the probability to have three aces in three cards,
this is (4/52)*(3/51)*(2/50)= 24/132600
- Next, there are 10 possibilities to distribute these three cards among five in a hand.

Therefore the probability is 240/132600 or 0.18% or 1 in 553.

Answer 6

There are 52 cards in a pack.
You're holding 5 cards that could be anything.
There are 4 aces in all. 3 aces can be selected in 4C3 ways.
2 more cards are left with you and they could come from 48 other cards.
They can be chosen in 48C2 ways.
All in all, 5 cards can be chosen from 52 cards in 52C5 ways.
The probability is..= 5C3 * 48C2/ 52C5
= 0.004340

That's your answer..hope it's right!

Answer 7

choose 5 from 52 cards
C(52;5)

There are 4 aces in 52 cards.
52-4=48

without aces C(48;5)


probability
1-[ C(48;5)/ C(52;5)]

Answer 8

The solution is 24/13^5 or slightly less than 1 in 15,385 (65 occurrences in a million)

Answer 9

for the first ace: 4 out of 52

second, 3 out of 51

third 2 out of 50

Answer 10

2598960 is the number of hands
to get 3 aces, you have
4* (49*48)/2

chance = 0.00181 or 0.181%