What is the probability of drawing 2 aces and 3 spades?

Answer 1

Hey,

to start,, i am not a math guy at all, i have a couple years of philosophy and psych but nothing to be proud of,, just a drunk (little too much drugs) guy whose bored on the internet and now really wants someone to finish what ive got so far.

So theres 3 possible outcomes when dealing the cards. "theres more than 3 looking at the suites, but only 3 as far as the probabilities go, take a second look if u dont know what i mean,,haha also if you dont u cant help me :( "

A A S S S

26/52* 25/51* 13/50* 12/49* 11/48 = .0035

A S A S S

26/52* 25/51* 24/50* 12/49* 11/48 = .0066

S S A A S

26/52* 25/51* 24/50* 23/49* 11/48 =.0127

1/4 of spade

1/4 of Ace

now you would look at scenario 3 and say that the prob of getting s,s,a is already--like another popular wikianswer shows,,

prob of #3 (ssa)

(1/4) (4/17) (13/50), =0.0153

and would already been thrown off..

but...u have to remember we have no specific order, which makes this question so difficult (for me at least)

the probability of the 3rd and 4th card being either a spade or heart is where its tough for me,

also that im doing this using my stickies and calculator on my macbook at 3am drunk would make it hard, was just interested in this question because i really enjoy Poker.

,,so that leaves us at the question,, how do we take the prob of the 3rd and 4th card being either heart or spade (ignoring the reacquiring probabilities, and understanding theres only 3 outcomes) and make a new probability. Even if i were to find those probs how do i inculde them into the 3 other probs i have on the scenarios.

Ill check this again, but if your wicked with your numbers could you please also email the solution to me ryen_00 at hotmail.com

...sorry to the question asker that i have no number for ya,,, i prob could have bullshitted an answer but i wanna know the real one now.

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WOKE UP SOBER,, I THINK ITS JUST AS EASY AS ADDING THE 3 POSSIBLE CASES TOGETHER

.0035+.0066+.0127= .0228

OR approx: 1/44