Probability question...?

Question

This is really a card game question but since no one there knew the answer I figured I'd ask it here. If 4 cards are dealt from a standard 52 card deck, what are the chances that the 4 card hand will contain at least one Ace and at least one 2. Also, what is the formula for figuring the anser out.

Thanks, anyone actually have an answer.

Just to clarify, I said at least on both, so there could be 2 or 3 aces or 2 or 3 2's, just as long as there is at least one of each.

I think NoName is onto it, but with a couple changes. First I think you meant 28.1% instead of 21.1%. But after the 28.1% of there being an Ace, then you would calculate one of the next 3 being a 2, which would be 47/51*46/50*45/49 which is about 77.9%, so a 22.1% chance of their being a 2. Then 28.1% times 22.1% is 6.2%. That looks right to me, anyone disagree?

I am sure volterwd is right, but would it be possible to translate that into something more understandable.

I absolutely believe that you are correct volter, what I meant was since not all of us have a Masters in Statistics what it be possible for you to break that down a little. I am sorry that my stupidity upset you, and I suggest you spend no more time dealing with someone as dumb as myself.

Answer 1

P(A and 2) = 1 + P(No A and No 2) - P(No A) - P(No 2)

The probabilities on the right are VERY easy to calculate

This is derrived from

P(A and B) = 1 - P[(A and B)^c] = 1 - P(A^c or B^c)
= 1 - [P(A^c) + P(B^c) - P(A^c and B^c)]
= 1 + P(A^c and B^c) - P(A^c) - P(B^c)

So by symmetry (i.e., A and B are equally likely)

= 1 + P(A^c and B^c) - 2P(A^c)

1+44*43*42*41 / (52*51*50*49) - 2*48*47*46*45 / (52*51*50*49)
= 0.06396158

EDIT

With all due respect i have a masters in statistics... its right.

The probability statements are mathematical in nature... so no... i only showed them to show you how i derived the result... but calculating it by cases is nearly impossible to do without error and very difficult to do at all. You can see by the previous answers how easy it is to get tripped up.

Ultimately the idea is to reduce it to something that is EASY to calculate.

As well... you could write a program to simulate such an event and run it 100,000 or so times... and still get a few digit accuracy.

EDIT

Sorry for the *********... if you give me 10 pts then i will draw venn diagrams to illustrate it... its a pain in the *** if i dont get the 10 pts...

Venn diagrams are the best i can do to help you understand what i did.

Answer 2

There are 4 aces so the chances of getting an ace as the 1'st card are 4 in 52.
There are also 4 2's and the chance of getting a 2 as the 2'nd card drawn in 4 in 51 (since there is an ace missing)
So the chances of getting an ace and a 2 on the first two cards is (4/52)*(4/51) = 16/2652. But the chances oif getting a 2 followed by an ace is exactly the same, so the overall probability is 32/2652.

See how the calculation is done? Good. Because **you** now have to go back and say what is the chance of ace-something-2, ace-something-something-2,
something-ace-2, something-ace-something-2,
and something-something-ace-2. (since those are all of the ways that an ace and a 2 can be drawn in the first four cards dealt)
Then you have to add up all of those probabilities to get the **total** probability for getting an ace and a 2 in the first four cards dealt.

Straightforward, just a lot of calculator time.

Have fun.


Doug

Hey, Dewd. When I do consulting I bill my time at $125/hr plus living and travel.

If you wanna cut into my drinking time, it'll cost ya ☺

Answer 3

REVISED
Combination of:
No Ace, No 2: 44X43X42X41 = 3258024
1 Ace, No 2 (4 ways): (4X44X43X42)X4 = 1271424
2 Ace, No 2 (6 ways): (4X3X44X43)X6 = 136224
3 Ace, No 2 (4 ways): (4X3X2X44)X4 = 4224
4 Ace, No 2: 4X3X2X1 = 24
No Ace, 1 2 (4 ways): (4X44X43X42)X4 = 1271424
No Ace, 2 2 (6 ways): (4X3X44X43)X6 = 136224
No Ace, 3 2 (4 ways): (4X3X2X44)X4 = 4224
No Ace: 4 2: 4X3X2X1 = 24

Total combination of NOT at least 1 Ace & 1 2 = 6081816
All possible combination = 52X51X50X49 = 6497400
Combinations of at least 1 Ace & 1 2 = 6497400 - 6081816 = 415584

Prob = 415584/6497400 = 6.4%

- Sorry, Will. Problem with original answer is that the when I calculated the prob. of No Ace (71.8%), this prob. includes possiblity of getting a 2 (but no ace). So when I combined this with prob. of "NO 2", there is some overlap.

Answer 4

Well I think it would go like this. You have 4 aces in a deck and you are looking for just one in a four card draw so 4/52 + 4/52 + 4/52 + 4/52 = 16/52, You would have the same odds for getting a two. Now to get both at the same time you multiply the probabilities or 16/52 * 16/52 = 256/2704 0r about 0.946 or 9.4%. At least if I remember my high school math.

Answer 5

Hi. One chance in 52/4 of getting an ace and 1 chance in 51/4 of getting a 2. (or vice versa, in either case one card has to be drawn first)

Answer 6

Probability of an ace = 1/13. Probability of a deuce, ditto. Compound probability is the product, 1/169, but since there are two ways of dealing (ace first, deuce first) you double this to get 2/169. There are some subtleties that change the number slightly.

Answer 7

It's called put down the cards and pick up a girl, move on with your life man.

Answer 8

NoName is correct.