probability of four 2's being dealt in a full deck...please help...

[ojsdaman]

hello all...i have a question that i cannot figure out...we were playing a four player game where the entire deck is dealt out and everyone gets 13 cards....what is the probability that one person gets all 4 of the 2's. I can figure out the odds that the first 4 cards are 2's but i dont know where to go from there to take in account that you have 13 chances to get the 4 2's.
Also in a side question. If you get 3 of the 2's, what is the probability that the person on your right gets the other 2. Thankyou for your help. I appreciate it. If you are explaining how to come to this please explain without the "choose" button on the calc. Thankyou

[slickpoppa]

here ya go (let me guess, you were playing as$hole):

[(4/52)*(3/51)*(2/50)*(1/49)*(48/48)*(47/47)*(46/46)*(45/45)*(44/44)*(43/43)*(42/42)*(41/41)*(40/40)] * (13!/(9!4!)) = 0.0026

The last 9 terms in the first part of the equation are obviously pointless, but they help show what is going on in the calculation. Notice that the first part of the equation in brackets is the chance of being dealt 1 combination of 4 2's plus 9 of any other type of card. The second part of the equation takes into account all of the combinations of such deals.

Further explanation on the 13!/(9!4!) term:
If each card were unique, there would be 13! possible combinations. However, for the purposes of this question, every card is not unique. There are just two types of cards that we are interested in: (1) the four 2's and (2) everything else. Since the difference between the two of clubs and two of hearts does not matter, we divide by 4!, which is the number ways that you can arrange the two's. We also divide by 9!, which is the number of ways that you can arrange "every other card" in your hand.

For a much better explanation of these concepts, you should look at any high school level statistics book.

[gaming_mouse]

hey slickpoppa,

i get...

(4 * (52 choose 9)) / (52 choose 13) = 0.0231747832

is the difference due to rounding on your part?

EDIT: I see you have an extra zero, actually. So one of us made an error.

[gaming_mouse]

Also, in case it was not clear:

I am using inclusion-exclusion with four events:

E1 = player 1 has 4 twos.
...
E4 = player 4 has 4 twos.

The first term gives an exact answer, since the events are mutually exclusive.

[slickpoppa]

I'm using my computers crappy calculator, so there is a good chance I screwed up typing in the numbers. I am pretty sure that my methodology is correct though.

Now that I think about it, your number seems pretty high. A 1/40 chance of getting all 4 of a particular card seems high, even if you are getting 13 cards.

[DougOzzzz]

[ QUOTE ]
hey slickpoppa,

i get...

(4 * (52 choose 9)) / (52 choose 13) = 0.0231747832

is the difference due to rounding on your part?

EDIT: I see you have an extra zero, actually. So one of us made an error.

[/ QUOTE ]

Isn't it 48C9 instead of 52C9 since the 4 2's are gone?

[gaming_mouse]

[ QUOTE ]
I'm using my computers crappy calculator, so there is a good chance I screwed up typing in the numbers. I am pretty sure that my methodology is correct though.

Now that I think about it, your number seems pretty high. A 1/40 chance of getting all 4 of a particular card seems high, even if you are getting 13 cards.

[/ QUOTE ]

Actually, we both made mistakes. You correctly calculated the chance that one specific player gets it. the correct answer is:

4*(48 choose 9)/(52 choose 13) = 0.0105642257

which is 4 times higher than your answer.

gm

[gaming_mouse]

[ QUOTE ]


Isn't it 48C9 instead of 52C9 since the 4 2's are gone?

[/ QUOTE ]

yep, just crossed posts w/ you.

[slickpoppa]

Ok, glad we sorted that out. I actually thought that he was asking what the probability of a particular person getting all 4 of them was. Confusion like that seems to happen often in this forum.

[slickpoppa]

Also, note that the OP asked for the answer without using the "choose" function.

GM, you should think about doing an FAQ for the probability forum that explains such things since your are usually the one that ends up answering all of the questions anyway.