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.

[gaming_mouse]

[ QUOTE ]
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.

[/ QUOTE ]

This isn't the greatest example of it, but I find that often people don't really know what their question is. They'll see AA versus KK, and wonder, What are the odds? But they don't really know if they're wondering what the chance of that happening on the next hand is, or what the chance of that happening _given_ that they were dealt AA or KK is, etc. It's almost like running into someone you know at a train station in a foreign country and asking, What are the chances?

[gaming_mouse]

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

[/ QUOTE ]

Yeah, didn't read that far.

[ QUOTE ]
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.

[/ QUOTE ]

Yes, but what I should REALLY start doing is stop answering so many of these questions and sticking to hand analysis posts, which is a much more profitable activity for me /images/graemlins/smile.gif

[slickpoppa]

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]

This isn't the greatest example of it, but I find that often people don't really know what their question is. They'll see AA versus KK, and wonder, What are the odds? But they don't really know if they're wondering what the chance of that happening on the next hand is, or what the chance of that happening _given_ that they were dealt AA or KK is, etc. It's almost like running into someone you know at a train station in a foreign country and asking, What are the chances?

[/ QUOTE ]

Yeah, like I said in my previous post, you might want to think about including stuff like that in a sticky for this forum. When most people ask questions, they just say "AA v. KK, what are the odds?" without specifying whether they mean "If I have KK, what is the probability of someone else having AA" or "What is the probability of one person having KK and another person having AA."

[ojsdaman]

thankyou all for your help...i appreciate it...but one thing..i was wondering what the chance of being dealt all 4 twos when each person gets 13 cards...does anyone know the chance that you get 3 of the 4 twos

[slickpoppa]

The probability of a particular person being dealt 3 twos is:

[(4/52)*(3/51)*(2/50)*(49/49)*(48/48)*(47/47)*(46/46)*(45/45)*(44/44)*(43/43)*(42/42)*(41/41)*(40/40)] * (13!/(10!3!)) = .0518

[ojsdaman]

slick- i am sorry but it has been so long since i was in school that i forgot what the ! means for the equation...it might be imaginary but i am not sure if i am remembering correctly..if it is imaginary then how do you do the equation without the !

[ojsdaman]

slick- also one more question...in your explanation of why you use 4! and 9! you phrased it as the ways each can be arranged...for example the 2's can be arranged 4 ways...is that because there are 4 2's and the 9! is because you have 9 other cards in your hand?

[slickpoppa]

! is the symbol for factorial. 9! = 9*8*7*6*5*4*3*2*1. In general, X! = X*(X-1)*(X-2)*......*1.

Since you don't already know that, it would probably be very diffucult for me to explain other probability concepts to you over this forum. If you are really interested in learning how to calculate things on your own, you should probably buy a book or find a good website that explains everything from scratch.

[BruceZ]

Here is a simple way to do this problem without combinatorics or factorials:

The probability of any particular card in your hand being the 2 /images/graemlins/club.gif is 1/52, and since you have 13 cards, the probability that the 2 /images/graemlins/club.gif is in your hand is 13/52.

Now given that the 2 /images/graemlins/club.gif is in your hand, each of the other cards in your hand has a probability of 1/51 of being the 2 /images/graemlins/diamond.gif, and since there are 12 other cards in your hand, the probability that the 2 /images/graemlins/diamond.gif is in your hand is 12/51.

Now given that the 2 /images/graemlins/club.gif and the 2 /images/graemlins/diamond.gif are in your hand, each of the other cards in your hand has a probability of 1/50 of being the 2 /images/graemlins/heart.gif, and since there are 11 other cards in your hand, the probability that the 2 /images/graemlins/heart.gif is in your hand is 11/50.

Now given that the 2 /images/graemlins/club.gif, 2 /images/graemlins/diamond.gif and the 2 /images/graemlins/heart.gif are in your hand, each of the other cards in your hand has a probability of 1/49 of being the 2 /images/graemlins/spade.gif, and since there are 10 other cards in your hand, the probability that the 2 /images/graemlins/spade.gif is in your hand is 10/49.

So all together, the probability that all four 2s are in your hand is the product of the above 4 probabilities, or 13/52 * 12/51 * 11/50 * 10/49 = 0.26% or 1 in 378.6. Since this is the probability that one particular hand has all four 2s, the probability that any of the 4 hands has all four 2s is 4 times this or 1.05% = 1 in 94.7.

I'm working on setting the above to music. /images/graemlins/grin.gif

[BruceZ]

[ QUOTE ]
I can figure out the odds that the first 4 cards are 2's

[/ QUOTE ]

That probability would be 4/52 * 3/51 * 2/50 * 1/49.


[ QUOTE ]
...but i dont know where to go from there to take in account that you have 13 chances to get the 4 2's.

[/ QUOTE ]

You actually have C(13,4) chances to get the four 2s, so you would just multiply by C(13,4). That's exactly what C(13,4) means - it is the number of ways to pick the 4 cards out of 13 which will be the 2s. It is equal to 13*12*11*10/(4*3*2*1). So multiplying this by the probability that the first 4 cards are 2s found above, we get 13*12*11*10/(52*51*50*49) = 0.26% or 1 in 378.6. This is the probabilty that you have all four 4s, so multiply this probability by 4 to get the probability that any of the four players has all four 2s. This would be 1.05% or 1 in 94.7.


[ QUOTE ]
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.

[/ QUOTE ]

1/3, since each of the other 3 players has an equal probability of getting the remaining 2.


[ QUOTE ]
If you are explaining how to come to this please explain without the "choose" button on the calc.

[/ QUOTE ]

See my other post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1820716&page=0&view=e xpanded&sb=5&o=14&fpart=) in this thread for how to do it without using choose.

If you understand combinations, the simplest way to do this is to just compute C(48,9)/C(52,13) = 0.26%. The total number of hands is C(52,13), and C(48,9) is the number hands with all four 2s, since we choose the remaining 9 cards from the 48 non-2s.

[ojsdaman]

Bruce- thankyou so much for your explanation.. that was very simple..i appreciate you taking the time to do that and it helps me a ton because that can be used for any card that you need specifically....