# Some probability in the game Big 2

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• Nov 11th 2007, 10:47 PM
chopet
Some probability in the game Big 2
Big 2 is a card game commonly played in Asia.
Big Two - Wikipedia, the free encyclopedia

There are 4 playeers and each is dealt 13 cards. The trump cards in this game is 2, hence the name Big 2. My question is:
What's the probability of 1 player dealt 4 '2's?

Method 1:
${{48 \choose 9} \over {52 \choose 13}} = {11 \over 4165}$

Method 2:
Using the ball-and urn model championed by William Feller,
we imagine 4 people as 4 urns, and the 4 '2's to be dealt as 4 balls:
So now, we are asking what's the probability of all 4 balls landing inside 1 urn:

The probability of 4 balls inside 1 urn(and not 3 others): $({1 \over 4})^4 = {1 \over 256}$

Since it can be any of the 4 urns, we multiply by 4 and get ${4 \over 256}$

Now, why don't the 2 methods tally?
Can anyone please explain?

Much appreciated!
• Nov 12th 2007, 06:57 AM
Plato
Quote:

Originally Posted by chopet
There are 4 playeers and each is dealt 13 cards. The trump cards in this game is 2, hence the name Big 2. My question is: What's the probability of 1 player dealt 4 '2's?

Method 1:
${{48 \choose 9} \over {52 \choose 13}} = {11 \over 4165}$

That is the correct probability if one hand, 13 cards, are dealt.
• Nov 15th 2007, 10:44 PM
chopet
Quote:

Originally Posted by Plato
That is the correct probability if one hand, 13 cards, are dealt.

I think the problem lies with Method 2, not Method 1.
I am treating each placement as independent, when in fact, they are not.
Or are they? Urgh. I am confused.