My friend and I play a Vietnamese card game called Thirteen (or Tien len).

I'd like to know, what are the chances of having at least two pairs after drawing thirteen cards?

Printable View

- Jun 12th 2009, 04:23 PMceasar_19134Trying to figure out a card game
My friend and I play a Vietnamese card game called Thirteen (or Tien len).

I'd like to know, what are the chances of having at least two pairs after drawing thirteen cards? - Jun 12th 2009, 05:05 PMHallsofIvy
The simplest way to do this is to ask the opposite questions:

1) What is the probability of drawing 13 cards with NO pair?

The first card can be anything. The second card can be any of the 48 cards that do not pair the first card: probability 48/51. The third card can be any of the 44 cards that do not match either of first two cards: probability 44/50. Do you see the pattern? For each card we subtract 4 from the numerator and 1 from the denominator. Multiplying all the fractions together, the denominator will be 51(50)(49)(48)...(39)= 51!/38! and the numerator is 48(44)(40)...(4)= 4^{12}(12)(11)(10)...(1)=4^{12}12!. The probability of getting 13 cards with no pair is $\displaystyle \frac{4^12 12! 38!}{51!}$.

2) What is the probability of drawing 13 cards with exactly one pair?

First calculate the probability that the first two cards pair and the other 11 do not. The first card can be anything. The second must match that- probability 3/51. The third can be anything that does not match those: probability 48/50. The fourth can be anything other than the first two or that: 44/49 and so on: (3/51)(48/50)(44/49)... (8/39).

The probability of drawing 13 cards with at most one pair is sum of those two and the probability of drawing 13 cards with**at least**two pair is 1 minus that probability. - Jun 12th 2009, 05:24 PMceasar_19134
My friend and I thought similar to you when we first looked at the problem. However, our numbers were a bit skewed when we calculated things out.

In response to you first point, you neglect to count the hands where players have three of a kind and no pair or four of a kind and no pair.

In response to the second point, your method suggest only one order of getting those cards. You could also not get a pair on the second draw, thus increasing your chances of getting a pair as now the next card can match any of the two drawn before it.

And in addition to all of that, you could draw a pair and then a three of kind or a four of a kind, thus increasing your odds even more. I just don't know how to manipulate probabilities in order to account for all of that. - Jun 12th 2009, 06:15 PMSoroban
Hello, ceasar_19134!

A variation of HallsofIvy's solution . . .

Quote:

My friend and I play a Vietnamese card game called Thirteen (or Tien len).

What is the probability of having at least two pairs after drawing thirteen cards?

The opposite of "at least two pairs" is "no pairs" or "one pair."

**No Pairs**

We must draw one of each of the 13 values.

There are: .$\displaystyle _4C_1 = 4$ ways to draw each value.

Hence, there are: .$\displaystyle 4^{13}$ hands with No Pairs.

**One Pair**

There are 13 choices for the value of the Pair.

There are: .$\displaystyle _4C_2 = 6$ ways to get the Pair.

The other 11 cards must not match the Pair or each other.

There are: .$\displaystyle _{12}C_{11} = 12$ choices for their values.

And: .$\displaystyle _4C_1 = 4$ ways to draw each value.

So there are are: .$\displaystyle 12\cdot4^{12}$ choices for the other 11 cards.

Hence, there are: .$\displaystyle 13\!\cdot\!6\!\cdot\!12\!\cdot\!4^{12}$ ways to get One Pair.

Then there are: .$\displaystyle 4^{13} + 936\!\cdot\!4^{12}\:=\:4^{13}(235)$ ways to get No Pairs or One Pair.

Hence: .$\displaystyle P(\text{No Pair or 1 Pair}) \;=\;\frac{4^{13}(235)} {_{52}C_{13}} $

Therefore: .$\displaystyle P(\text{at least 2 Pairs}) \;=\;1 - \frac{4^{13}(235)}{_{52}C_{13}} $