# Question about cards (challenging problem)

1. Jun 16, 2010

### njama

Hello!

I have been doing a research about probability of facing a larger pair when holding let's say a KK. So the possible pairs larger than KK are AA (total 2_C_4=6).

Now if I play with 2 opponents the probability of getting larger pair is:

$$\frac{6*{48 \choose 2}}{{50 \choose 2}{48 \choose 2} \div 2!}}=$$

$$=0,0097959183673469387755102040816327$$

But now there is similar approach which is much efficient and good approximation:

Here is the idea:

You suppose that you play individually by two of the players.

$$P(A_1 \cup A_2)= P(A_1) + P(A_2) - P(A_1 \cap A_2)$$

where $$P(A_1)$$ is the probability of getting larger pair of the first opponent and $$P(A_2)$$ of the second opponent.

Now because $$P(A_1 \cap A_2)$$ is very small you can just compute $$P(A_1) + P(A_2)$$ and you will get:

$$0.009795918367346938775510204081633$$ which is a almost perfect approx.

But now the problem is that I want to subtract $$P(A_1 \cap A_2)$$ so that I can get the correct answer.

Now because $$P(A_1)=P(A_2)=\frac{6}{{50 \choose 2}}=0.0048979591836734693877551020408163$$

$$2*P(A_1)=0.009795918367346938775510204081633$$

Now the error is about $$10^{-34}$$, so $$P(A_1 \cap A_2) \approx 10^{-34}$$

Now I only need to find $$P(A_1 \cap A_2)$$, that is "what is the probability that same pair of AA will come two the opponents"?

For ex. what is the probability that A(spades) A (diamond) will come to both opponents?

Here is my reasoning:

$$P(A_1 \cap A_2) \approx (\frac{4}{50} * \frac{3}{49})^2 \approx 0,00002399$$

or the correct one $$\frac{(120}{(50*49)^2}=0,00001999$$

which is not even close to $$10^{-34}$$.

Where is my error?

Last edited: Jun 16, 2010
2. Jun 17, 2010

### njama

bump!

Anybody, what I am doing wrong?

3. Jun 25, 2010

### techmologist

Hi, njama

$$\frac{6*{48 \choose 2}}{{50 \choose 2}{48 \choose 2} \div 2!}}=$$
$$P(A_1)=P(A_2)=\frac{6}{{50 \choose 2}}$$