Question about cards (challenging problem)

[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?

Thanks in advance.

 

[njama]

bump!

Anybody, what I am doing wrong?

 

[techmologist]

Hi, njama

Your first solution,

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

is exactly twice the probability P(A_1):

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

So I don't think your first solution gives the probability you are looking for, that at least one opponent has a higher pair. The very small difference of 10^-34 between your first answer and P(A_1) + P(A_2) is probably due to rounding error. That could happen when you multiply by and then divide by 48 choose 2. Your second approach, using P(A_1 or A_2) = P(A_1) + P(A_2) - P(A_1 and A_2) is the right way to go about it. I don't understand how you are calculating P(A_1 and A_2), though.

Is this question about hold'em poker, where each player gets two cards and there is a round of betting before the "flop" (first three community cards) are dealt?