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?