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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:

[tex]\frac{6*{48 \choose 2}}{{50 \choose 2}{48 \choose 2} \div 2!}}=[/tex]

[tex]=0,0097959183673469387755102040816327[/tex]

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.

[tex]P(A_1 \cup A_2)= P(A_1) + P(A_2) - P(A_1 \cap A_2)[/tex]

where [tex]P(A_1)[/tex] is the probability of getting larger pair of the first opponent and [tex]P(A_2)[/tex] of the second opponent.

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

[tex]0.009795918367346938775510204081633[/tex] which is a almost perfect approx.

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

Now because [tex]P(A_1)=P(A_2)=\frac{6}{{50 \choose 2}}=0.0048979591836734693877551020408163[/tex]

[tex]2*P(A_1)=0.009795918367346938775510204081633[/tex]

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

Now I only need to find [tex]P(A_1 \cap A_2)[/tex], 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:

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

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

which is not even close to [tex]10^{-34}[/tex].

Where is my error?

Thanks in advance.

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# Question about cards (challenging problem)

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