cards

Suppose we wish to indicate the position of the key card in the following manner: 

Find the card of lowest rank, and use its suit to determine the position of the key card:

C=1, H=2, S=3, D=4.   

Clearly if there are multiple cards of the same rank, this will skew the probabilities towards lower positions.  But by how much?

Suppose we wish to indicate the position of the key card in the following manner: 

Find the card(s) of lowest rank, and use its suit to determine the position of the key card:

C=1, H=2, S=3, D=4.   

Clearly if there are multiple cards of the same rank, this will skew the probabilities towards lower positions.  But by how much?

After some computation, we see that the probabilities of each position, two three digits, are as follows: 

Pr(1) = 0.282

Pr(2) = 0.260

Pr(3) = 0.240

Pr(4) = 0.221

Notation: $r_i$ is the probability that the least rank is $i$ (mod 4).

$s_i$ is the probability that the earliest suit of the least rank is $i$

$t_i$ is the probability that least rank plus earliest suit of the least rank (mod 4) is $i$.

So it appears that these calculations are indeed correct: 

4:hands with 4 of the same suit: $4 \binom{13}{4}$

3,1:  3 of one suit, 1 of another: $4 \binom{13}{3}3\binom{13}{1}$

2,2:  2 of one suit, 2 of another: $\binom{4}{2}\binom{13}{2}\binom{13}{2}$

2,1,1: 2 of one suit, 1 each of two other suits: $4 \binom{3}{2}\binom{13}{2}.13.13$

1,1,1,1: 1 of each suit: $13^4$

Probability of position 1: 

\[ (4 \binom{13}{3}3\binom{13}{1}/3  + 4 \binom{3}{2}\binom{13}{2}.13.13/3 )/\binom{52}{4}\]

Probability of position 2:

\[ (4 \binom{13}{3}3\binom{13}{1}/3+ \binom{4}{2}\binom{13}{2}\binom{13}{2)2/3+13^4)/\binom{52}{4} \]

Probability of position 4: 

$$ (4 \binom{13}{4} + 6 \binom{13}{2}\binom{13}{2}/3 +4.3.\binom{13}{2}.13.13/3 )/\binom{52}{4}$$