A Game of cards with Mr. 11 ...

Number of ways in which I can pick two cards and distribute between Krishna and Satvik $= {52 \choose 2} 2!$

Number of ways in which the cards can be of equal value $= 13 {4 \choose 2} 2!$

So, number of ways the cards can be chosen so that the distribution is unequal $= {52 \choose 2} 2! - 13 {4 \choose 2} 2!$

Now, for every distribution where Satvik gets a card greater than Krishna, there exists a distribution consisting of the same cards such that Krishna is greater than Satvik, just by switching the cards.

So, half of the time, Krishna's card is greater than Satvik's

So, required number of ways $= \frac{1}{2}\left ( {52 \choose 2} 2! - 13 {4 \choose 2} 2! \right )$

Evaluating it gives $1248$

The answer is $1248 \pmod{11} = \boxed{5}$

Let's say that for ease, Satvik has card of $\spadesuit$ . Then there are $13$ choices for his card.

For each card, Krishna will have a card greater than that number, being of any suit (4 choices of suits).

img See the the number of ways in which this happens is

$48+44+40+...+4\\ =4\times (1+2+3+...+12) \\= 4\times \dfrac{12\times 13}{2} \\ = \quad\mathbf{312}$

This was for $1$ suit of Satvik, so for all $4$ suits , there will be $312\times 4=1248$ ways.

Thus, answer = $1248 \pmod{11} = \boxed{5}$

Firstly, there is are $52\times 51$ ways for Satvik and Krishna to choose cards. For any card that Krishna has, there is a $\frac { 3 }{ 51 }$ chance for Satvik to have a card of the same value.

Then, notice that the chance of Krishna having a card strictly greater than Satvik's must be the same as the chance of Satvik having a card strictly greater than Satvik's card, with the probability of both events happening being $(1-\frac { 3 }{ 51 } )(\frac { 1 }{ 2 } )=\frac { 24 }{ 51 }$ .

By multiplying the probability to the number of ways for cards to be picked, we get the following:

$K\quad =\quad \frac { 24 }{ 51 } \times \quad 51\quad \times \quad 52\\ \qquad =\quad 24\quad \times \quad 52\\ \qquad \equiv \quad 2\quad \times \quad 8\\ \qquad \equiv \quad \boxed{ 5 } \quad (mod\quad 11)$

Let $k$ be the number on Krishna's card. We know that there are 4 ways of getting a card with a value $k$ .

Since Krishna's card is higher than Satvik's, $k$ cannot have a value of $1$ . The number of cards lower than $k$ is $4(k-1)$ . Given these, the number of ways in which Krishna's card is greater than Satvik's is:

$\sum\limits_{k=2}^{13} 4(4)(k-1)$

$=16\sum\limits_{k=2}^{13} k-1$

$=16\sum\limits_{i=1}^{12} i$

$=\frac{16(12)(12+1)}{2}=1248$

$1248 \equiv 5\mod 11$