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# For the first bullet point, My first approach to this problem is as

follows:

A bridge hand consists of 13 cards, so that maximum number of pairs that a hand can consist of is 6.

There are 13 ranks in a standard deck of 52 cards.

Examples of different ways to pair (S, H) in a bridge may be the following:

1S 1H, 2S 2H, 3S 3H, 4S 4H, 5S 5H, 6S 6H, 7S or

1S 2H, 2S 3H, 3S 4H, 4S 5H, 5S 6H, 6S 7H, 7S , etc.

Therefore, I believe all possible values of the pair (S, H) would be 13 x 6 = 78.

However, I believe we must be more specific than that and so my second approach is as follows:

There are 4 suites in total and we only want spades and hearts, so there are (4 choose 2) ways to acquire that.

If we only take all the cards of 2 suites, there are 26 cards of those in total and 13 cards to choose for a bridge hand. Thus, there are (26 choose 13) ways to choose the cards in the hand.

Therefore, all possible values of the pair (S, H) would be (4 choose 2) x (26 choose 13) ?

For the second bullet point,

P (S = s, H = h) = P(S=s) and P(H=h).

I am unsure as to how we can generalize the way to find the probabilities of any s and h.

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