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Super super math problems in senior high school
In all cases, there are C (40,2 2) * c (38 38,2) = 548,340 species.

(First, choose 2 cards from 40 cards to one of them, and then choose 2 cards from 38 cards to the next one. )

1. Pairing is equal:

There are 60 kinds of C (10,1) * c (4,2) =.

(First, choose a number from 1 to 10, and then choose two cards from the four cards of this number to give one of them, and the other one has no choice. )

2. In the case of equal points:

These cards are the same and different.

(1) If they are the same, for example, {2,6} pair {2,6}

There are * * * such cases: C (10/0,2) * a (4,2) * a (4,2) = 5480.

(Since they are not a pair, first choose two numbers from 1 to 10, and then choose two cards from each number in order for the two families. )

(2) The points are the same, but the cards are different.

All 0 points: {1, 9}, {2,8}, {3,7}, {4,6} * *.

They are all 1 min: {1, 10}, {2,9}, {3,8}, {4,7}, {5,6} * *.

It's all 2 o'clock: {2 10}, {3,9}, {4,8}, {5,7} * *.

All three points: {1, 2}, {3 10}, {4,9}, {5,8}, {6,7} * *.

They are all four points: {1, 3}, {4 10}, {5,9}, {6,8} * *.

They are all five points: {1, 4}, {2,3}, {5, 10}, {6,9}, {7,8} * *.

They are all six points: {1, 5}, {2,4}, {6, 10}, {7,9} * *.

All 7 o'clock: {1, 6}, {2,5}, {3,4}, {7,0/0}, {8,9} * *.

8 o'clock: {1, 7}, {2,6}, {3,5}, {8, 10} * *.

All 9 o'clock: {1, 8}, {2,7}, {3,6}, {4,5}, {9, 10} * *.

Even numbers * * include c (5,1) * a (4,2) * c (4,1) * c (4,65438.

(First, choose one from five even numbers, then choose two from four types in sequence for two people, and then two people choose one card from each of the four cards of two numbers of this type, four times, four times at a time 1).

Odd numbers * * include c (51) * a (5,2) * c (41) * c (41) * c (41) * c (465438).

(First, choose one of the five odd numbers, then choose two of the five types in order to give them to two people, and then two people choose one card from each of the four cards of the two numbers of this type, four times, four times at a time 1).

In short, the dealer and easycard are equal in size, including pairs and points.

Add up to 60+5480+15360+25600 = 46500 species.

The rest * * * is 548340-46500=50 1840.

These situations are based on the principle of symmetry. The two families should have equal opportunities.

So everyone has 250,920 situations to win, and the bookmakers of equal size will win.

Therefore, the category of Ying Zhuang is 250920+46500=297420.

The probability is 297420/548340=4957/9 139.

I only calculated the probability of winning the bet once.

Without considering the doubling and the size of the bet, if we do this, we can calculate which side will benefit from this long-term gambling. That's an expectation and it's not difficult to calculate.