A new type of poker called the razorback six has been created by your finite instructor. This game is similar to regular poker except each player receives 6 cards instead of the traditional 5. He has also invented a few new hands. Find the probability of obtaining the following 4 hands if you are dealt 6 cards. Round to 9 decimal places.

A. Danny Tanner's Full House- this hand contains 4 of one type of card and 2 of another. (EX: 4 kings and 2 fours)

B. The Double Triple- this hand contains 3 of two types of cards (ex: 3 fives and 3 jacks)

C. The Third Wheel- Any hand that contains the King Queen and Jack of a single suit (ex: K, Q, J of spades, 2 of hearts, 7 of clubs, and 4 of diamonds)

D. The Bachelor's Party- This hand contains the King and Jack of 3 suits

This is obviously a homework question, so I'll help you with the first one, but you'll need to show some effort on your part for the remaining three questions if you want more help.

To calculate the probability of the Danny Tanner's Full House (DTFH), we need to first calculate the total number of possible ways you could be dealt such a hand (irrespective of the order in which the cards were dealt), then divide by the total number of possible hands that could be dealt.

P(DTFH) = number of possible DTFH hands / total number of possible razorback poker hands

The number of possible DTFH hands can be computed as the number of possible four-of-a-kind combinations times the total number of ways of pairing the remaining two cards. The first part is easy; there are only 13 ways of obtaining four of a kind (four aces, four twos, for threes, etc.). For the second part, you need to compute the number of ways of choosing two cards out of four suits (4C2 = 4!/(2!*(4-2)!) = 6), then multiply that by the total number of possible remaining denominations (of which there are 12; the deck started with 13, but you used up one of the denominations with the four-of-a-kind portion of the hand).

Thus the numerator for P(Danny Tanner) is 13*6*12 = 936.

The denominator is the total number of possible hands, which is 52C6 = 20358520.

So the probability is 936/20358520 = 0.000045976.

Now can you use similar techniques to do the other three parts?