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 Cribbage 29 hand and some Cribbage statistics
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 Cribbage 29 hand odds
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We already showed the math of calculation of 29 cribbage hand. Now we received a request from our user Mark M Maciejewski to show the calculation of the second highest hand 28hand. Here it is. The second highest score is 28 (any of 10,J,Q,K and 5555 in hand and starter excepting for the known 29 hands) The calculation is similar to 29hand calculation.
First of all we should remember, what upon dealing a player receives 6 cards, 2 of them will be discarded to the crib later. Let's count the number of hands for 28 points
The base of calculation is a combinatorics' formula for calculating the number of the all possible combinations of r elements selected from the set of k total elements:
C_{(k,r)} = k! / (r! * (kr)!)
where k! means k factorial i.e. k! = 1*2*3*....*k;
Approach 1
Let X be any 10point card (we have 16 10point cards, 10,J,Q,K)
First, the good hand can contain all 5s plus any X as a starter: 5555 X, total 16
Second, the good hand can have 3 5s plus any X in hand and one 5 as a starter.
C_{(4,3)} = 4! / (3! * (43)!) = 4, and again multiply on 16 as any of X, total 4 * 16 = 64
We need to substract known 29hand combination, we know there are 4 of them. So the final number will be 16 + 64  4 = 80  4 = 76 good hands
Now the similar to 29hand calculation. We select 6 cards from the 52card pack
C_{(52,6)} = 52! /(6! * (526)!) = 52! / (6!*46!) = 52*51*50*49*48*47 / 720 = 20358520  the total number of 6 card hands
The good hands: as shown above there are 76 good hands, 4 cards in hand and one starter card. So exclude 5 required cards from the set  we can select our free 2 cards from the set of 525 = 47 cards.
It will give the next formula:
76 * C_{(47,2)} = 76 * 47!/ (2! * (472)!) = 76 * 47 * 46 / 2 = 82156  the number of the good 6card cribbage hands.
Now we have 82156 from 20358520 without a cut card.
A cut card we can pick 1 from the other 46 cards not in our hands, exactly one card for each of our defined good combinations, so the final calculation is
82156 / 20358520 / 46 = 1786 / 20358520 = 8,77e5
Therefore, the probability of getting a 28 hand in 6 card cribbage is 0.0000877 or about 1 to 11399. Amaizingly, it is more than 1 / 15,028, stated everywhere. Any thoughts? The different approach calculating below.
Approach 2
Let's score together the hand and a starter card for each possible case. Xcard means 10,J,Q,K  total 16 cards; Acard means not 5, not X (A,2,3,4,6,7,8,9)  total 32 cards

1) Four 5s in hand and X card as a starter. Other 2 cards in hand are A  cards. C_{(4,4)}  four 5s, C_{(16,1)}  X card as starter, C_{(32,2)}  Acards  free 2 cards in hand.
C_{(4,4)} * C_{(32,2)} * C_{(16,1)} = 1 * (32 * 31 / 2) * 16 = 7936 hands+upcard 
2) Four 5s in hand, one X card in hand, one Acard in hand, one X card as starter
C_{(4,4)} * C_{(16,1)} * C_{(32,1)} * C_{(15,1)} = 1 * 16 * 32 * 15 = 7680 hands+upcard 
3) Four 5s in hand, two X cards in hand, one X card as starter
C_{(4,4)} * C_{(16,2)} * C_{(14,1)} = 1 * 16 * 15 / 2 * 14 = 1680 hands+upcard 
4) Three 5s in hand, one X card in hand, two Acards in hand, one 5 as starter
C_{(4,3)} * C_{(16,1)} * C_{(32,2)} * 1 = 4 * 16 * 32 * 31 / 2 = 31744 hands+upcard 
5) Three 5s in hand, two X cards in hand, one Acard in hand, one 5 as starter
C_{(4,3)} * C_{(16,2)} * C_{(32,1)} * 1 = 4 * 16 * 15 / 2 * 32 = 15360 hands+upcard 
6) Three 5s in hand, three X cards in hand, one 5 as starter
C_{(4,3)} * C_{(16,3)} * 1 = 4 * 16 * 15 * 14 / (3*2) = 2240 hands+upcard
Subtract the number of hands+upcards that give 29 points. It is a subset of above items 4,5,6. From the calculation of 29 hand we know, that it is C_{(4,3)} * C_{(47,2)} =
4 * 47 * 46 / 2 = 4324
This way the number of good hands+upcards is 66640  4324 = 62316
Let's calculate the number of all 6card hands with upcard. It is C_{(52,6)} * C_{(46,1)} = 52*51*50*49*48*47 / (6*5*4*3*2*1) * 46 = 936491920.
The final propability will be our number of good hands+upcard divided by the number of all hands+upcard, i.e. 62316 / 936491920 = 6,65e5
Or about 1 to 15028