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 28-hand. 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 29-hand 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! * (k-r)!)
where k! means k factorial i.e. k! = 1*2*3*....*k;

### Approach 1

Let X be any 10-point card (we have 16 10-point 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! * (4-3)!) = 4, and again multiply on 16 as any of X, total 4 * 16 = 64
We need to substract known 29-hand 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 29-hand calculation. We select 6 cards from the 52-card pack

C(52,6) = 52! /(6! * (52-6)!) = 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 52-5 = 47 cards.
It will give the next formula:
76 * C(47,2) = 76 * 47!/ (2! * (47-2)!) = 76 * 47 * 46 / 2 = 82156 - the number of the good 6-card 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 = 8,77e-5

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. X-card means 10,J,Q,K - total 16 cards; A-card 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) - A-cards - 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 A-card 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 A-cards 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 A-card 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
So that total 7936+7680+1680+31744+15360+2240 = 66640 hands+upcards

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 6-card 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 = 6,65e-5