Yet another card game problem. Also it's prob-based again. There are only three things asked this time:
A) What are the odds of having the perfect cribbage hand? (In layman's terms: having three 5's and the matching Jack of the 4th 5 flipped, this resulting in a 15-16 quad-5 Jack = 16 + 12 + 1 = 29!)
B) What is the highest number of points scorable in a round? (cut jack, pegging, hand, crib)
C) What are the odds of this round's occurrence?
Posts: 1363 | Location: Lowell, MA, USA | Registered: 06-03-02
This I sort of exchanged for the stumper. It probably would have been solved in a day because of prime-generating programs... it used to be harder. Cribbage is a card game where when you earn points, you move your peg that many holes down the board. The first one to 120 points wins. It's much more complex than that, though.
Posts: 1363 | Location: Lowell, MA, USA | Registered: 06-03-02
Part A is easy. First of all, I'll assume you mean the odds of a given player getting this hand (as opposed to either player getting this hand). I'll call the start card (the face-up card) Card 1, and the given player's cards Card 2, Card 3, Card 4, and Card 5. I'll abbreviate the cards C1, C2, C3, C4, and C5 for convienience, and I'll use the abbreviaton HN for the jack of the same suit as C1 (this is short for his nobs, as it is called in the game).
Using the product rule for a chain of events, the probability of getting the perfect cribbage hand is
P(C1=5) × P(C2=5 or HN) × P(C3=5 or HN) × P(C4=5 or HN) × P(C5=5 or HN)
= 4/52 × 4/51 × 3/50 × 2/49 × 1/48
= 1/3,248,700
However, it is possible to get a higher score if you count his heels, which is a 2-point bonus the dealer gets if the start card is a jack. The probability of the dealer getting this hand (worth 30 points) is
P(C1=J) × P(C2=5) × P(C3=5) × P(C4=5) × P(C5=5)
= 4/52 × 4/51 × 3/50 × 2/49 × 1/48
= 1/3,248,700
which is exactly the same probability.
Either way, I'm still not done yet, since each player gets two extra cards which they discard before counting their hand. This means I still have to multiply the probability by the number of 4-card combinations in a 6-card hand:
By the way, while we're all thinking about Parts B and C, here's a trivia question. Experienced cribbage players have an inside joke: when they get a hand with 0 points, they often say they have a 19-point hand. Can you explain the joke?
This message has been edited. Last edited by: silenteuphony, 05-26-04 03:31 AM
09-30-02, 03:28 PM silenteuphony Okay, here's a first attempt at Part B. First of all, when you say "highest number of points scorable in a round," I assume you mean by one player (the dealer), not the combined scores. I'm also going to assume both players play so that they maximize the dealer's score.
Starter Card: 5
Non-dealer's hand: 4 4 6 10
Dealer's hand: 4 4 6 6
Crib: 5 5 5 J (his nobs)
PLAY (ND=nondealer, DL=dealer):
ND 4
DL 4 (8 and a pair for 2 pts)
ND 4 (12 and three of a kind for 6 pts)
DL 4 (16 and four of a kind for 12 pts)
ND 10 (26 and 1 pt for last)
DL 6
ND 6 (12 and a pair for 2 pts)
DL 6 (18 and three of a kind for 6 pts plus 1 pt for last)
So far, the nondealer has 9 pts, and the dealer has 21 pts.
SHOW:
ND Three 15's and a double run for 14 pts
DL Four 15's and a quadruple run for 24 pts
Crib Eight 15's, four of a kind, and his nobs for 29 pts
This brings the non-dealer's total to 23 pts, and the dealer's total to 74 pts!
So if I'm right, 74 pts is the highest score a player can get in one round. I'll save part C until I see if someone comes up with a higher scoring round.
09-30-02, 08:39 PM WiteoutKing From what I've played, a cut Jack is worth one peg to the dealer. Otherwise, the rest looks rather soundproof. Tell ya what I'm gonna do. Remember the Scrabble thread? Let's see who can get the highest total hand/peg/deal.
So, s.e. you have 74 pts.
10-01-02, 08:38 AM silenteuphony I think I found a higher total for Part B.
Starter Card: 5
Nondealer's hand: 6 6 7 7
Dealer's hand: 6 6 7 7
Crib: 5 5 5 J (his nobs)
PLAY
ND 7
DL 7 (14 and a pair for 2)
ND 7 (21 and three of a kind for 6)
DL 7 (28, four of a kind for 12, and 1 for last)
ND 6
DL 6 (12 and a pair for 2)
ND 6 (18 and three of a kind for 6)
DL 6 (24, four of a kind for 12, and 1 for last)
SHOW
ND Quadruple run for 16
DL Quadruple run for 16
CRIB Eight 15's, four of a kind, and his nobs for 29
This gives the nondealer 28 pts, and the dealer 75 pts!
10-01-02, 09:00 AM silenteuphony Starter Card: 5
Nondealer's hand: 3 3 4 4
Dealer's hand: 3 3 4 4
Crib: 5 5 5 J (his nobs)
PLAY:
ND 4
DL 4 (8 and pair for 2)
ND 4 (12 and three of a kind for 6)
DL 4 (16 and four of a kind for 12)
ND 3 (19)
DL 3 (22 and pair for 2)
ND 3 (25 and three of a kind for 6)
DL 3 (28, four of a kind for 12, and last for 1)
SHOW:
ND Two 15's and a quadruple run for 20
DL Two 15's and a quadruple run for 20
CRIB Eight 15's, four of a kind, and his nobs for 29
This gives the nondealer 32 pts, and the dealer 78 pts!
10-01-02, 06:44 PM JoMS 19 is the smallest "impossible" points score in a cribbage hand. So, "19" is a joking reference to a zero hand. Except that I nearly had one, once! Playing a tournament at a local club, I was dealt Jack, Queen, King, King of Hearts (an extra king of hearts had crept into the pack). If another king had turned up, I'd have had: three 3-runs (9 points) three-of-a-kind (6 points) a four-card flush (4 points) - total 19 big grin Unfortunately for this story, the turnover card was a three, so I only made 12 - they let me keep the flush.
10-01-02, 08:39 PM WiteoutKing Umm... you can't have a 4-card flush if there are only three different valued cards. Also, if you all prefer, we could open it up to 3- and 4-mand cribbage which involves dealing 5 instead of 6, and btw, s.e., yes, I meant 6-card cribbage.
Leaderboard:
silenteuphony - 78
10-01-02, 10:13 PM silenteuphony Perhaps I'm being presumptuous, but I'm going to answer Part C corresponding to my last answer for Part B; I believe the 78 pt. record will stand. First let's establish some useful notation and procedures. Let S be the Starter card, D1 thru D6 be the dealer's cards, and N1 thru N6 be the nondealer's cards. Let Ck,n be the number of k-object combinations in a larger set of n objects (cards, in this case).
From probability, we have the formula Ck,n = n! ÷ [k! × (n-k)!]
Once I specify that S is a 5 and the crib is the 29 pt. hand consisting of his nobs and the three remaining 5's (which I have already specified in Part B), then we can drop the cumbersome notation "5 or HN" and just refer to the four crib cards as CR, bearing in mind that there are exactly four of them in the deck.
Now we're ready to do some math. Keep in mind I'm drawing from a deck without replacement, so you'll see the available number of cards (the denominator in the fractional probabilities) decrease by one with each card drawn, and the same goes for the numerator for each type of card (e.g. 4 threes, then 3, then 2, then 1). Using the product rule for a chain of events, and assuming an arbitrary order of deal (First S, then the 3's, then the 4's, then the CR's or crib cards), here is the probability of getting the 78 pt. round:
Ah, but we're not done yet. I said assuming an arbitrary order of deal, but I can't really assume that. There's no problem assuming S is dealt first, since it's separate from the two hands, but within each hand, we can't assume anything about the order of the cards dealt. So now I need to multiply by answer by additional number to reflect the different orders the cards could be dealt in.
The 3's may not be dealt first, so I need to multiply by C2,6 = 15 for both hands to reflect the different ways the two 3's could be dealt in each six-card hand. This leaves four cards in each hand, so I need to multiply by C2,4 = 6 for both hands to reflect the different ways the two 4's could be dealt in each hand. Finally, there are only two cards left in each hand for the two crib cards, so I don't have to worry about them. So the final answer is
