Log of Cribbage Questions

Now that my age matches the Answer, I am going to start logging my answers to random cribbage queries directed to me. After all, no one else seems inclined to do it (sorry, Gordon...not that you'll read this log...but then that's my point?).

16 May 2011

My family played cribbage the other day. Both had the exact same hand, 2 kings 1 queen, 1 ten, a seven, and a six. They both threw the 7 and 6 into the kitty.

What are the odds this could happen? We are curious two know!

Probably not as low as you'd think for two reasons. The first is related to the birthday paradox: the chance that two players get any specific hand at the same time is vanishingly small, but the chance that their hands are identical is much higher. The second is the fact that most players of reasonable skill, when faced with the same set of six cards, are actually fairly likely to discard the same two.

Let's start by bounding the probability of getting the same hand. A good fraction of hands contain six different ranks, so we'll consider only such hands. The chance of one player being dealt such a hand is

(52 / 52) (48 / 51) (44 / 50) (40 / 49) (36 / 48) (32 / 47)

or about 34.5% (odds better than 1 in 3). The factors above represent the cards. The first card can be anything. The second's rank cannot match that of the first (48 of the remaining 51 cards). And so forth.

If a player has such a hand, the chance that the other player receives a hand with matching ranks is

(18 / 46) (15 / 45) (12 / 44) (9 / 43) (6 / 42) (3 / 41)

or about 0.00778% (odds better than 1 in 12849). Here again the factors represent the cards. The first card must be drawn from one of the 18 in the remaining deck that match the first player's six cards. The second must be one of the fifteen that match one of the other cards in the first player's hand. And so forth.

The chance overall is then about 0.00269% (odds better than 1 in 37216).

Before we move on, let's make the calculation a little more accurate by considering hands with two matching cards. Part of the reason I ignored those originally was that the chance of the two players' hands in such cases drops quickly. Intuitively, this reduction occurs because all four of the matching rank must end up in the hands, whereas only two of four from each rank need be drawn in the six-different-ranks case. Numerically, the chance of the second player drawing a hand matching the first player's hand containing one matching rank is 1/9th of the 0.00778% probability just calculated: 0.000865% (odds better than 1 in 115640).

This type of hand (with one pair and four other ranks) is actually more common, and makes up 48.6% of draws. Multiplying this factor with the chance of hand matching, we get another 0.000420% chance to match, which we add to our previous value of 0.00269% to get 0.00311% (odds better than 1 in 32187).

Other types of hands won't change this result significantly.

Now let's consider the chance of the discards being the same. There are 15 ways to pick two cards out of six, so if both players pick at random, the chance that they'll match is 1 in 15, assuming again hands of six different cards. But players don't choose randomly, they discard according to their playing strategies.

It's possible to follow strategies that reduce the probability to zero, but these strategies are not good ones. For example, if one player always discards the cards of highest rank, and the other always discards those of lowest rank, the discards will never match if the two players have the same hand.

I'd guess that for most hands, the number of "reasonably good" discard choices--those that might be considered by a player of moderate skill--might number three or four. So the chance that the two player pick the same one is then more like 1 in 3 or 1 in 4.

Let's pick a nice round number and say that the overall chance is then about 1 in 100,000 (32187 times 3.11).

Given an average of about 20 hands per game (not calculated--just pulled at random from the web, but sounds reasonable in my experience), we can expect to play about 5,000 games before we have such an experience.

11 May 2013

My cribbage buddy ... and I have a question for you. Is it possible for the players in a 2 person game to be dealt cards which produce 0 points in both players hands , and the crib hand ---with the turn card included ?

We laid out cards experimenting ---and we were able to create "0" points in all 3 hands ---except that when we included the turn card, we always got some points.

Your analysis would be appreciated

The answer depends on whether the 0-point objective depends on which cards each player puts in the crib or is guaranteed for any choice that they might make.

If we interpret the question as meaning that no points are possible regardless of which cards are put into the crib, the answer is no, such an event is impossible.

Theorem: It is not possible to have both players be dealt cribbage hands and an upcard turned in a way that, regardless of what the players discard into the crib, no points are scored by any of the three hands. (That is, there is no combination of 13 cards that achieves this guarantee.)

Proof: 13 cards. Clearly we cannot have any matching ranks, since if we did, those two cards could end up in the same hand (or in a hand and as upcard). Therefore, if we are to find a 13-card combination that meets our goal, we must have exactly one card of each rank. But now consider two cards that form a 15, say 6-9, 7-8, or 5-10. Those two cards pose the same problem as a pair: they might end up in the same hand, or as upcard and in a hand. So we can not actually pick 13 cards without creating point opportunities that make our choice of 13 not meet the goal.

Hope that's clear.

On the other hand, if we assume a slightly weaker variant, in which neither player has any points in their initial hand of six cards, but the crib's score depends on the two discards (with some possibility of 0 points all around), constructing such an event is possible, as follows.

Consider one player's hand. Of the six cards in the player's hand, no two cards can the same rank (no pairs).

Neither can four of them have the same suit (no flushes).

So pick three of the cards and make them some suit (say, hearts). Let the other three cards come from another suit (say, clubs). Now you only have to worry about runs, 15s, and his nobs.

For the other player's hand, swap the suits (those cards aren't used yet). For the up card, pick a diamond (no player has jack of diamonds, even if they have a jack).

So: if you can use the guidelines above to come up with one hand with 0 points (7 cards total, counting the upcard), you can construct the other hand. And all you need to think about is runs and 15s.

As a concrete example, consider the following:

Pick any four distinct ranks for the crib, then all get 0. But if players toss in ranks in common, crib has points.

[ The authors of the question came up with a 0-point combination themselves, but their combination was a slightly weaker variant in that it required, I believe, that at least one player deliberately break a scoring combination from their original hand when discarding into the crib. ]

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