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Should I avoid pairing my opponents first lay in cribbage (singles)? I ask because more often than not, he seems to have a third.

2

Let's look at some statistics. Let's assume that you have only a single one of some card value 'x' and 3 other cards. Let's also assume that your opponent has a single one of that card value 'x' (since he laid the first of the 'x' cards). Assuming also that he chose 4 random cards of the 6 he was dealt (not likely, but makes the math simpler), he has 3 chances to have second of that 'x' valued card from a remaining deck of 45 cards (52 - Your 4 - The 2 you threw into the crib - Your opponent's 1 'x' card), 2 of which are value 'x'. The chance of him not getting the 'x' valued card in his 3 more chances is (43/45)*(42/44)*(41/43) or .870 meaning that he has only a 13.0% chance of having a third card of value 'x' and therefore being able to get that 3-of-a-kind. Because you get 2 points for the pair and your opponent gets 6 points (aka you lose 6 points) for the 3-of-a-kind, playing to get the pair nets you (2*.870 - 6*.130) or +.960 points.

If we assume that if he is given a pair of 'x' value cards in his 6 cards, then he will keep them (a reasonable strategy), then the chances of him not getting the 'x' valued card in his now 5 more chances is (43/45)*(42/44)*(41/43)*(40/42)*(39/41) or .788 meaning that he now has a 21.2% chance of having a third card of value 'x' and therefore being able to get that 3-of-a-kind. This nets you +.304 points, a lesser amount than before, but still positive.

In both these scenarios, it is unlikely that your opponent will have that third 'x' card and either way you get a net gain of points over your opponent. Others may have strategies to provide, but these are the raw statistics of your scenario.

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I'd say go for it as long as you're not also setting them up for a possible 15 or something. Like if they play an 8, your 8 goes past 15 and you're safe. But if they play a 4, and then you play a 4, then they could have either another 4 or a 7 to capitalize on the situation.

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Probably Not

I'll work out all the math below, but the part that is difficult to capture in the math, is the strategy of your opponent. Since your opponent is leading, they are tossing cards into a crib that they won't get. So, it is possible they'd break up a pair. Or not. In the below, I assume they never toss a pair into the crib.

Consider the following typical beginner strategy for first card play. Lead with a pair card if you have a pair, otherwise lead with your highest card. If that is your opponent's strategy, then anytime they lead a 3, they have a pair, and you don't want to match it. But, if they lead a face card, you may want to match it. You can compute the exact probabilities using the techniques below.

However, what if their strategy is "if I have a pair, lead from the pair. If I don't have a pair, lead with any random card." Then, the odds of your opponent having the third card is roughly 54%. (Details below. Note, this is significantly higher than B-Rad computed, because they didn't take into account the conditional probabilities.)

In this case, it also does not pay to match. The expected outcomes are:

  • Don't match (you score 0 points)
  • Match (you score 2 points, and with a 45% chance your opponent will score 6. Or an expected gain of 2 - 0.54*6 = -1.2 points)

So, "on average" you lose 1.2 points matching.

Long Story Short: If you know your opponent well, you may want to catch a match, but if you don't know your opponent well, probably safer to never match.

To compute the above claim (and hopefully, you'll see how to adapt the below computation to better reflect your opponent's strategy, which does impact the numbers) we have to make a number of assumptions about the strategy of your opponent. It assumes that if your opponent has 2 pairs, they flip a coin to determine which card to lead. It doesn't count triples (that's a low probability issue, so it's ignored) It assumes that your opponent always keeps a pair if he has one. Note, also, in the computations below, I include the cut card. I'm assuming the cut card is not the card in question. The probabilities change significantly if the cut card is the card in question.

Here's what you want to compute:

P( player 1 has a pair of card x given player 1's first play was an x, you have an x, and cut card is not an x)

This is an example of a conditional probability. If we use P1 to represent player 1, and P2 to represent player 2, we obtain:

P( P1 has pair of x | P1 lead x, P2 has x, cut card not an x)

    P(P1 has a pair of x, P2 has an x, P1 leads an x, cut card not an x)
=  ---------------------------------------------------------------------
             P(P1 leads an x, and P2 has an x, cut card not an x)

To compute these turns out to be a bit complicated. The numerator really needs to be broken up into three cases (actually more than 3, there are a number of edge cases (like triples) that I'm not addressing, but they should not affect the computation too much)

  P(P1 has a pair of x, P2 has an x, P1 leads an x, cut card not an x)
   = P(P1 has a pair of x & no other pairs, P2 has an x, P1 leads an x, cut card not x)
    + P(P1 has a pair of x & one other pair, P2 has an x, P1 leads an x, cut card not x)
    + P(P1 has a pair of x, and 2 other pairs, and leads an x, P2 has an x, cut card not x)

      Choose(4,2) * 4^4 * choose(12,4) * choose(2,1)*choose(44,5)*choose(39,1)
   =  + Choose(4,2)^2 * choose(12,1) * 4^2 * choose(11,2) * (1/2) * choose(2,1)*choose(44,5) * choose(39,1)
      + Choose(4,2)^3 * (1/3) * choose(12,2)^2 * choose(39,1)
     -------------------------------------------------------------------
          choose(52,6)*choose(46,6)*choose(40,1)

   = 0.0105

Now, the denominator of the original expression is as follows:

   P(P1 leads an x, and P2 has an x, cut not x)
   = P(P1 has a pair of x, leads an x, and P2 has an x, cut not x)
     + P(P1 does not have a pair of x, leads an x, and P2 has an x, cut not x)
   = 0.0105 + Choose(4,1)*4^5*choose(12,5) * (Choose(3,1)*choose(43,5) + choose(3,2)*choose(43,4))/6*choose(38,1)
                ---------------------------------------------------------------------------------------------
                               Choose(52,6) * Choose(46,6) * Choose(40,1)
   = 0.0105 + 0.00878

Dividing the numerator by the denominator you get 54%.

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