It was Labor Day weekend, 1999. I was taking a break from arranging furniture at my new apartment to play a best-of-three match with Broncos_3peat, one of the highest rated cribbers at MSN Gaming Zone. I'd won the first game, and had gotten off to an early lead in the second. I was feeling pretty good about the match, but then, as is so often the case, my hands died down the stretch. By the time we reached Fourth Street, Broncos_3peat had a sizeable lead. Some nifty pegging with a flush hand as pone got him to 119, while my meager hand and crib only brought me to 105. Now it was his deal, and things were looking bleak. I consoled myself with the knowledge that I could still take the match by winning the third game. But against a player of Broncos_3peat's caliber, you've got to seize every chance, even the slim ones.

I was dealt 5-6-8-9-10-10. I tossed 5-6, leaving me with 8-9-10-10, the only holding that gave me a shot at the 16 points I needed. My eyes lit up when the cut was...a 10. Not quite the full 16, but close enough to give me a fighting chance.

PONE  (105):
Sche11

	8-9-10-10
	play:
	crib:	5-6 ?-?		cut:	10
	?-?-?-?

DEALER  (119*):
Broncos_3peat

Now the problem was how to handle the play. I needed to peg at least one point while preventing Broncos_3peat from pegging two, a tall order indeed. All my opponent needed was a single go plus last card, assuming he didn't get a pair, run, 15 or 31 first. Playing the percentages would minimize my risk of giving up pairs and 15s, but I knew I'd have to get at least one go to win. Looking at my hand, I imagined three ways that might happen.

1. I get the first two goes, leaving dealer with only one point for last card. This would probably require a mistake on my opponent's part, similar to the following:

¹⁰  _K?  ^10 (30-1)    ₄  ⁹  _K  ^8 (31-2)    _6 (6-1)

2. I get the first go with the two 10s, and dealer's remaining cards total 13 or less, giving him only one go. For example:

¹⁰  _K  ^10 (30-1)    ₂  ^8!  ₃  ⁹  _4 (26-1)

3. Dealer gets trapped into playing two successive cards on the same go, letting me steal last card as pone. Since my lowest card was an 8, I figured I had an unusually good chance of pulling this off.

Choosing the correct lead was easy. I just counted the losers: all the possible cards that would let Broncos_3peat score. If I led a 10, he could score if he had a 5 or a 10. Since I'd tossed 5-6 to the crib, held two 10s and cut a third, that left three 5s and one 10, for a total of four losers. Likewise there were six losers on an 9 lead (three 6s and three 9s) and seven on an 8 lead (four 7s and three 8s). Naturally I led the 10. My opponent played a 2. Whew! I'd passed the first hurdle (he had no 5 or 10). Now what should I play second?

PONE  (105):
Sche11

	8-9-10-10
	play:
	10  2
	crib:	5-6 ?-?		cut:	10
	2-?-?-?

DEALER  (119*):
Broncos_3peat

Obviously, playing the 9 here was out of the question (it would make the count 21), so the choice came down to the 8 or the second 10. My first thought was to play the 8, making the count 20. If dealer replied with an A or 2, I'd get a 31-2. If he replied with a 3 through 7, he'd get a go and could well be forced to play another low card on top of that (hopefully one that didn't make a pair or 31!). In that case, I'd have a safe lead (my other 10) on the second play series, and my point for last card would give me enough to win with my 15 point hand.

The problem with playing the 8 is that it loses if dealer has an 8 or a ten-card. Remember if dealer gets this go, and doesn't have to play an additional card on top of that, then he's won, since he'll still be guaranteed last card at the end of the play. A ten-card would make the count 30, and even if he has a trapped A, the resulting 31-2 would win the game immediately. I counted the losers: three outstanding 8s and twelve outstanding Js, Qs and Ks (he obviously didn't have a 10), making fifteen total.

What would happen if I played my second 10 instead? By making the count 22, I'd block out any ten-cards dealer might have. I'd still lose on an 8 or 9, but since there were only six of those available to him, this was mathematically less likely. So I played the 10.

(Having said that, I wouldn't quarrel if you played the 8 here, provided you approached the decision the right way. Perhaps you'd prefer taking an additional risk here, knowing that if your trap succeeded, you'd be home free thereafter.)

Dealer played a 7, getting a go. It turned out he also had an A, which he was obliged to play on top of the 7. My ploy worked! I'd stolen last card. Now if I could just keep him from pegging on my next card, victory would be mine.

PONE  (105):
Sche11

	8-9-10-10
	play:
	10  2  10  7  A (30-1)
	crib:	5-6 ?-?		cut:	10
	A-2-7-?

DEALER  (119*):
Broncos_3peat

I had a new decision to make: lead the 8, or lead the 9. Once again it was time to count the losers. The 8 lead ostensibly has six of them (three 7s and three 8s), but if my opponent had an 8, why didn't he play it on my second 10, making the count 30? His ace would than have won the game, unless I had one of the three remaining aces, in which case my maximum hand would be twelve points, not enough to win even with the 31-2. I knew from experience that Broncos_3peat is an expert player, so I'm sure he would have made the percentage play in that case. Therefore I assumed he had no 8, leaving the three remaining 7s as losers.

Leading the 9 would lose if dealer had a 6 or a 9. But obviously if he had a 9, he'd have played it on my second 10 for a 31-2. That left the three remaining 6s (I'd tossed one to the crib) as losers.

Percentage-wise, the 8 and 9 leads were thus identical. Could I apply logic as a tiebreaker? Let's review the situation. Dealer has A-2-7 on the table, and I was concerned that his fourth card might be a 6 or 7. I looked for a clue in his play of a 2 on my 10 lead. Since he also had an A, the 2 play was perfectly safe. If I'd scored on it, he'd have retaliated for either a 15-2 or a run of three, winning in either case. It seemed a good play from A-2-6-7, but a tad less likely from A-2-7-7, since in the latter case he also would have had a safe play with the 7 (if I pegged on that, he'd triple me with his second 7). Even holding A-2-7-7, though, I could see why my opponent might want to unload one of his low cards early (they can easily get trapped into runs on a high go). After thinking about it, I could find no reason why my opponent's play sequence would favor him having a 6 over a 7.

More compelling was the likelihood that a player of Broncos_3peat's caliber, in a situation where he only needed to peg two points, would try to hold four different cards to maximize his scoring chances. (If you don't understand why this is so, consider that holding A-2-7-7 as dealer, you'd score two points on pone's A, 2, 7 or 8 lead. But holding A-2-6-7, you'd score on pone's A, 2, 6, 7, 8 or 9 lead.) This made A-2-6-7 a more likely candidate than A-2-7-7. So I led the 8.

PONE  (105):
Sche11

	8-9-10-10
	play:
	10  2  10  7  A (30-1)    8  6  9 (23-1)
	crib:	5-6 ?-?		cut:	10
	A-2-6-7

DEALER  (119*):
Broncos_3peat
 

Sure enough, he was holding a 6. My 9 made last card, and my hand score got me to exactly 121 while my opponent languished in the stinkhole. As the blue victory flag waved on my screen, he walked away from his computer shaking his head. Or at least that's how I imagine it — you never know on the Internet.

Obviously I was lucky this time, getting both a favorable cut and a favorable match-up against dealer's hand. But as you can see, making the wrong discard or leading the wrong card on either play series would have cost me the game. And that's probably what would have happened if this had been played a couple of years earlier, before I started studying cribbage seriously. Boy, is it satisfying when your hard work actually pays off in a one-point victory!

Playing the percentages and applying logic won't win every endgame. But in the long run, it'll get you more than your fair share of great victories.

- July 2000