August 8, 2020
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| Total votes: 189 | |||||
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Rosemarie44 2052 votes Joined: March 2016 |
Saturday 3:16 AM
These end game positions always seem to throw me as I am not playing cribbage very often. let's see what others choose for retention and discard. |
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JQT 4143 votes Joined: October 2008 |
Saturday 3:32 AM
Pone needs Eight Points to WIN, and we could try to play DEFENSE here, but Ten Points is average for Pone, and usually about Two Points of that Ten comes from pegging.
Thus, what we definitely should have more CONTROL over is our own forward movement, since we need just Three Points Pegging as Dealer to WIN, and the average is closer to 3.5 Points. And having been dealt no Jacks, we have 4 DIV 46 equals 0.087 or nearly a 9% chance of Cutting Mr. Jack! A Jack Cut solves all of our problems today, since Dealer shall *always* peg at least One Hole - that is, unless Pone pegs out first! And Pone would have to jump through magical hoops to peg Eight Points without us getting one lousy hole in the interim. So, how can we peg Three Holes (assuming that 91% of the time, we shall NOT Cut a Jack) with four of these six cards? Good question! The PAIR of Tens is the least likely rank to show up in Pones Hand, as we were dealt HALF of these. Thus, I am initially inclined to Keep (2 3 Q K) and Toss (T T). The points and even the Cards in our Crib have no bearing whatsoever, and even if we held only a combine total of Two Points, we shall always either peg that Third Point or lose if we don't peg out FIRST. And so we could even do something as silly as Toss (3 Q) and still be assured of having enough points, should Pone falter. But back to reality: should we jettison the PAIR of Tens? One or both Tens could help us in at least two ways: If we hold a PAIR until last, we may see Pone run out of cards, and then PAIR ourselves. This however is extremely unlikely with a Deuce and a Trey, so that plan would require Keep (3 T T Q) or even Keep (T T Q K)! I don't like either of these. Or, we might Keep (2 3 T Q) and use the Ten Card in order to possible "trap" a Jack and peg a RUN. While this is unlikely, we need as many "unlikelys' stacked together in order to pull out a VICTORY today, so let's try Keep (2 3 T Q) and Toss (T K). Our only burden is pegging those Three Holes, and I'm not so confident we can do this any large percentage of the time today. But our goal is to at least attempt to maximize our possibilities. |
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dec 6358 votes Joined: April 2008 |
Saturday 4:00 AM
I am thinking opponent more then not likely to Keep a King in their hand. I will keep four different rankings a nd pairing is my first strat. Next hope to peg on an end run. Last hope to only got a minimal hand with no cut here for them. dec |
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thelawnet 262 votes Joined: January 2020 |
Saturday 4:02 AM
the possibilities today:
50% ish, pone has 8 points, and wants to stop us pegging 3 points at all costs 20% ish pone has 6 or possibly 7 and needs to peg 2 or maybe 1 20% ish pone has 4 or 5 and needs to peg probably 4 or 3 So MOST of the time we will be trying to peg 3 points. And the rest of the time it won't necessarily be obvious if pone has made it or not So the best hand is clearly the one that makes it easiest to score 3 points. this is 23qk, since the t's are bad as we have two of them and less likely to be paired. ttqk or similar is a terrible pegging hand because court cards are bad. if pone was a bit further away, then we might prefer 23TT here, which has the best defensive potential. But as 50% of the time we are on offense and there's no doubt whatsoever that 23qk is best for that, we'll hold the 23qk, and the slightly better defensive properties of 23tt when pone needs to peg 2 or 4, will have to be passed over. some pegging stats here, which will be reasonably valid for us in that pone holds a normal hand in this spot - he can't pick a pegging hand, which would invalidate all assumptions, as he might at 119-119 or similar. thelawnet says: er, that link https://docs.google.com/spreadsheets/d/1jfN0AVM_Rreqs-o0ZIRcEBws1wrdsszrOp7cj-Fm0Qo/edit?usp=sharing thelawnet says: note re pegging:
5 XXX is worth 6 points, or 8+ with a pair/3oak
so for example
j-k-q and pone scores 4
then
3-5-Q-T-2 and we score 1
meanwhile
j-3-k-2-5 for 1 for pone
Q-t-k
if pone has 55xx then he's home, and in that case
x-k-5-3 go
x-2-5-q 1
So it seems that although e.g., Q-Q-Q-A is a risk for us to lose, it seems that we should pair a ten card led, and reply with a 2 or 3 if we can't pair it
the reason being that if we get to round two we may score a run for 3 points if pone is holding a jack. E.g., K-J-Q
In the case of the common A4 XX hand, then pone is short, and if he leads the Ace then we can reply with the 2. In this case if a 3 comes back we pair it. If it's a 2 back then we score 15-2
There doesn't seem any sense in replying with the X in that pone probably scores 15-2, following which we don't have a reply. james500 says: Hi Law, hope the weather's as nice where you are as it is here. Thank you for your post on Reddit yesterday, plenty of information to digest. |
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mrob2199 1435 votes Joined: February 2009 |
Saturday 4:26 AM
Not the ideal pegging hand here -obviously we are keeping 2-3 but what picture cards to go with them? When I am dealer in a situation like this -needing to peg 3-I like to keep the pair and see if I can parlay them into 3 points at the end-it may prove to be tougher while holding a 2-3,but I can foresee a few scenarios where pone leads an ace or 4,and we respond with a 2 or 3,and they try to get away from trouble by playing a facecard-we then dump our other little card and hope the line has to use up both his remaining cards for the go |
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Mark6 702 votes Joined: June 2020 |
Saturday 5:40 AM
Wasn’t sure about which to toss K or Q . Love to see JQT , lawnet and mrob have 3 different solutions . Shows how complex this game is . Mark6 says: See now I should have tossed both 10s . Only thought it through half way |
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james500 3923 votes Joined: June 2013 |
Saturday 6:02 AM
Very likely to be a lone voice today, probably I'm guilty of over thinking it.
Since I've been dealt a pair of tens it's less likely my opponent has any, so they should be safe to peg with. I don't imagine Pone'd want to keep a King either, so again, if I keep mine it should be safe from being paired. With X-X-X in hand, I'll want something different as an escape card. Often those on here warn of the dangers of a lone small card, so my fourth card will be the less small of the 2-3 non-X's. |
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Jazzselke 2586 votes Joined: March 2009 |
Saturday 6:53 AM
Hopefully we can pair a Q or K, or work some magic with the 23. If we needed more than 3 points would hold the 10s. |
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Eolus619 1342 votes Joined: June 2020 |
Saturday 7:18 AM
Picking up on JQT comment from yesterday about extending the game one hand is not a viable option today. So it seems the way forward involves pairing options to help achieve 3 pegging points. Do I keep my 10s to try to create the pair that way or spread out my cards in hopes of pairing a pond played card ? Decided to go with trying to pair my own tens at the end of the sequence. Losing by two @ the end will be frustrating to say the least Eolus619 says: actually it better be pone who played the card ..yikes JQT says: When we need to peg three holes as Dealer, we should remember that if we can score with our first card played with a PAIR or (15-2), we shall *still* always get either a "go" or Last Card, and thus WIN, as long as Pone does not peg out FIRST. A PAIRS Royal by Pone could jeopardize our initial PAIRING, since now the score could be (120-119) and Pone has two cards and we have three cards remaining. Still, we MUST score if and when we can. If we can grab an early (15-2), we are assured of winning, since our final "go" or Last Card will put us out. Whether we should hold (2 3 T T), (2 3 T Q), or (2 3 Q K) is still a bit unclear, but getting an early two points is for certain a great path toward VICTORY. As Dealer here, we should never give up, because we can always score those needed Three Points with our FINAL card played, either by (15-2), PAIRING our Opponent, or PAIRING ourselves, all plus One Point for Last Card. Eolus619 says: copied your comment t to me note book ....thanks |
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ILostMyJob 3565 votes Joined: September 2013 |
Saturday 9:16 AM
I don't expect pone to have 10s, and I do expect aggressive pegging defense. So I tried to maximize the chance to either pair a card or sneak in a 31-2. |
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Gougie00 5730 votes Joined: March 2008 |
Saturday 9:20 AM
One of the each. The opponent is unlikely to have a 10, so they get tossed into the crib. Obviously in attack mode and will pair anything. I can afford yielding a triple.
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zeke76 1396 votes Joined: August 2018 |
Saturday 9:27 AM
Went for a mix ... I doubt I can pair my own 10. Why I keep a 10 I’m not sure now that I’m writing. Mark6 says: 666 Votes JQT says: I kept the Ten and Queen for the same silly reason: of course a Queen and King can 'trap' a Jack every bit as good as a Ten and Queen, so what was I thinking? It's clear that the Queen and King shall be more frequently 'seen' from Pone. Yet I am also re-thinking the whole Deuce-Trey combo here, and I wonder if the likes of something such as Keep (3 T T K) might actually work more reliably, as it would (A) Push the Count up toward Thirty-One more quickly, and (B) perhaps allow us to self-PAIR the Ten Cards. Whether this idea might be more reliable than hoping to PAIR one of the unique cards (2 3 Q K), I am not certain, but I am having second thoughts. Pegging Three Holes with four of these six cards is no easy task! zeke76 says: Mark—This must be Nero’s discard. |
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cribbagepogo 3251 votes Joined: October 2007 |
Saturday 9:37 AM
In spite of all the brain power I am against. Think of this, we want to peg out. If I keep 2 10's what are the chances that pone leads one? Very small. Pone will not throw a K into my crib unless he is a rank amateur. No need to kill my crib. Three points good, four would have been disaster. |
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warquaker 238 votes Joined: July 2020 |
Saturday 9:41 AM
Grunch: needing to peg out I keep four unmatched cards hoping to pair and then the automatic one for last card as dealer gives me the three points that I need for game. thelawnet says: Regarding your post yesterday, essentially there is a correct strategy for both players at every score, many of which will of course be the same. I am not sure why Hal's crib gives win% as w2, w3,but as a matter of provable mathematics, dealer has one single exact % chance of winning at every score pair. Therefore for a given position such as 95-100 with a given hand, we can exactly mathematically calculate (albeit that it computationally expensive) the optimal strategy which is the one that if the other player also plays optimally maximises the chance of winning. Given fixed percentage win chances from any score pair, the problem then becomes the weighted average of the winning chance for every possible score combination after our current hand. It follows that this number should simply be W Vs. L, otherwise the program is by definition not playing optimally, since whereas today's position clearly has a set of binary outcomes either 0 or 1 because it must be over in one deal, the more complicated weighted average at scores further out would value a position like 100-95* as say 0.6, such that the program then considers the possible outcomes as a continuous distribution of numbers between 0 and 1, with which it attempts to maximise its pay-off. warquaker says: the lawnet: thank you for your comprehensive and considered response. JQT says: Starting at (0-0), the number of deals needed to win is probabilistic, not deterministic. Otherwise, all accurately-played games would end in N deals, say N=7. Clearly, most games end in 7, 8, or 9 deals, but it depends on probability. The program is looking ahead N deals and using statistical, probabilistic methods, and then it shows us a few of the more likely outcomes, such as Win in one or two deals, or W1 and W2. It might take more deals, but the software is showing us just two of the most likely number of deals and the associated winning and losing percentages for each. They do not add up to 100% because all other possible number of deals are not shown due to limited space, and low chance of occurrence. The odds are only pseudo-deterministic at a given relative score before we see our cards: one the cards are dealt, and now we "see" six (and we do NOT see forty-six) cards, now Cribbage essentially becomes probabilistic in nature. Eolus619 says: now that JQT and thelawnet have commented on “style “ i feel much better about the input you received warquaker says: JQT: thanks for fleshing out just what Halscrib shows. thelawnet says: Hi, JQT while it's true that games could last 6, 7,8,9,10, or possibly even more deals, that doesn't change the fact that both before and after the deal there is a single binary probability of winning that can be proven mathematically. That is to say, if we consider positions 120-120* 119-120*, etc., the game is completely certain to end, regardless of what the cards are. If we iterate over the all combinations of cards, and then simultaneously optimize both players' decision, we get a number p between 0 and 1 that reflects the exact chance of dealer (pone = 1 - d) winning from that score. If we continue this process at scores further from 121, we will find scores where the game might not end this turn. At those scores the process is still the same - iterate over all possible deals, and then simultaneously optimize the players' respective choices, but this time there will be some outcomes that result in 1, some that result in 0, and some in an outcome somewhere in between. Maximizing the value of this is still a matter of the sum of the product of the value of each node (now not always 0 and 1 but also say 0.43845843) by the probability of it happening based on optimal player space.
It therefore is not correct to 'look ahead', because the number of deals is not at all relevant to the clear goal of the program, which is to maximize the chance of winning. Since we know and have proven by exhaustive analysis the chance of winning from all terminal positions, then from all non-terminal positions we simply do not care and nor should we be interested in the number of deals remaining, because we know that from say 100-100 we have a certain p of winning, and whether this p occurs in 1, 2, 3, or even 4 (!?) deals shouldn't concern us at all, only whether a certain outcome (in terms of score after the game) is a good one. As far as the difference between 'before' and 'after' the cards goes, it doesn't really change the process. The simultaneous optimization process still applies. As pone we do not know what cards dealer has and vice versa, but we can solve (with exact accuracy) the problem by considering the full set of all trillions of deals and simultaneously optimizing for a given score based on both players mutually solving for the information they have about the other (the other player's cards are not shared, but do feed into the pegging decision tree, one card at a time). Having solved the full set of deals, when we hold a hand such as 23TTQK, then it comes a question of considering the subset of all possible deals where dealer holds 23TTQK, considering pone hands. Since both players' strategy have been calculated for all possible deals, we iterate over each one, given the conditional probability that dealer holds 23TTQK, and find the exact % chance that dealer wins. |
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SallyAnn3 908 votes Joined: March 2020 |
Saturday 9:43 AM
Keep what I need to go out....lead a 10 in case they are desperate enough to need 2 points...if they do, I'm out :) |
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warquaker 238 votes Joined: July 2020 |
Saturday 1:43 PM
We know the cut but when we discarded we did not. His Heels saves the day 8.7% of the time. |
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Ras2829 5154 votes Joined: November 2008 |
Saturday 1:53 PM
Dealer can often get in a large pair to close the count. If not scoring two pegs with deuce or trey, still have the pair in reserve. mrob2199 says it all for me. Just an echo this day! JQT says: This is a very common and reliable tactic, to self-PAIR our own cards, and I am beginning to wonder whether it might actually work and work better if we Keep (3 T T K) to drive the Count up toward Thirty-One more rapidly. And due to BIAS for the pegging battle, the King might be Pone's treasured (and thus most commonly-held) "safe" card here. |
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Coeurdelion 5595 votes Joined: October 2007 |
Saturday 2:40 PM
We need 3pts to peg out. I think I'll try 2-3-Q-K. |
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HalscribCLX 5318 votes Joined: February 2008 |
Saturday 3:09 PM
At 118*-113 playing an Optimal strategy (cautious offense) the Peg Out %s are:
Optimal_______Peg Out % 2-3-10-10_______58.7 2-3-10-Q________50.0 2-3-10-K________48.9 2-3-Q-K_________48.1 So I'll select Q-K to discard. After the 8 cut I'll play Defense to the lead. warquaker says: Is this answer definitive or would anyone disagree with this? JQT says: Pone can 'duck and dodge' (or at least try to avoid) the (15-2) and any potential RUNS, but what Pone cannot deny is the Thirty-One ceiling on the Count, and so Pone may very well have to "play out" all of his or her cards thus allowing us to self-PAIR and WIN, a very common and reliable tactic. I even wondered (above, late in the day) whether Keep (3 T T K) might even do this more reliably, but the program says NO, and I am inclined to trust it. It's just that against very strong human players, I think it might be crucial to drive the Count up more quickly, and the Deuce and Trey don't seem best for this. But Pone will want to avoid handing us an early (15-2) as that clinches a WIN for us, so out of four 'safe' leads, namely A, 2, 3, 4, if we retain (2 3 T T), we have HALF of these bases covered. I've lost any number of games in such similar positions against this program when it is the Dealer at Hole 118 and it self-PAIRS with exactly such tactics, so I feel it's definitive. warquaker says: If it's good enough for you JQT then it is more than sufficient for me. thelawnet says: the answer can't possibly be definitive because it prioritises pegging out, whereas there is only one goal, which is winning. It might still be correct, but we can't really assume that.
In general there is one correct hold, and then one correct reply to each possible lead by pone. However it doesn't seem that the program is necessarily finding it. There is some indication here about improvements in the program's endgame play http://web.archive.org/web/20160125190941/http://www.halscrib.com/update/about
however endgame play has a correct strategy and this is evidently an approximation, which might or might not be correct. |
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