I just played a game where the table host enabled "difficult scoring mode". What does that mean? It certainly meant that I didn't win when I thought I ought to. Any ideas?
"Difficult" scoring mode
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RicardoRix
- Posts: 2109
- Joined: 29 April 2012, 23:43
Re: "Difficult" scoring mode
When difficult scoring mode is enabled you need to check the size of the card sets does not exceed 3. As sets increase, the points decrease which may cause confusion for those new to this scoring mode.
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PurplishCat
- Posts: 11
- Joined: 29 November 2016, 05:02
Re: "Difficult" scoring mode
More specifically (from Wikipedia): The more difficult scoring scheme gives a player one point for one card of a color, four points for two cards, eight points for three cards, seven points for four cards, six points for five cards, and five points for six or more cards of the same color. This arrangement can lead to usually desirable wild cards being a liability for players.
That really should be in the help for the game, as the current description is very vague.
That really should be in the help for the game, as the current description is very vague.
Re: "Difficult" scoring mode
Thus, the ideal scenario at the end of the game is to possess exactly 9 cards (3 sets of 3 cards in a color) plus a lot of +2-cards.
Playing in mode "difficult", you usally want to take only a single card in each round. +2-cards are extremely important.
Playing in mode "difficult", you usally want to take only a single card in each round. +2-cards are extremely important.
Re: "Difficult" scoring mode
I tried to figure out how the scoring is done after my friends and I played in difficult scoring mode tonight. With a wild card present, it did not score the way we expected when one of us had to draw a card he didn’t want, and we could not figure out how his score was calculated at the end of the game. I think I tried 2-3 different theories and got scores that were a few below what he actually got.
We didn’t want to spend the time to try to figure it out, since we wanted to get more games in before the end of our evening, and I haven’t found anything online that addressed this, including the brief description of difficult scoring included in the rules for the game at BGA, so when I saw this post, I thought I’d reply under it, despite being a bit on the older side. Maybe someone else has figured this out?
I don’t remember his exact hand, but if you apply the wild card to blue, which was the largest set of cards, he scored a 5. Add 6 for his second set and an 8 for his 3rd set, that gets us 19. The remaining cards had point values of 4, 1, and 1 (2 card set, 1 card set, 1 card set). That’s -6. That’s a total of 13. Only I’m pretty sure he final score was 17 (I don’t have it in front of me anymore; point is, it was more than 13). So clearly we aren’t applying the wild card to his sets the way BGA is, but it is not at all transparent how BGA does this. It’s almost as if the wild card improved his hand somehow instead of hurting it, but we can’t figure out how.
If wild cards are applied in whatever way they best help your score (which is how I would imagine it is intended, given the way it works on the other scoring methoed), but they add another card of a chosen color to your hand, it *can’t* help you if you already have 3 sets of 3 or more colors. In other words, if you have 5 blue, 4 yellow, and 3 green, adding a wild card doesn’t help any of these. And you can’t add it to another card, say a single red, because now you’ve decreased your score by another 3 points. So, in this scenario, a wild card does seem bad and it would only appear to help if you somehow managed to take so few cards that the wild card takes your 3rd set up from 2 to 3 cards.
So, even if we pulled the wild card from my friend’s hand, I don’t see how he could have a score so much higher than what we calculated. This of course assumes it is added wherever it does the least damage, which to ours was the blue set, since putting it there meant no impact on his score (you can’t go back under 5 points once you max out the color). And if it had applied to one of his other sets, it would have lowered his score by at least one point.
Anyway, you get the idea. It makes no sense to me. Can someone explain it?
Recap:
7 BLUE (5 points) | 5 BROWN (6 points) | 3 YELLOW (8 points) | 2 RED (-4 points) | 1 GREEN (-1 points) | 1 ORANGE (-1 points) | 1 WILD (? points)
If the wild card is applied to blue, it effectively looks like this (no extra points, because you can’t go any lower than 5 after hitting 8):
8 BLUE (5 points) | 5 BROWN (6 points) | 3 YELLOW (8 points) | 2 RED (-4 points) | 1 GREEN (-1 points) | 1 ORANGE (-1 points)
This is a total of 19 - 6, or 13. This is a few points lower than actual (17?)
Why do it this way? Well, if the wild card is applied to the set where it does the most good or, conversely, the least damage, then there is no better location to put it than blue, where no more points can be taken away. Brown would lose a point. Yellow would lose a point. Red would lose 4 points. Either green or orange would lose 3 points.
We didn’t want to spend the time to try to figure it out, since we wanted to get more games in before the end of our evening, and I haven’t found anything online that addressed this, including the brief description of difficult scoring included in the rules for the game at BGA, so when I saw this post, I thought I’d reply under it, despite being a bit on the older side. Maybe someone else has figured this out?
I don’t remember his exact hand, but if you apply the wild card to blue, which was the largest set of cards, he scored a 5. Add 6 for his second set and an 8 for his 3rd set, that gets us 19. The remaining cards had point values of 4, 1, and 1 (2 card set, 1 card set, 1 card set). That’s -6. That’s a total of 13. Only I’m pretty sure he final score was 17 (I don’t have it in front of me anymore; point is, it was more than 13). So clearly we aren’t applying the wild card to his sets the way BGA is, but it is not at all transparent how BGA does this. It’s almost as if the wild card improved his hand somehow instead of hurting it, but we can’t figure out how.
If wild cards are applied in whatever way they best help your score (which is how I would imagine it is intended, given the way it works on the other scoring methoed), but they add another card of a chosen color to your hand, it *can’t* help you if you already have 3 sets of 3 or more colors. In other words, if you have 5 blue, 4 yellow, and 3 green, adding a wild card doesn’t help any of these. And you can’t add it to another card, say a single red, because now you’ve decreased your score by another 3 points. So, in this scenario, a wild card does seem bad and it would only appear to help if you somehow managed to take so few cards that the wild card takes your 3rd set up from 2 to 3 cards.
So, even if we pulled the wild card from my friend’s hand, I don’t see how he could have a score so much higher than what we calculated. This of course assumes it is added wherever it does the least damage, which to ours was the blue set, since putting it there meant no impact on his score (you can’t go back under 5 points once you max out the color). And if it had applied to one of his other sets, it would have lowered his score by at least one point.
Anyway, you get the idea. It makes no sense to me. Can someone explain it?
Recap:
7 BLUE (5 points) | 5 BROWN (6 points) | 3 YELLOW (8 points) | 2 RED (-4 points) | 1 GREEN (-1 points) | 1 ORANGE (-1 points) | 1 WILD (? points)
If the wild card is applied to blue, it effectively looks like this (no extra points, because you can’t go any lower than 5 after hitting 8):
8 BLUE (5 points) | 5 BROWN (6 points) | 3 YELLOW (8 points) | 2 RED (-4 points) | 1 GREEN (-1 points) | 1 ORANGE (-1 points)
This is a total of 19 - 6, or 13. This is a few points lower than actual (17?)
Why do it this way? Well, if the wild card is applied to the set where it does the most good or, conversely, the least damage, then there is no better location to put it than blue, where no more points can be taken away. Brown would lose a point. Yellow would lose a point. Red would lose 4 points. Either green or orange would lose 3 points.
Re: "Difficult" scoring mode
Your problem was a nice puzzle to solve - with a solution that surprised me.
I assume that you are talking about game #113777880. One player had 7 blue, 3 green, 3 pink, 2 yellow, 1 brown, 1 orange and 2 jokers. He got +17 points in his final score.
If you count the two jokers as blue (in order to avoid the jokers to count negative), then you get 15 points (+5 for blue, +8 for green, +8 for pink, -4 for yellow and -1 each for brown and orange). +21-6=+15.
However, there is a better way to choose the colors of the jokers and it is: 1 joker is blue, 1 joker is yellow (!). This solution is unique and gives +17 points!
Now you have 8 blue, 3 green, 3 pink, 3 yellow, 1 brown and 1 orange. The trick is to count blue as one of your negative colors! From the rule book: "Each player chooses which 3 of his colors score plus points. All other colors score minus points for the player."
So we get 17 points (+8 each for green, pink and yellow, -5 for blue, -1 each for brown and orange). +24-7=+17.
@epski: The numbers in your post are very similar to the numbers of game #113777880 but you have some differences (you talk about 5 brown and only 1 joker). If you mean another game the trick for optimal joker placement surely will be the same. You do not have to count blue as positive only because it is the color with maximum number of cards.
I assume that you are talking about game #113777880. One player had 7 blue, 3 green, 3 pink, 2 yellow, 1 brown, 1 orange and 2 jokers. He got +17 points in his final score.
If you count the two jokers as blue (in order to avoid the jokers to count negative), then you get 15 points (+5 for blue, +8 for green, +8 for pink, -4 for yellow and -1 each for brown and orange). +21-6=+15.
However, there is a better way to choose the colors of the jokers and it is: 1 joker is blue, 1 joker is yellow (!). This solution is unique and gives +17 points!
Now you have 8 blue, 3 green, 3 pink, 3 yellow, 1 brown and 1 orange. The trick is to count blue as one of your negative colors! From the rule book: "Each player chooses which 3 of his colors score plus points. All other colors score minus points for the player."
So we get 17 points (+8 each for green, pink and yellow, -5 for blue, -1 each for brown and orange). +24-7=+17.
@epski: The numbers in your post are very similar to the numbers of game #113777880 but you have some differences (you talk about 5 brown and only 1 joker). If you mean another game the trick for optimal joker placement surely will be the same. You do not have to count blue as positive only because it is the color with maximum number of cards.
Re: "Difficult" scoring mode
I just looked at your post again. You seem to quote another game of yours.epski wrote: ↑24 September 2020, 05:15
7 BLUE (5 points) | 5 BROWN (6 points) | 3 YELLOW (8 points) | 2 RED (-4 points) | 1 GREEN (-1 points) | 1 ORANGE (-1 points) | 1 WILD (? points)
If the wild card is applied to blue, it effectively looks like this (no extra points, because you can’t go any lower than 5 after hitting 8):
8 BLUE (5 points) | 5 BROWN (6 points) | 3 YELLOW (8 points) | 2 RED (-4 points) | 1 GREEN (-1 points) | 1 ORANGE (-1 points)
This is a total of 19 - 6, or 13. This is a few points lower than actual (17?)
If you apply the wild card to RED and choose the positive colors to be brown, yellow and red, then you obtain:
7 BLUE (-5 points) | 5 BROWN (6 points) | 3 YELLOW (8 points) | 3 RED (8 points) | 1 GREEN (-1 points) | 1 ORANGE (-1 points)
This gives a total of 15 points which is more than your expected 13. Please check with the actual game of yours - the total should be +15 if you gave the correct totals of each color.
The trick is that counting 3 red positive changes your score from -4 to +8 which gains you 12 points. But now you have to change a positive color to negative and you choose the blue one: Instead of +5 you get -5 which loses you 10 points. So in total the joker gained you +2 points.
Re: "Difficult" scoring mode
Thank you for that explanation. I hadn't even considered the possibility that the wild cards would flip colors that were not in the 3 sets of higher cards. I'll have to look at this again, and keep it in mind for future games!
I'm not sure what you mean about quoting another game I played. You had the right one. I did that from memory, as I didn't know at the time how to find games I'd played with friends (they don't appear to be listed in notifications).
So, I remembered incorrectly. The final cards for elugardo were:
7Bl | 3Gr | 3Pi | 2Ye | 1Br | 1 Or | 2 W
Treating blue as a liability worth -5 (max points for taking 6 or more cards) and applying 1 wild card to yellow and 1 to blue gives me the following scoring:
8Bl | 3Gr | 3Pi | 3Ye | 1Br | 1 Or
-5 | 8 | 8 | 8 | -1 | -1 = 17
Man, nice work. This gives me a deeper appreciation for thoughtful scoring at the end of an in-person game, so I don't miss a better score than what I may initially see.
I'm not sure what you mean about quoting another game I played. You had the right one. I did that from memory, as I didn't know at the time how to find games I'd played with friends (they don't appear to be listed in notifications).
So, I remembered incorrectly. The final cards for elugardo were:
7Bl | 3Gr | 3Pi | 2Ye | 1Br | 1 Or | 2 W
Treating blue as a liability worth -5 (max points for taking 6 or more cards) and applying 1 wild card to yellow and 1 to blue gives me the following scoring:
8Bl | 3Gr | 3Pi | 3Ye | 1Br | 1 Or
-5 | 8 | 8 | 8 | -1 | -1 = 17
Man, nice work. This gives me a deeper appreciation for thoughtful scoring at the end of an in-person game, so I don't miss a better score than what I may initially see.