Poll

No votes (0%)
2 votes (40%)
1 vote (20%)
No votes (0%)
No votes (0%)
1 vote (20%)
1 vote (20%)
1 vote (20%)
No votes (0%)
1 vote (20%)

5 members have voted

ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
June 3rd, 2018 at 6:14:38 PM permalink
Quote: beachbumbabs


W, 2, 3, 6, T
W, A,2,3,4



Doesn't W, A, 2, 3, 4 qualify with (6)-A-3 / 2-4?


A quick way to determine your hand:

If you have no joker, it's easy - add up the points of all five cards, and if any pair adds up to the corresponding "modulo 10" value, use that hand. It does not matter if there are multiple ways to make a qualifying hand, as all of the 2-card hands will have the same value.

If you have a joker, it's a little more complicated. It is the highest of:
(a) Add up the points of the other four cards, and if any one card has the sum's "modulo 10" value, use that card and the joker value that gives you the highest two-card total (3, if your card is 5-7; 6, if it is 2-4; 9, if it is anything else)
(b) For each pair of non-joker cards, if they add up to 4, 7, 11, 14, or 17, use the value of the other two cards (for example, with Joker-2-3-5-6, (3)-2-5/3-6 is worth 9, and (9)-5-6/2-3 is worth 5.
gordonm888
Administrator
gordonm888 
Joined: Feb 18, 2015
  • Threads: 49
  • Posts: 3185
June 3rd, 2018 at 6:32:07 PM permalink
Bab's first answer is correct. One of only two correct answers I have been able to find so far (but I have not been using a spreadsheet or computer.) Congrats !

I believe Bab's 2nd answer is not correct.
Bab's 2nd answer was (W,A,2,3,4) but, unfortunately, 4+3+W =10


This is harder than it looks.

Edit: I believe I have found 5 correct answers (without using a computer) by employing a mathematical insight.
Last edited by: gordonm888 on Jun 3, 2018
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
beachbumbabs
Administrator
beachbumbabs
Joined: May 21, 2013
  • Threads: 99
  • Posts: 14232
June 3rd, 2018 at 6:40:38 PM permalink
Quote: gordonm888

Bab's first answer is correct. One of only two correct answers I have been able to find so far (but I have not been using a spreadsheet or computer.) Congrats !

I believe Bab's 2nd answer is not correct.

Bab's 2nd answer was (W,A,2,3,4) but, unfortunately, 4+3+W =10


This is harder than it looks.



Yeah, darn. (6)A3 does qualify.

Added that one too fast. Was trying to find one specific with an Ace in it. Hmmm.
If the House lost every hand, they wouldn't deal the game.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4922
Thanks for this post from:
beachbumbabs
June 3rd, 2018 at 7:03:35 PM permalink
Quote: beachbumbabs

Added that one too fast. Was trying to find one specific with an Ace in it. Hmmm.


There is only one - Joker, A, 2, 4, 8

Note that any hand with Joker, Ace, and any of 3, 6, or 10 can qualify (with (6)-A-3, (3)-A-6, and (9)-A-10, respectively).
This leaves 2, 4, 5, 7, 8, and 9 as the possible cards that can be included; there are (6)C(3) = 20 such hands.

A 2 4 5 qualifies with (3)-2-5
A 2 4 7 qualifies with (9)-4-7
A 2 4 8 does not qualify
A 2 4 9 qualifies with (9)-2-9
A 2 5 7 qualifies with A-2-7
A 2 5 8 qualifies with (3)-2-5
A 2 5 9 qualifies with (3)-2-5
A 2 7 8 qualifies with A-2-7
A 2 7 9 qualifies with A-2-7
A 2 8 9 qualifies with (3)-8-9
A 4 5 7 qualifies with A-4-5
A 4 5 8 qualifies with A-4-5
A 4 5 9 qualifies with A-4-5
A 4 7 8 qualifies with (9)-4-7
A 4 7 9 qualifies with 4-7-9
A 4 8 9 qualifies with (3)-8-9
A 5 7 8 qualifies with 5-7-8
A 5 7 9 qualifies with (6)-5-7
A 5 8 9 qualifies with (3)-8-9
A 7 8 9 qualifies with (3)-8-9

gordonm888
Administrator
gordonm888 
Joined: Feb 18, 2015
  • Threads: 49
  • Posts: 3185
June 3rd, 2018 at 9:29:42 PM permalink
I had not found the solution that ThatDonGuy came up with. Congrats to Don.



Here's the solutions that I did in my head:

W + any 4 of the following 5 cards: 2,3,6,9,10
W,2,3,6,9
W,2,3,6,10 (Bab's solution)
W,2,3,9,10
W,2,6,9,10
W,3,6,9,10

My thoughts went like this: we want to avoid any 3 cards that sum to 10 or 20. 10 is 1mod(3) and 20 is 2mod(3).

Now the wild card is 3,6, or 9 so it is always 0mod(3). If the other cards are also of the form 0mod(3) then they (combined with the wild card) can only sum to a total that is 0mod(3) and thus cannot be combined in any way to make a sum of 10 or 20. So, I started with W,3,6,9 because 3,6 and 9 are all 0mod(3). But I needed a fifth card so, by inspection, I determined that W,3,6,9,2 and W,3,6,9,10 were the only possibilities that worked. And then I realized that 2+10=12 and 12 is also 0mod(3).

From there, it was a short leap to see that any 4 numbers of the set (2,3,6,9,10) could be combined with a Joker and not qualify.

Of course, none of that deduction indicates that other solutions could not exist, but I suspected that those 5 solutions were all there were. ThatDonGuy has found an additional solution. And maybe there are more.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 34
  • Posts: 2424
June 4th, 2018 at 8:56:41 AM permalink
Quote: gordonm888


W + any 4 of the following 5 cards: 2,3,6,9,10
W,2,3,6,9
W,2,3,6,10 (Bab's solution)
W,2,3,9,10
W,2,6,9,10
W,3,6,9,10


I think those with W29 would add up to 20. It seems you need to numbers
(a) Two of them don't add up to 10x+1, 10x+4, 10x+7 or
(b) obviously any three (not the joker) to 10x.

The method I've tried runs as follows.
If your first card is a 10 then you can't have a second card that is 1 4 or 7 (as this with the joker would qualify).
Use a similar idea for all the other possible starting cards.
10147
9258
8369
74 10
6158
5269
43710
3148
2 59
13610

Then consider all starting 2-cards 10 9, 10 8, 10 6, 10 5, 9 7.....
This gives a list of numbers which can't be considered.
10 9 (1 4 7 2 5 8) leaving 10 9 6 3 as possible.
10 8 (1 4 7 3 6 9) leaving 10 8 5 2 - but W25 adds up to 10 (this can be seen by seeing whether any numbers rejected by using 5 then 2 are already in the numbers used).
Continue this idea gives
10 9 6 3
10 6 3 2
9 7 6 3
8 4 2 1
7 6 3 2
edit - removed one erroneous entry 8541 as 541 add up to 10!
Last edited by: charliepatrick on Jun 4, 2018
Wizard
Administrator
Wizard 
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23283
June 4th, 2018 at 9:31:23 AM permalink
These come by way of my program. If I'm missing any, I must have a programming error.


Joker1111
Joker1112
Joker1114
Joker1115
Joker1117
Joker1119
Joker1124
Joker1144
Joker1149
Joker1155
Joker1157
Joker1179
Joker1199
Joker1248
Joker1288
Joker1444
Joker1448
Joker1449
Joker1499
Joker1555
Joker1557
Joker1558
Joker1588
Joker1788
Joker1799
Joker1888
Joker1999
Joker2333
Joker2336
Joker2337
Joker23310
Joker2366
Joker2367
Joker23610
Joker231010
Joker2466
Joker2666
Joker2667
Joker26610
Joker261010
Joker2788
Joker2888
Joker3333
Joker3335
Joker3336
Joker3337
Joker3339
Joker33310
Joker3355
Joker3357
Joker33510
Joker3366
Joker3367
Joker3369
Joker33610
Joker3379
Joker3399
Joker33910
Joker331010
Joker3555
Joker3557
Joker351010
Joker3666
Joker3667
Joker3669
Joker36610
Joker3679
Joker3699
Joker36910
Joker361010
Joker3799
Joker3999
Joker39910
Joker391010
Joker4444
Joker4445
Joker4446
Joker4448
Joker4449
Joker4455
Joker4458
Joker4466
Joker4469
Joker4499
Joker4555
Joker4558
Joker4666
Joker4669
Joker4699
Joker4999
Joker5555
Joker5557
Joker5558
Joker5588
Joker5888
Joker58810
Joker581010
Joker6666
Joker6667
Joker6669
Joker66610
Joker6679
Joker6699
Joker66910
Joker661010
Joker6799
Joker6999
Joker69910
Joker691010
Joker7888
Joker7999
Joker8888
Joker88810
Joker881010
Joker9999
Joker99910
Joker991010



Joker1248
Joker2367
Joker23610
Joker3679
Joker36910
It's not whether you win or lose; it's whether or not you had a good bet.
gordonm888
Administrator
gordonm888 
Joined: Feb 18, 2015
  • Threads: 49
  • Posts: 3185
June 4th, 2018 at 10:04:07 AM permalink
Ah, Charlie and Wiz are both right. Three of my solutions were in error because I didn't notice that any five cards with a W,2,9 in it will qualify.

Based on their list, an unexpected conclusion is that if your hand has a Joker and a five and no duplicate ranks, the hand will always qualify (will always have 3 cards that add up to either 10 or 20.)

There is a weird mathematics here that probably no one has ever looked at. Of course, it is probably utterly useless.
Last edited by: gordonm888 on Jun 4, 2018
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 34
  • Posts: 2424
June 4th, 2018 at 10:21:29 AM permalink
Quote: Wizard

These come by way of my program. If I'm missing any, I must have a programming error.
...Full list (including duplicate ranks)...Short list (no duplicate ranks)....


I agree with the number in your long and short lists.
terapined
terapined
Joined: Dec 1, 2012
  • Threads: 86
  • Posts: 5796
June 4th, 2018 at 11:02:14 AM permalink
I caught an error but I see error was caught, BBB 2nd spoiler, o deleted
"Everybody's bragging and drinking that wine, I can tell the Queen of Diamonds by the way she shines, Come to Daddy on an inside straight, I got no chance of losing this time" -Grateful Dead- "Loser"

  • Jump to: