## Poll

 Yes -- Simple games for simple minds. No votes (0%) Call me maybe -- Depends on the Wiz's math. 2 votes (40%) No -- Probably better bets available. 1 vote (20%) I'll stick to Flip-It, thank you very much. No votes (0%) I'm not sure I trust the Cambodian police. No votes (0%) Nathan for moderator. 1 vote (20%) Total eclipse reminder -- 04/08/2024 1 vote (20%) Congratulations on your bike ride yesterday, Wiz! 1 vote (20%) The Winn casino in Bavet looks familiar. No votes (0%) Vegas Golden Nights! 1 vote (20%)

5 members have voted

ThatDonGuy
Joined: Jun 22, 2011
• 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

Joined: Feb 18, 2015
• 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
Joined: May 21, 2013
• 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
Joined: Jun 22, 2011
• Posts: 4922
Thanks for this post from:
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

Joined: Feb 18, 2015
• Posts: 3185
June 3rd, 2018 at 9:29:42 PM permalink

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
Joined: Jun 17, 2011
• 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.
 10 1 4 7 9 2 5 8 8 3 6 9 7 4 10 6 1 5 8 5 2 6 9 4 3 7 10 3 1 4 8 2 5 9 1 3 6 10

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

Joined: Oct 14, 2009
• 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.

 Joker 1 1 1 1 Joker 1 1 1 2 Joker 1 1 1 4 Joker 1 1 1 5 Joker 1 1 1 7 Joker 1 1 1 9 Joker 1 1 2 4 Joker 1 1 4 4 Joker 1 1 4 9 Joker 1 1 5 5 Joker 1 1 5 7 Joker 1 1 7 9 Joker 1 1 9 9 Joker 1 2 4 8 Joker 1 2 8 8 Joker 1 4 4 4 Joker 1 4 4 8 Joker 1 4 4 9 Joker 1 4 9 9 Joker 1 5 5 5 Joker 1 5 5 7 Joker 1 5 5 8 Joker 1 5 8 8 Joker 1 7 8 8 Joker 1 7 9 9 Joker 1 8 8 8 Joker 1 9 9 9 Joker 2 3 3 3 Joker 2 3 3 6 Joker 2 3 3 7 Joker 2 3 3 10 Joker 2 3 6 6 Joker 2 3 6 7 Joker 2 3 6 10 Joker 2 3 10 10 Joker 2 4 6 6 Joker 2 6 6 6 Joker 2 6 6 7 Joker 2 6 6 10 Joker 2 6 10 10 Joker 2 7 8 8 Joker 2 8 8 8 Joker 3 3 3 3 Joker 3 3 3 5 Joker 3 3 3 6 Joker 3 3 3 7 Joker 3 3 3 9 Joker 3 3 3 10 Joker 3 3 5 5 Joker 3 3 5 7 Joker 3 3 5 10 Joker 3 3 6 6 Joker 3 3 6 7 Joker 3 3 6 9 Joker 3 3 6 10 Joker 3 3 7 9 Joker 3 3 9 9 Joker 3 3 9 10 Joker 3 3 10 10 Joker 3 5 5 5 Joker 3 5 5 7 Joker 3 5 10 10 Joker 3 6 6 6 Joker 3 6 6 7 Joker 3 6 6 9 Joker 3 6 6 10 Joker 3 6 7 9 Joker 3 6 9 9 Joker 3 6 9 10 Joker 3 6 10 10 Joker 3 7 9 9 Joker 3 9 9 9 Joker 3 9 9 10 Joker 3 9 10 10 Joker 4 4 4 4 Joker 4 4 4 5 Joker 4 4 4 6 Joker 4 4 4 8 Joker 4 4 4 9 Joker 4 4 5 5 Joker 4 4 5 8 Joker 4 4 6 6 Joker 4 4 6 9 Joker 4 4 9 9 Joker 4 5 5 5 Joker 4 5 5 8 Joker 4 6 6 6 Joker 4 6 6 9 Joker 4 6 9 9 Joker 4 9 9 9 Joker 5 5 5 5 Joker 5 5 5 7 Joker 5 5 5 8 Joker 5 5 8 8 Joker 5 8 8 8 Joker 5 8 8 10 Joker 5 8 10 10 Joker 6 6 6 6 Joker 6 6 6 7 Joker 6 6 6 9 Joker 6 6 6 10 Joker 6 6 7 9 Joker 6 6 9 9 Joker 6 6 9 10 Joker 6 6 10 10 Joker 6 7 9 9 Joker 6 9 9 9 Joker 6 9 9 10 Joker 6 9 10 10 Joker 7 8 8 8 Joker 7 9 9 9 Joker 8 8 8 8 Joker 8 8 8 10 Joker 8 8 10 10 Joker 9 9 9 9 Joker 9 9 9 10 Joker 9 9 10 10

 Joker 1 2 4 8 Joker 2 3 6 7 Joker 2 3 6 10 Joker 3 6 7 9 Joker 3 6 9 10
It's not whether you win or lose; it's whether or not you had a good bet.
gordonm888

Joined: Feb 18, 2015
• 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
Joined: Jun 17, 2011
• 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
Joined: Dec 1, 2012