Pai Gow Challenge

Here's a fun challenge for anyone who likes Pai Gow tiles.

Try to come up with a distribution of the 32 tiles into 8 low and high hand settings which collectively have the fewest points.

Guidelines:

- Use a standard 32-tile deck
- Deal 4 tiles to 8 positions
- Split each 4-tile combination into a low and high hand using the Traditional Way (as described on paigow.com) or a viable way
- For example, do not deal 4-4-6-6 or 9-10-10-11 to a position and set it as 0/0
- Count a Pair as 12 points
- Count a Wong as 11 points
- Count a Gong as 10 points

A distribution with the most points would be one where all 8 positions receive two pairs, for a total of 192 points.

But what might be the lowest possible total number of points, and what is an example of a distribution which achieves that total?

Here's an example which sums to 70 points:

5 H7 H6 H8 = 3/3 (6 points)
GJ H7 L0 H8 = 5/6 (11 points)
L8 L6 H4 L2 = 0/Gong (10 points)
L7 9 L4 H6 = 3/3 (6 points)
5 11 H0 H2 = 3/5 (8 points)
GJ L0 11 H4 = 5/6 (11 points)
L4 H0 L2 H2 = 4/4 (8 points)
L7 L8 9 L6 = 5/5 (10 points)

This is gonna take me a while.

Edit -- with my luck in tiles, I could just go play eight $25 hands and report my results.

I came up with a solution that equals yours JB, can't get lower yet but I am going to tinker a bit more. It seems very efficient and I would be surprised if it could be reduced much further if any.

My solution:

Quote: JB

Here's a fun challenge for anyone who likes Pai Gow tiles.

Try to come up with a distribution of the 32 tiles into 8 low and high hand settings which collectively have the fewest points.

Guidelines:

- Use a standard 32-tile deck
- Deal 4 tiles to 8 positions
- Split each 4-tile combination into a low and high hand using the Traditional Way (as described on paigow.com) or a viable way
- For example, do not deal 4-4-6-6 or 9-10-10-11 to a position and set it as 0/0
- Count a Pair as 12 points
- Count a Wong as 11 points
- Count a Gong as 10 points

A distribution with the most points would be one where all 8 positions receive two pairs, for a total of 192 points.

But what might be the lowest possible total number of points, and what is an example of a distribution which achieves that total?

Here's an example which sums to 70 points:

5 H7 H6 H8 = 3/3 (6 points)
GJ H7 L0 H8 = 5/6 (11 points)
L8 L6 H4 L2 = 0/Gong (10 points)
L7 9 L4 H6 = 3/3 (6 points)
5 11 H0 H2 = 3/5 (8 points)
GJ L0 11 H4 = 5/6 (11 points)
L4 H0 L2 H2 = 4/4 (8 points)
L7 L8 9 L6 = 5/5 (10 points)

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Without using a computer, I have come up with a distribution that sums to 50 points. I will share it later if nobody can find a better (worse) distribution.

Best I could do was also 50.

Quote: sodawater

Best I could do was also 50.

Show Spoiler

Lol I set my 4/4 hands with gee joon as 3/5 but it still works if you correct them to 4/4

I posted this last night soon before going to bed. After posting it, I realized it wasn't as challenging as it initially seemed. If you add up all the tiles (as singletons, counting 10s as 0, 11s as 1, 12s as 2) they add up to 150 points. So the goal is to have as many "wraparound" combinations as possible, each of which reduces the total points by 10. I think the maximum quantity of wraparound combinations you can make is 10, leaving 50 points.

My first attempt, the 50-point solution I put in the spoiler, I started setting 1/2 and 2/2 and 2/3 and 2/4 and and 3/4 and 2/5 hands but was stuck when I got to the Gee Joon tiles, which work best with a 7, 8, or 9. What I ended up doing was keeping them as a pair, with 9&11 in the low hand, and despite the pair the total was 50 points. On my second attempt I rearranged things and came up with the same solution as Gazreal (before seeing his post), which was four sets of similar low & high hands.

I'm sure there are plenty of ways to achieve the 50-point total, but I now doubt that any further reduction is possible.

Quote: JB

I posted this last night soon before going to bed. After posting it, I realized it wasn't as challenging as it initially seemed. If you add up all the tiles (as singletons, counting 10s as 0, 11s as 1, 12s as 2) they add up to 150 points. So the goal is to have as many "wraparound" combinations as possible, each of which reduces the total points by 10. I think the maximum quantity of wraparound combinations you can make is 10, leaving 50 points.

My first attempt, the 50-point solution I put in the spoiler, I started setting 1/2 and 2/2 and 2/3 and 2/4 and and 3/4 and 2/5 hands but was stuck when I got to the Gee Joon tiles, which work best with a 7, 8, or 9. What I ended up doing was keeping them as a pair, with 9&11 in the low hand, and despite the pair the total was 50 points. On my second attempt I rearranged things and came up with the same solution as Gazreal (before seeing his post), which was four sets of similar low & high hands.

I'm sure there are plenty of ways to achieve the 50-point total, but I now doubt that any further reduction is possible.

I wonder if there is a way to get to 50 pts without pairing both 9s with both 11s. that seems to be the most essential part.