Pai gow tiles - probabilities

I have been pondering probabilities related to receiving certain hands at pai gow.
There are 960 possible two-hand values.
There are 3,620 unique four-tile hands which can be configured in numerous ways to result in 35,960 possible four-tile combinations.
This consists of:

120 two-pair hands, one configurations each combination = 120. This can be determined with combin(16,2)
1,680 one-pair hands, four configurations each combination = 6,720. This can be determined with 16 x combin(15,2) x 4
1,820 no-pair hands, sixteen configurations each combination = 29,120. This can be determined with combin(16,4) x 16

The probability of holding two pairs is 120 / 35,960 = 0.00333704116
The probability of holding one pair is 6,720 / 35,960 = 0.186874305
The probability of holding no pair is 29,120 / 35,960 = 0.809788654

I like to determine the probability that the dealer would have a higher pair when I was holding a pair.
The probability of any particular pair being held is 420 / 35,960 = 0.011679644
So I just count my tile rank to quickly determine the probability of being beaten by a higher pair.

I have a question on the math used to determine some other probabilities

I see on the Wizard's page Bonanza Pai Gow
The probability of being dealt a wong is 2,704 / 35,960 = 0.0751946
I would like to know the equation by which this is determined.

There are eight possible ways to form a wong and sixteen possible ways to form a gong.
What would be the equation to determine the probability of a gong. Would it be twice as many or four times?
Tthere are also sixteen ways to form high nine so I think the probability would be the same as for the gong.

One more question unrelated to probability:
I read somewhere that the average high hand is nine with gor and the average low hand is five with gor.
Are these the median or the mean?

Well, since nobody responded, I will just use this space to store some random statistics for my own enjoyment!
This is helping me to learn the optimal strategy for each hand value, and maybe it will help other interested people.

There are 960 differently valued four-tile hands possible in pai gow.
Each occurs with a varying degree of frequency anywhere from 1 to 256 ways.
The frequency above 1 is always a multiple of 2, as there are two of each value tile.
The hand valuation only considers the numeric value of each tile and does not consider tile ranking or how the hand will be played.
Many of the 960 hand values have an only way to play and
many have multiple decisions based on ranking and other strategy factors.

The frequency of hand values breaks down as follows...
256 combinations
15 hand values
3,840 hands
10.67% of total.

128 combinations
80 hand values
10,240 hands
28.47% of total.

64 combinations
150 hand values
9,600 hands
26.69% of total

32 combinations
204 hand values
6,528 hands
18.15% of total

16 combinations
262 hand values
4,192 hands
11.65% of total

8 combinations
162 hand values
1,296 hands
3.6 % of total

4 combinations
51 hand values
204 hands
0.56% of total

2 combinations
24 hand values
48 hands
0.13% of total

1 combination
12 hand values
12 hands
0.03 % of total

This represents the 960 hand values comprising the 35,960 possible four-tile hands.
By focusing attention on the hands which are statistically most likely to appear,
one can increase their proficiency at the strategy of the game.
There is an excellent resource of sortable tables outlining the hand values in ascending, descending and numerical order
Pokerstrategy.us

Quote: WongBo

I see on the Wizard's page Bonanza Pai Gow
The probability of being dealt a wong is 2,704 / 35,960 = 0.0751946
I would like to know the equation by which this is determined.

I believe this should be the probability of being dealt a wong but no pair.

A wong has to have one of the two 9 tiles (which I will refer to as 63 and 54) and one of the four Day or Teen tiles (11a, 11b, 66a, 66b).
A hand with one 9 and one Day/Teen has 2 x 4 x C(26,2) possibilities - but that includes the 13 remaining pairs, so the number of 4-tile hands with a wong but no pair = 2 x 4 x (325 - 13) = 2496.
A hand with two 9s has a pair, so ignore those.
Each hand with one 9 and two Day/Teens (which have to be different - otherwise there is a pair) has 26 remaining tiles that can be the fourth tile, so there are 2 (possibilities for the 9) x 4 (possible non-pair groups of 2 for the Day/Teens) x 26 = 208.
A hand where all four tiles are 9 and/or Day/Teen is either:
(a) four Day/Teens, but this is not a wong;
(b) three Day/Teens and a 9, but you have to have a pair of Day/Teens in that;
(c) both 9s and two Day/Teens, but the two 9s are a pair.
Therefore, the total number of wongs without pairs = 2496 + 208 = 2704.

Calculating gongs is similar, but the 8 group consists of four tiles instead of two (44a, 44b, 53, 62).

I find this post interesting and I’m surprised you know the math of this game. Recently, I’ve been learning about Pai Gow Tiles. One probability has bothered me for some time.

What is the probability of a dealer high hand of 4? On Wizard website, it shows a probability of roughly 1/26 depending on the house way, but I have no idea how to handle this problem. Could you please take a look?

Make a program that arranges all 30,000+ possible hands via the house way. Count how many have a dealer high hand of 4. That's it.

There are only 3,620 unique four-tile hands for both the dealer and the player together, so there are not many of them, but I don’t know how many ways of arrange them to make each legal for the dealer. Also, I don’t know programing. Is it possible to use Excel to calculate out the probability?

Quote: aceside

There are only possible 3,620 unique four-tile hands for both the dealer and the player together. Among these, some are illegal. So there are not many of them, but I don’t know programing. Is it possible to use Excel the calculate the probability out?
link to original post

I don’t think you can use excel to replicate the complexity of the nightmare that is Pai Gow Tiles house way

How about you invest some time on this game? I have a few ideas to renovate this game. I am thinking to learn Visual Basic programming to combine it into Excel to handle this problem.

If you need me to explain you the rules and house way, I would be happy to do it. This game seems hard, but really does not have many combinations.

Quote: aceside

How about you invest some time on this game?
link to original post

No thanks I have to go outside and live life

I’ve almost figured out this part using excel.

When I looked at this years ago I created a program to analyse, for any Player hand, how often that hand would win/tie/lose against all possible House hands for the three ways to set the Player hand. It created a line of output for each Player hand, which I then put into a spreadsheet. From this you could deduce the correct (optimal) way to play any Player hand. I then tried to develop simple strategies and see the cost of them for the Player.
Coding note: I used a four digit number for the value of hands, Pairs=20xx,Wongs=11xx,Gongs=10xx,otherwise 9xx through 0xx (where xx is the tile number and all Player zeroes are 000.)

I’ve thought about this probability problem. To calculate the dealer hand probability of a high hand of 4, we do not need any information about these win/loss/tie rates. All we need is the house way rule. There are only these five possible low/high combinations:

0/4, 1/4, 2/4, 3/4, and 4/4.

Of all 3620 possible unique 4-tile combinations, there are only about 90 of these satisfy the above requirement. I will try to list all of them all soon.