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Just now on JRM, I was a ko player and got dealt ten tiles of one suit - I am guessing that the probability of that isn't very high...

I think the probability would be

(3*(36 choose 10)*(100 choose 3))/(136 choose 13)

=0.00025488...

Or about 1 in 4000.

Iapetus wrote:I think the probability would be

(3*(36 choose 10)*(100 choose 3))/(136 choose 13)

=0.00025488...

Or about 1 in 4000.

Iapetus' number for chin'itsu-grade hands sounds good. Conversely, a hon'itsu hand of 10/13 tiles would be:

(3*(64 choose 10)*(72 choose 3))/(136 choose 13) == 5.6 % (one in 19)

However, this is an approximation as there are cases that overlap, with some of the hands that have more than 7 jihai (alone, that is 0.468 %).

What's the 3* for?

You have to set your exclusions.

One suit vs rest.
Second suit vs rest.
Third suit vs rest.

In my example, it then becomes :

One suit and honours vs rest.
Second suit and honours vs rest.
Third suit and honours vs rest.

In both cases you have to add those together, thus 3x. In the hon'itsu case, there is some overlap.

I would like to follow these mathematical calculations but I am not familiar with the usage of the word "choose" in this context.
Can you be more specific with the actual numbers you are multiplying and dividing in these calculations.

By "choose", he means Combination.

"64 choose 10" = "number of possible sub-sets of 10 distinct elements from a set of 64 elements" = 64! / (10! × (64 - 10)!)

Choose refeers to stuff in combinatorics where out of Y tiles, you want X amount, "Y choose X" (usually written with two-line parthenses, Y on top of X) is a simplified mathematical representation of the factorial operations Y! / (X! * (Y-X)!). Each factorial number is the product of multiplying every integer from 1 to itself ( 1 * 2 * 3 * 4 * ... * N ). 6 choose 4 == 6! / (4! * 2!) == 720 / (24 * 2) == 15.

A single choose operation is symmetrical, as choosing 4 of 6 is the same as "not choosing", or "choosing to negate" 2 of 6. It's easy to visualize choosing 2 of 6 items as being 6 * 5 {*4*3*2*1} / 2 {*4*3*2*1} == 15, but the same formula as above applies. When combined with other choose operations or operations, then things get funky.

The other issue with factorials as that they explode in value faster than exponentially. The math curves would be ranked x^x > n^x > n! > x^n. A million cubed is 10^18. A million factorial is 5.56 million digits long. Still, with only 136 tiles to worry about, that leaves numbers to be calculated with systems able to accept values in the 136! range... of 3.659 * 10^232, without rounding them arbitrarily to values.


To be fair and answer your question, writing out every value that the choosing gives is both exhaustive and trivial, because a lot of it divides itself out of the way.

Senechal wrote: "Y choose X" is a simplified mathematical representation of the factorial operations Y! / (X! * (Y-X)!).

This definition says it all. I am very familiar with the term factorial. Thanks.

I assume that this calculation still takes a lot of button pushing on a calculator even though some of it divides itself out of the way......or is there now a really quick shortcut available that I don't know about.
3*(36 choose 10)*(100 choose 3))/(136 choose 13)=0.00025488...

(3)(36!)(100!)(13!)(123!)
-------------------------------
(10!)(26!)(3!)(97!)(136!)

Amazing how a seemingly simple question concerning a probability turns into an educational math lesson. Great Stuff!
(Anyone want to learn how to solve 1 equation with 2 unknowns.......just kidding....er....maybe)

or2az wrote:Amazing how a seemingly simple question concerning a probability turns into an educational math lesson. Great Stuff!
(Anyone want to learn how to solve 1 equation with 2 unknowns.......just kidding....er....maybe)

Mahjong probability is anything but simple because the probabilities themselves are dynamic.