mathematical probability question
Moderator: Shirluban
mathematical probability question
From previous posts, I'm aware that some of you are familiar with probability calculations.http://reachmahjong.com/en/forum/viewto ... =5&t=53301
Care to indulge?
Mathematically speaking,
If a box contains 136 tiles, 4 each of 34 different, what is the probability of drawing 14 tiles at random and having 3 identical tiles in the 14.
Re: mathematical probability question
If my math isn't too rusty, the odds for three of a specific tile to be in the dead wall should be 0.326%; for any of the tiles, I think 11.1%. Are you also interested in the details of how to get that result?
Re: mathematical probability question
I am. Rusty is better than no idea.
.326% is just about 13 out of 4000, and naturally, 11.1% is slightly better than 1 out of 10. Sounds reasonable, I guess.
.326% is just about 13 out of 4000, and naturally, 11.1% is slightly better than 1 out of 10. Sounds reasonable, I guess.
Re: mathematical probability question
In cases like these, "I have a deck, I want to draw X cards and end up with Y of the same type", you can use the hypergeometric distribution (https://en.wikipedia.org/wiki/Hypergeom ... Definition); in it,
- N is how many cards/tiles/whatever you have in the deck
- K is how many tiles there are of the type you want to draw
- n is how many tiles you draw
- k is how many tiles of the specific type you want to draw
In this case:
- N = 136
- K = 4 (since each tile has four copies)
- n = 14
- k = 3 (since I want three tiles, out of four, of the same type)
The hypergeometric distribution gives you the chances of having three of the same specific tile (it being 1p, red dragon, whatever, but a specific one); since you have 34 tile types, the chance to have three of any one type is that number time 34.
- N is how many cards/tiles/whatever you have in the deck
- K is how many tiles there are of the type you want to draw
- n is how many tiles you draw
- k is how many tiles of the specific type you want to draw
In this case:
- N = 136
- K = 4 (since each tile has four copies)
- n = 14
- k = 3 (since I want three tiles, out of four, of the same type)
The hypergeometric distribution gives you the chances of having three of the same specific tile (it being 1p, red dragon, whatever, but a specific one); since you have 34 tile types, the chance to have three of any one type is that number time 34.
Re: mathematical probability question
Very nice! After refreshing my memory on what a binomial coefficient is, and following that example with the red and green marbles, I got it. Thanks.
As I stated in that other probability thread,
That 11% figure is a little hard to believe. That means that in every hanchan, on average, there are 3 identical tiles in the dead wall in one of the hands played.
As I stated in that other probability thread,
(good thing most of those factorials cancel each other out!)Amazing how a seemingly simple question concerning a probability turns into an educational math lesson. Great Stuff!
That 11% figure is a little hard to believe. That means that in every hanchan, on average, there are 3 identical tiles in the dead wall in one of the hands played.
Re: mathematical probability question
Slight (Veeeeeeery slight) overstimation because of double counting. (ie, the cases in which there are three copies each of two tiles are counted twice), but also slight understimation because we're looking for the cases where there are exactly three copies of a tile in the dead wall, so the cases where there are four of them are not counted.
In any case, it may be counter-intuitive. But yeah, in just under 1 of each 9 hands there's a dead triple. Usually, though, it's a tile no one cares about, and it isn't noticed. Also note that this includes dora indicators. Remeber Iwao's dora 12 hand that got head-bumped?
In any case, it may be counter-intuitive. But yeah, in just under 1 of each 9 hands there's a dead triple. Usually, though, it's a tile no one cares about, and it isn't noticed. Also note that this includes dora indicators. Remeber Iwao's dora 12 hand that got head-bumped?
Re: mathematical probability question
I don't think I know this.Referee wrote:Remeber Iwao's dora 12 hand that got head-bumped?
Re: mathematical probability question
It's from Akagi. Washizu Iwao has in his hand. When two kan calls reveal dora indicators of , he has 6 dora. When he draws another for a concealed kong and another is revealed as the dora indicator, he now has 12 dora.
When Akagi tosses his winning tile, Washizu goes totally bananas calling ron, thinking he's won, but Yasouka, Akagi's partner, sitting to his right, also calls ron and gets the win via the head-bump rule.
When Akagi tosses his winning tile, Washizu goes totally bananas calling ron, thinking he's won, but Yasouka, Akagi's partner, sitting to his right, also calls ron and gets the win via the head-bump rule.
The only catch here, relating to probability, is that there is no actual dead wall in Washizu mahjong. The supplemental tiles are drawn at random from the remaining batch of tiles, so the three were not actually in a 14 tile dead wall.RON - "The discarded tile completes my hand, and I win". Any player can call this. You say this, show your hand, and do a little dance if you wish. Note that if multiple people call Ron, and you are playing with the Head Bump rule, the player who would have taken their next turn first wins. So, if South discards, and East and North can both win, North actually wins.
So if 2 players are able to call Ron on the same tile, the person first after the player who played the tile, will win.
Trivia: Hard of Hearing trophy
The name of this trophy comes from a scene where Akagi asks "Are you deaf, Washizu Iwao?" because Washizu went so crazy after apparently winning he didn't hear Yasuoka declare a head bump.
Win a mahjong hand with a Head Bump and win the trophy.