Deal Master Pro

[wyman]

About 27 1/2 - 1 against, I think. Not overly damning.

 

If this is correct (haven't checked), it's way way less damning given that they've used the deals in many club games.

[BunnyGo]

If this is correct (haven't checked), it's way way less damning given that they've used the deals in many club games.

 

Yeah, I guess this should happen about once a month (for a daily game), and if it happened twice or three times that often I wouldn't be the least be surprised. The most damning "evidence" so far is Han's suggestion that spot cards have a "memory". I have not looked at the hand data produced over thousands (or millions) of trials, but I doubt the data address this issue.

 

It'd be interesting to see a study of correlation between successive hands.

[PrecisionL]

About 27 1/2 - 1 against, I think. Not overly damning.

 

8221 / 8311 / 8320: f(x) = 0.42% or 4.2 times in 1000 hands for only one 8-card suit.

 

f(x) 3 times = (0.0042)**3 = 73 / 10**9 or 7.3 times in 100,000,000 hands which is highly suspect.

 

[Edited 11/1: This calculation is for being dealt three 8-card suits in 3 deals in a row.]

 

Is this within 3 sigmas of the mean? We can't calculate the range without knowing the variance of the program.

Edited November 2, 2011 by PrecisionL

[BunnyGo]

8221 / 8311 / 8320: f(x) = 0.42% or 4.2 times in 1000 hands for only one 8-card suit.

 

f(x) 3 times = (0.0042)**3 = 73 / 10**9 or 7.3 times in 100,000,000 hands which is highly suspect.

 

Is this within 3 sigmas of the mean? We can't calculate the range without knowing the variance of the program.

 

 

Your 8 card suit number is just for one player. I don't know how off the top of my head how to handle the correlation between hands, but it is certainly better than 1-200 against having some player have an 8 card suit on any given deal.

[Lobowolf]

The probability of a single hand having an 8 card suit (not just 8221 8320 or 8311) is about .5%. The probability of a single deal having one is around 2%. The probability of at least 3 deals having one, with 36 deals in play, are 1-(probability of 0 deals + probability of 1 deal + probability of 2 deals). On the rough but really close approximation that each event is a 2% occurrence, that's 1-(.4832+.3550+.1268), or 3.5%.

[BunnyGo]

The probability of a single hand having an 8 card suit (not just 8221 8320 or 8311) is about .5%. The probability of a single deal having one is around 2%. The probability of at least 3 deals having one, with 36 deals in play, are 1-(probability of 0 deals + probability of 1 deal + probability of 2 deals). On the rough but really close approximation that each event is a 2% occurrence, that's 1-(.4832+.3550+.1268), or 3.5%.

 

 

Could you sketch the derivation of the 2% number?

[Lobowolf]

I pretty much cheated to keep the math simple - I took the figures for an 8-card suit out of the encyclopedia (.4668) (8221+8311+8320+8410+8500 = .1924+.1186+.1085+.0452+.0031), and I quadrupled it. So my 0.5% overstates the figure in a couple of respects - first, it's bigger than .4668% (and given that we're going to be raising things to the 36th power, maybe that's too much rounding), and second, quadrupling the 1-hand percentage isn't precise, either, as they're not independent events. The fact that one hand doesn't have an 8-card suit surely must decrease the chance that another hand in the same deal does.

 

However, as balanced against that, I also wasn't taking into consideration any more freakish deals, the existence of which would further my main point - that the reported occurrence wasn't that unusual. For instance, if you add in the possibility of a nine-card suit, the probability that will see such a long (8 or 9) suit on a particular hand goes *over* .5% (and that says nothing of the 10-, 11-, 12-, and 13-card suits).

 

I wasn't aiming for precision, but I don't think that the 'pure' numbers would lead to a different conclusion - what happened at the table wasn't all *that* odd.

[BunnyGo]

I pretty much cheated to keep the math simple - I took the figures for an 8-card suit out of the encyclopedia (.4668) (8221+8311+8320+8410+8500 = .1924+.1186+.1085+.0452+.0031), and I quadrupled it. So my 0.5% overstates the figure in a couple of respects - first, it's bigger than .4668% (and given that we're going to be raising things to the 36th power, maybe that's too much rounding), and second, quadrupling the 1-hand percentage isn't precise, either, as they're not independent events. The fact that one hand doesn't have an 8-card suit surely must decrease the chance that another hand in the same deal does.

 

However, as balanced against that, I also wasn't taking into consideration any more freakish deals, the existence of which would further my main point - that the reported occurrence wasn't that unusual. For instance, if you add in the possibility of a nine-card suit, the probability that will see such a long (8 or 9) suit on a particular hand goes *over* .5% (and that says nothing of the 10-, 11-, 12-, and 13-card suits).

 

I wasn't aiming for precision, but I don't think that the 'pure' numbers would lead to a different conclusion - what happened at the table wasn't all *that* odd.

 

Ok, that's about what I'd done too, but I was hoping you'd done the precision.

 

I couldn't tell how much error was introduced that way.

[hrothgar]

Ok, that's about what I'd done too, but I was hoping you'd done the precision.

 

I couldn't tell how much error was introduced that way.

 

Too brain fried to work out the conditional probabilities analytically however I coded up a quick Monte Carlo simulation for length >= 8 in at least one of the four hands.

I ended up with 1.978%

 

Please note: this sim calculates the % of 8+ card suits.

Also, if a single deal includes two 8+ card hands it will only count once.

 

I attached the MATLAB code for anyone who cares

 

 

clear all
clc

Rank = zeros(52,1);
Rank = nominal(Rank);
Deal = dataset(Rank);

Deal.Rank(1:4,1) = 'A';
Deal.Rank(5:8,1) = 'K';
Deal.Rank(9:12,1) = 'Q';
Deal.Rank(13:16,1) = 'J';
Deal.Rank(17:20,1) = '10';
Deal.Rank(21:24,1) = '9';
Deal.Rank(25:28,1) = '8';
Deal.Rank(29:32,1) = '7';
Deal.Rank(33:36,1) = '6';
Deal.Rank(37:40,1) = '5';
Deal.Rank(41:44,1) = '4';
Deal.Rank(45:48,1) = '3';
Deal.Rank(49:52,1) = '2';

Suit = zeros(52,1);
Suit = nominal(Suit);

for i = 0 : 12
   
   Suit(4*i + 1) = 'S';
   Suit(4*i + 2)  = 'H';
   Suit(4*i + 3) = 'D';
   Suit(4*i + 4) = 'C';
   
end

Deal.Suit = Suit

%% shuffle

simlength = 100000;
count = 0;
length_criteria = 8;

for i = 1:simlength
   
   index = randperm(52);
   Deal = Deal(index,:);
   
   foo1 = summary(Deal(1:13,2));
   foo2 = summary(Deal(14:26,2));
   foo3 = summary(Deal(27:39,2));
   foo4 = summary(Deal(40:end,2));
   
   if max(foo1.Variables.Data.Counts) >= length_criteria
       count = count +1;
   elseif max(foo2.Variables.Data.Counts) >= length_criteria
       count = count +1; 
   elseif max(foo3.Variables.Data.Counts) >= length_criteria
       count = count +1; 
   elseif max(foo4.Variables.Data.Counts) >= length_criteria
       count = count +1;   
       
   end
   
end

count /1000

[kfay]

Yeah, I guess this should happen about once a month (for a daily game), and if it happened twice or three times that often I wouldn't be the least be surprised. The most damning "evidence" so far is Han's suggestion that spot cards have a "memory". I have not looked at the hand data produced over thousands (or millions) of trials, but I doubt the data address this issue.

 

It'd be interesting to see a study of correlation between successive hands.

 

April Fools?!?!

 

Seriously, idk how many of Han's posts I've seen replies like this for.

 

I thought it was blatantly obvious that Han was joking. Of course you can find whatever pattern you want for a series of 5 deals. Give me 32 deals and I'll find some pattern to connect all the hands, I guarantee you. I also don't think Han spends his spare time looking through deals to see which hand holds the 5 of clubs.

 

Maybe you're being sarcastic too. In that case... well played.

[Trinidad]

Too brain fried to work out the conditional probabilities analytically however I coded up a quick Monte Carlo simulation for length >= 8 in any of the four hands.

I ended up with 1.978%

 

Please note: this sim calculates the % of 8+ card suits.

Also, if a single deal includes two 8+ card hands it will only count once.

 

I attached the MATLAB code for anyone who cares

 

What a fine example of circular reasoning: You use a random deal generator to check whether a random deal generator delivers the correct amount of 8 card suits. It is like saying that snow is white (or red) because it has the same color as snow which we know is white (or red).

 

If people won't accept that the random generator in Deal Master Pro is good enough how are you going to convince them that it is by using a random generator in Matlab?

 

And most likely, your random generator in Matlab is not better than the random generator Deal Master Pro uses.

 

Rik

[hrothgar]

If people won't accept that the random generator in Deal Master Pro is good enough how are you going to convince them that it is by using a random generator in Matlab?

 

And most likely, your random generator in Matlab is not better than the random generator Deal Master Pro uses.

 

Rik

 

Out of curiosity, do you have even the slightest clue

 

1. What MATLAB is?

2. What random number generators are used inside the product?

 

I'm only asking because from the outside looking in, your comments look astoundingly ignorant (Even by the standards to which we are accustomed on this forum)

 

In any case, I don't know what RNG Deal Master Pro uses, not do I know how they implement their hand generation algorithm.

I suspect that their implementation is very different than the one I used.

 

With this said and done, having multiple independent implementations deliver consistent results is often seen as a good thing...

[wyman]

Gah, someone just compute the probability of an 8+ card suit appearing in a hand directly. This should not be that hard. I will try to do so if I get some time today.

[PrecisionL]

Gah, someone just compute the probability of an 8+ card suit appearing in a hand directly. This should not be that hard. I will try to do so if I get some time today.

 

Read the postings, it has been done already. 0.42% if you ignore 8-4 and 8-5 variations

[wyman]

Read the postings, it has been done already. 0.42% if you ignore 8-4 and 8-5 variations

 

Please excuse that my comment was imprecise.

 

I meant (and I thought this was clear): someone please compute directly the probability of an 8+ card suit appearing in a _deal_.

 

Also, there's no reason to ignore 8-4 or 8-5 variations. And we should include the (small) probability of 9, 10, 11, 12, and 13 card suits appearing in a deal as well.

 

edit: this should be just a big PIE but I think there are some subtleties I ran into when I tried to do it in 2 seconds.

[BunnyGo]

Out of curiosity, do you have even the slightest clue

 

1. What MATLAB is?

2. What random number generators are used inside the product?

 

I'm only asking because from the outside looking in, your comments look astoundingly ignorant (Even by the standards to which we are accustomed on this forum)

 

In any case, I don't know what RNG Deal Master Pro uses, not do I know how they implement their hand generation algorithm.

I suspect that their implementation is very different than the one I used.

 

With this said and done, having multiple independent implementations deliver consistent results is often seen as a good thing...

 

Do you work at the Mathworks, Hrothgar? Seems like you're in the area.

[hrothgar]

Do you work at the Mathworks, Hrothgar? Seems like you're in the area.

 

Yeap. For anyone who ever wants to hear my melodious voice, feel free to check out:

 

http://www.mathworks.com/company/events/webinars/wbnr59911.html?id=59911&p1=923401052&p2=923401070

 

for an exciting discussion of the pros and cons of sequential feature selection as opposed to ridge regression

[inquiry]

Gah, someone just compute the probability of an 8+ card suit appearing in a hand directly. This should not be that hard. I will try to do so if I get some time today.

 

The answer is 0.505%

 

This is because there are 3,209,923,136 possible combinations of hands with 8xxx, 9xxx, 10xxx, 11xxx. 12xxx. and 13000 hand types out of a total possibility of 635,013,559,600.

 

To figure this, there are 4 way you can hold a 13 card suit.

2,028 ways to hold 12,1,0,0 pattern,

73,008 ways to hold 11,2,0,0 pattern

158,184 ways to hold 11,1,1,0

etc....

[wyman]

The answer is 0.505%

 

This is because there are 3,209,923,136 possible combinations of hands with 8xxx, 9xxx, 10xxx, 11xxx. 12xxx. and 13000 hand types out of a total possibility of 635,013,559,600.

 

To figure this, there are 4 way you can hold a 13 card suit.

2,028 ways to hold 12,1,0,0 pattern,

73,008 ways to hold 11,2,0,0 pattern

158,184 ways to hold 11,1,1,0

etc....

 

:sigh:

[hrothgar]

The answer is 0.505%

 

This is because there are 3,209,923,136 possible combinations of hands with 8xxx, 9xxx, 10xxx, 11xxx. 12xxx. and 13000 hand types out of a total possibility of 635,013,559,600.

 

To figure this, there are 4 way you can hold a 13 card suit.

2,028 ways to hold 12,1,0,0 pattern,

73,008 ways to hold 11,2,0,0 pattern

158,184 ways to hold 11,1,1,0

etc....

 

I believe that Wyman was talking about the chance within a deal rather than the chance within a hand, in which case you need to worry about conditional probabilities...

(Which is why I went the Monte Carlo route)

[Trinidad]

Out of curiosity, do you have even the slightest clue

 

1. What MATLAB is?

Yes, it is a very nice program that could for example be used to calculate analytically what the probability is that a given player will get exactly 8 hearts when he gets 13 from a pack of 52, without the use of silly simulations. It is even capable of calculating the probability that a given player will get 9, 10, 11, 12 or 13 hearts.

2. What random number generators are used inside the product?

Yes. To the best of my knowledge, but I may be wrong, they use a very nice mt19937ar Mersenne Twister. They come in 32, 64 and 128 bit variations. As I wrote earlier, you will need 96 bits. Unless you have an expensive computer, your computer will not be able to give you the required 96 bits.

I'm only asking because from the outside looking in, your comments look astoundingly ignorant (Even by the standards to which we are accustomed on this forum)

It is nice to know why you are asking. Personnally, I am not sure why I am responding to your comments. However, I do know why I won't comment on the quality of your comments.

In any case, I don't know what RNG Deal Master Pro uses, not do I know how they implement their hand generation algorithm.

I suspect that their implementation is very different than the one I used.

 

With this said and done, having multiple independent implementations deliver consistent results is often seen as a good thing...

[sarcasm]Of course it is always good to have an independent observation to check. I regularly check whether our atom clock is still running fine by comparing it to my wrist watch.[/sarcasm]

 

Rik

[Trinidad]

I believe that Wyman was talking about the chance within a deal rather than the chance within a hand, in which case you need to worry about conditional probabilities...

(Which is why I went the Monte Carlo route)

Given that 0.505% << 100%, do you think that you will be far off by stating that the probability that South gets an 8 card or longer suit is approx. 0.25 x the probability that any player will have an 8 card or longer suit?

 

Rik

[semeai]

Getting exact numbers is fairly annoying. A bit of work gets us upper and lower bounds.

 

Upper bound: 4*(probability south has an 8 or more card suit) = 4*.00505 = .02020 (using Ben's work)

 

Lower bound: 4*(probability south has an exactly 8 card suit) - 6*(probability south and east each have an exactly 8 card suit) = .01867 - .00031 = .01836

 

So it's between 1.83% and 2.03%. Good enough?

 

If what we really wanted to do was reproduce hrothgar's program's number, sorry.

 

How to get the probability south and east each have an exactly 8 card suit:

Multiply the following three lines and divide by (52 choose 13)*(39 choose 13)

4*3 // ways to choose south's and east's long suits

(13 choose 8)^2 // the two 8 card suits

sum from n=0 to 5 of (5 choose n)*(26 choose [5-n])*([26+n] choose 5)

/* in the sum the first term is the ways to choose south's cards in east's long suit

the second term is the ways to choose south's remaining cards

the third term is the ways to choose east's remaining cards */

[PrecisionL]

Getting exact numbers is fairly annoying. A bit of work gets us upper and lower bounds.

 

Upper bound: 4*(probability south has an 8 or more card suit) = 4*.00505 = .02020 (using Ben's work)

 

Lower bound: 4*(probability south has an exactly 8 card suit) - 6*(probability south and east each have an exactly 8 card suit) = .01867 - .00031 = .01836

 

So it's between 1.83% and 2.03%. Good enough?

 

If what we really wanted to do was reproduce hrothgar's program's number, sorry.

 

How to get the probability south and east each have an exactly 8 card suit:

Multiply the following three lines and divide by (52 choose 13)*(39 choose 13)

4*3 // ways to choose south's and east's long suits

(13 choose 8)^2 // the two 8 card suits

sum from n=0 to 5 of (5 choose n)*(26 choose [5-n])*([26+n] choose 5)

/* in the sum the first term is the ways to choose south's cards in east's long suit

the second term is the ways to choose south's remaining cards

the third term is the ways to choose east's remaining cards */

 

Very nice! A later question in these postings: What is the probability of having Three 8-card suits in 36 deals???

[Siegmund]

Very nice! A later question in these postings: What is the probability of having Three 8-card suits in 36 deals???

 

The expected number of 8+card suits in 36 deals is .7279 (expectations add, even though the 4 hands in a given deal aren't independent, so we just multiply the chance of an 8+ card suit in one hand by 144). The number of successes for a rare event in a large number of trials is approximately Poisson, so the chance of 3 or more successes is approximately 3.76%.

 

The Poisson approximation is imperfect, and (since if one player has an 8+ card suit, the chance of another player also having an 8+ card suit is slightly elevated) the extreme cases happen a bit more often than predicted by the Poisson assumption. But not by much. The right answer is surely under 4% and I wouldn't be surprised if it were under 3.8%.

 

I could swear I saw an estimate very close to 3.76% posted way back at the beginning of this thread. But nobody seemed willing to believe it. Sigh.

 

At any rate, the previously stated "should happen about once a month at an every-day club" is right on the money, and it doesnt appear there is any evidence of a bad dealer.