[BunnyGo]

hrothgar, on 2011-November-01, 06:11, said:

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]

BunnyGo, on 2011-November-01, 10:44, said:

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...52&p2=923401070

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

[inquiry]

wyman, on 2011-November-01, 08:05, said:

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]

inquiry, on 2011-November-01, 13:21, said:

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]

inquiry, on 2011-November-01, 13:21, said:

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]

hrothgar, on 2011-November-01, 06:11, said:

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.

hrothgar, on 2011-November-01, 06:11, said:

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.

hrothgar, on 2011-November-01, 06:11, said:

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.

hrothgar, on 2011-November-01, 06:11, said:

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]

hrothgar, on 2011-November-01, 13:31, said:

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]

semeai, on 2011-November-01, 16:04, said:

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]

Quote

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.

[Cascade]

Probabilities of the longest and shortest suits in a four-hand deal:

Longest Suit
4 0.0292944677905751
5 0.402336782879317
6 0.422075366415914
7 0.126442683507377
8 0.0183051683243252
9 0.00147821697795984
10 6.58450191573011E-5
11 1.45628565027458E-6
12 1.27745286982186E-8
13 2.51963123452264E-11

Shortest Suit
0 0.183766327186213
1 0.609953117477308
2 0.205349135658076
3 0.000931419678402451

These numbers were computed by considering every possible four hand distribution and calculating the probability of those four-hand distributions. The longest and shortest suits were noted and the probabilities summed accordingly.

[Cascade]

Binomial Distribution for the number of four-hand deals with an 8-card suit and an 8-card or longer suit in a session of 36 deals.

	
	8-card	        8-card or longer
	0.018305168	0.019850699
0	0.514226158	0.485870394
1	0.345186568	0.354247273
2	0.112639097	0.125553303
3	0.023803677	0.028818368
4	0.003661805	0.004815118
5	0.000436991	0.000624123
6	4.20998E-05	6.53077E-05
7	3.36434E-06	5.66854E-06
8	2.27408E-07	4.16162E-07
9	1.31922E-08	2.62218E-08
10	6.6417E-10	1.43387E-09
11	2.92723E-11	6.86394E-11
12	1.13714E-12	2.89612E-12
13	3.91452E-14	1.08285E-13
14	1.19916E-15	3.60289E-15
15	3.27947E-17	1.0702E-16
16	8.02603E-19	2.84478E-18
17	1.76067E-20	6.77818E-20
18	3.46543E-22	1.44903E-21
19	6.12172E-24	2.78023E-23
20	9.70263E-26	4.78611E-25
21	1.37844E-27	7.38528E-27
22	1.75248E-29	1.01981E-28
23	1.98907E-31	1.2572E-30
24	2.009E-33	1.37917E-32
25	1.79812E-35	1.34074E-34
26	1.41852E-37	1.14881E-36
27	9.79644E-40	8.6172E-39
28	5.87152E-42	5.60963E-41
29	3.02023E-44	3.13408E-43
30	1.31405E-46	1.48105E-45
31	4.74242E-49	5.80555E-48
32	1.38171E-51	1.83716E-50
33	3.12292E-54	4.51E-53
34	5.13807E-57	8.0594E-56
35	5.47469E-60	9.32713E-59
36	2.83566E-63	5.24722E-62

[Cascade]

So the probability in 36 deals of have three or more deals with at least one eight-card or longer suits is:

1 - 0.485870394 - 0.354247273 - 0.125553303 = 0.034329029

Edit:

The probability in 36 deals of having three or more deals with at least one eight-card suit (precisely eight) is:

1 - 0.514226158 - 0.345186568 - 0.112639097 = 0.027948178

[semeai]

Awesome data, Cascade!

Cascade's exact answer of 1.985...% is pleasantly close to hrothgar's simulation answer of 1.978...%.

[PrecisionL]

semeai, on 2011-November-02, 13:35, said:

Awesome data, Cascade!

Cascade's exact answer of 1.985...% is pleasantly close to hrothgar's simulation answer of 1.978...%.

From bridge odds complete, by Frost, Kibler, Telfer, and Traub, 2nd edition, 1971:

f(x) 8-cards (exactly) = 0.117% (Specific suit, one player); 0.47% (One player, any suit); 1.87% (The Deal, any suit)

And, 8+ cards = 1.87 + 0.148 + 0.0066 + 0.00015 = 2.025 % (8 or more cards in a suit).

[wyman]

Great.

Cascade's answer is for 8 cards or longer (which is really what we should use here, since had you seen an 8, a 10, and an 8, you'd also be surprised, but I'm happy to use PrecisionL's low estimate).

So let p be the probability you want to use (whether it be 0.0187 or 0.01985 or whatever). Then the odds of seeing 3 or more deals is 1 - the probability of seeing exactly 0 - the probability of seeing exactly 1 - the probability of seeing exactly 2

Prob of seeing exactly N in 36 boards = 36CN * p^N * (1-p)^(36-N)

So your computation should look like this, when p=0.0187
http://www.wolframal.../?i=1+-+Sum+of+[Binomial[36%2Cn]+*+0.0187^n+*+%281-0.0187%29^%2836-n%29]+for+n+from+0+to+2

You get 0.0295, which is about 3%, so once in 33 sessions (or, as others have said, ~once a month for a daily game). If you were to include 9+ card suits, you get 3.4%, as Cascade mentions above.

So, this provides no compelling evidence that Dealmaster's RNG is broken. Also, keep in mind that as you look at the daily games to see what seems normal, you're running experiments. So you might say "oh my god, we got 2 days in a week where we got 8 card suits 3 times! What are the odds?" But if you've been watching the output for a year, you've actually run that experiment (look at one week, count days where we get 3 8-card suits) 52 times (and arguably more, since you can choose to start your week on Monday, or Thursday, or whenever you want). It's a dangerous game (see comic strip below).

http://xkcd.com/882/

Cheers.

[semeai]

PrecisionL, on 2011-November-02, 16:03, said:

And, 8+ cards = 1.87 + 0.148 + 0.0066 + 0.00015 = 2.025 % (8 or more cards in a suit).

This is off slightly; you're overcounting. One player could hold an eight card suit and one player could hold a nine card suit, for example.

[mrdct]

Sorry if it's been covered by the mathematicians already, but if one hand has an 8-card suit that must signicantly increases the chances that one of the other three hands also holds an 8+ card suit. Kind of like the Sith they follow the Rule of Two.

Notwithstanding that, the prevailing view seems to be that a set of 36-boards containing three 8-card suits has a probability of around 3.5% so it's quite reasonable to expect it to come up once every 30 sets or so. The OP has indicated that his sample is 60 sets so if this is the only occurence, it's actually happening less frequently than expected; but of course only 60 sets of boards is surely an inadequate sample size to draw any firm conclusions.

It's been a good 20 years since I last played bridge with hand-dealt cards so I can't really remember if hands were flatter back in the days of shuffling and dealing, but my local club has been using Deal Master Pro for years and I've never heard anyone complaining that the hands seem more distributional than expected.

[Cascade]

PrecisionL, on 2011-November-02, 16:03, said:

From bridge odds complete, by Frost, Kibler, Telfer, and Traub, 2nd edition, 1971:

f(x) 8-cards (exactly) = 0.117% (Specific suit, one player); 0.47% (One player, any suit); 1.87% (The Deal, any suit)

The bolded (my editing) number is slightly different than what I calculated. I have double checked my calculation and think it is correct. Can anyone confirm?

[PrecisionL]

Cascade, on 2011-November-02, 18:52, said:

The bolded (my editing) number is slightly different than what I calculated. I have double checked my calculation and think it is correct. Can anyone confirm?

Your calculations for 9 - 13 card suits agree very well with my reference.

For 0 - 7 card suits I find poor agreement.

(Before electronic calculators I had calculated the HCP frequencies and found exact agreement with the authors.)