What is happening here?

Are you really going to post 192 sets of 52 cards and manually count and color code them all like this?

You've already messed up:

1. The first card was Club:Five. You skipped it and listed the second card first, which was Spade:Seven

2. You missed two back to back clubs in the first set you posted.

3. I didn't actually check you for accuracy. Those are just 2 things I happened to notice, so I don't know if there are other errors.


Also, there are all the problems that were being discussed in post numbers 669 through 673.



What you're doing is going to be useless.

Thank you for your input, I am not sure , this is at least a start , it should give us a basic idea off the numbers when the second test results are posted.

Yes, I agree with Lego05 and NewOldGuy. My post just above was attempting to put it into words that everyone would understand.

Some keys to remember:

1. Design your "test(s)" ahead of time [no fair designing your tests after perusing the data]

2. Design simple yet powerful tests

3. Make sure you have enough data to fairly test the hypotheses before performing any tests

4. Perform tests programmatically [doing anything by hand is silly and verboten]

5. Repeat your tests on a sufficient number of similarly-generated datasets.

I guess this is the way to go. This is probably how it should be done.



The "runs test" I suggested would maybe be the next best way. I had no idea it was called a "runs test". I was just making up a way to test for what you wanted to test for. I had no idea it was already named after Abraham Wald and Jacob Wolfowitz.



And obviously, whatever you're going to do, you have to set forth with specificity the test you are going to run before you look at the output you will be testing. And, again, it would be so much easier, more efficient, faster and more accurate to code (or download) a program to do the counting rather than doing the counting manually.

Some of the colours would of merged , I will have to change the way I do it slightly. I do not think I will have to all the sets, I am hoping the numbers shows an anomaly.


Results of 10,000 deck shuffle, picking 1st card from each deck.
Club-Ace : 179
Club-Two : 200
Club-Three : 171
Club-Four : 190
Club-Five : 182
Club-Six : 193
Club-Seven : 178
Club-Eight : 194
Club-Nine : 195
Club-Ten : 198
Club-Jack : 174
Club-Queen : 194
Club-King : 202
Diamond-Ace : 204
Diamond-Two : 198
Diamond-Three : 183
Diamond-Four : 172
Diamond-Five : 177
Diamond-Six : 192
Diamond-Seven : 196
Diamond-Eight : 185
Diamond-Nine : 172
Diamond-Ten : 182
Diamond-Jack : 199
Diamond-Queen : 181
Diamond-King : 218
Heart-Ace : 205
Heart-Two : 217
Heart-Three : 188
Heart-Four : 206
Heart-Five : 190
Heart-Six : 202
Heart-Seven : 208
Heart-Eight : 193
Heart-Nine : 194
Heart-Ten : 194
Heart-Jack : 180
Heart-Queen : 156
Heart-King : 191
Spade-Ace : 200
Spade-Two : 217
Spade-Three : 180
Spade-Four : 206
Spade-Five : 192
Spade-Six : 184
Spade-Seven : 207
Spade-Eight : 201
Spade-Nine : 191
Spade-Ten : 191
Spade-Jack : 208
Spade-Queen : 190
Spade-King : 200


Ok I am going to bed, goodnight all and thanks for the help and advice.

Hope ur dreams are ostensible.

No you can't.

So what are you expecting to see from this? No matter what the numbers are, I don't think you understand how to evaluate their significance. Viewing them and thinking about them is not going to tell you anything even if you think something doesn't "look" right. Randomness does not ever "look" right. Randomness looks random, and humans truly aren't able to recognize what that should look like.

What is the standard deviation for the count of each card? How many of them should be within 1 or 2 or 3 standard deviations? Do you know how to do a chi-squared test or a T test?

I'm sure you can now look these up, but I'm also sure you don't know any of the answers at the time you read this.

I really haven't followed all of the multiple versions of the tests that are going to be run, and I have no doubt about how they will turn out. But I will say that I have run a chi-squared test on the results that were posted above, where they looked at the 1st card for each of 10,000 shuffles. Each card should have an equal probability of being chosen, and the test confirms that there was no significant deviation from that expectation (p=.777 for those who care). In other words, the cards came out just about how one would expect them to come out.

Because you haven't read all of his earlier posts in this thread, you may not know you're doing all of this work, very generously, for someone who genuinely doesn't understand or accept that 26/52 = 2/4 = 1/2 = 50%, etc, and who thinks that most answers for any probability question is, specifically, ?/52 or ?/4 or ?/2 or ?%, etc.

Whatever your results, as soon as you tell him that an answer for any run is 156/10,000 he will neither understand nor accept that this is "equal" to 192/10,000, or 1.92/100, and that it does instead prove that, whatever his belief is, he is right and that the probability is ?/10,000 but not (?/100)/100.

As you want to help him, you might consider starting him off with a level one basic probability course, or, indeed, an even more basic course in how numbers work and are used in real life, perhaps using 100 matchsticks.

You are a good man, EvilGreebo.

FYP.

I haven't gotten back to the code yet, I had errands to tend to yesterday, but my plan was to work it like this:

(btw anyone know on average how many tables are running on stars? Average across the whole day - rough guess?)

Pick a random number between 1 and 500 ("active tables") and pull that many decks off the stack. Pick the top card from the last deck picked. Randomly sort the decks removed from the stack and send them back to the shuffler to be reshuffled (which, I stress, is, I'm SURE, not how they would do it at stars because there's no freaking point to it and its just extra effort)

Rinse and repeat the above 10,000 times.

Publish same output as from first run - the list of cards picked and the count of suit/values.

The program probably COULD be made to do that. But it won't be done by me. The source code will be available though to anyone who wishes to play with it..


I'll probably link my google drive to my work computer and then I can have the program copy the output of a million runs to that output folder on my drive, it won't matter to me. I'm just providing sample data.



Oh I've seen it. I noticed his ace count is 4/52 too.

I'm only providing data. I'm not doing the testing for him.

For that matter - I'm too damn rusty at probability and statistics to even try - but I know this forum is full of people who are able to evaluate his testing methods and try to correct him.

My aim here is simply to provide data so that he can put his money where his mouth is.

I am not quite sure what you are saying there. We need the first experiment you did, but with random choice of deck from the queue rather than in order. Making sure that once one is picked, the deck goes back tot he end of the queue shuffled.
The results hopefully show the numbers to be difference.
In your first test there is about 170-200 average of each value over 10,000 hands.

I am expecting a different average in the second test. I am expected some value will be greater than, less than or equal to the first test averages.

so you are expecting 340/2 to 400/2 ?

All the best.

Somebody who understands the difference between finite probability and infinite probability.

No, that would be the same as the first results run twice.


170-200/{A}

170-200/{B}


{a}+{b} = 340 - 400 / {A,B}

Finite probably is not without boundaries i.e 1/2 , 2 is the boundary

Infinite probability is without boundaries i.e 1/ ? , ?/2 , ? has no boundary


added- Just in case you did not understand,

You could spent an infinite amount of time trying to get the 1/?

You could spend an infinite amount of time trying to get ?/2 because we do not know if the value is there.

You sure? You may want to check your math ?/2 times, and then look it over manually, and perhaps use a mechanical calculator to be sure, and then mail a postcard in with anything you see that is interesting.

All the best.

Random choice from the queue yes - but every other deck in the queue cannot just be ignored either.
Lets say our target player is table 1 seat 1 and they get the first card from the first deck.

Now tables 2, 3, 4, 5, 7, 23, 95, 2835 and 28953 ask for decks. Those tables get decks 2, 3, 4, 5, 6, 7, 8, 9 and 10. Finally our table 1 comes back for a deck - they get deck 11.

You still have decks 2-10 to be dealt with. Some of them will return to the queue (which again, is a stupid notion) before the deck table 1 got, some will return after, because some tables will have longer hands than others.

Again, its a stupid thing to do, sending the virtual deck back to the queue. It's far more efficient to make a new deck out of nothing each time you start a shuffle and to discard it into nothing when done, but you're calling the shots, so I'll go ahead and preserve the object references and recycle them, even though it's the dumb way to do it and I'll bet good money Stars DOES NOT do that.
I'll be shocked if they do.

It is the same thing, creating a new shuffled deck is the same as placing the re-shuffled deck, and picking random decks is the same as receiving deck one , then deck 11, there is no specific order the decks are assigned to tables, it is the same thing.


If you want to discard the used deck and create a new deck at the end of the queue, no problem it is the same. The main parameter of the second test is to make sure the decks are randomly chosen rather than order.

I'm glad we agree on that. Previously it seemed to me that it was important to you that we recycle them.

No that isn't important, I just slightly misunderstood you sorry.

Ok then the plan will be to simply discard all "used" decks and add new ones to the queue as old ones are pulled.

Yes that's right.