# The Naked Scientists Forum

### Author Topic: Is this sequence a binomial distribution?  (Read 9793 times)

#### Thebox

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##### Is this sequence a binomial distribution?
« on: 13/04/2015 19:44:02 »
Hello, if I had  100 different sequences of sets of numbers that were placed in a que system , the first sequence being the first from the line in  distributing , the sequence distributed 10 numbers to five receivers on one outlet, and the same to another outlet,  2 numbers of the sequence to each individual receiver in a corresponding order.

In ten of the sequences, outlet one, receiver 3, would receive 5 numbers 5's if only number 1 outlet was in operation, and outlet 2 was dormant,

when we run the sequence a second time, outlet 2 is in operation, and the distribution ratio is defined by each individual receiver time of action on both outlets, making the sequence more random and not probabilities of the original sequence of outlet one, receiver 3, of receiving the 5, 5/100

So basically the distribution of the 100 sequences becomes time based distribution.

Am I correct in thinking that outlet one, receiver 3, the probabilities of now getting the 5, is now unaccountable?

a standard dice 1-6, has a 1/6 chance of any number per role

2 dice would have a 6^2 chance of rolling the same number

If I was betting on one dice , and you was betting on the other dice, each time we role we both have 1/6 chance of hitting our betted number.

However , your sequence of rolls would be bimodal different to my roles. Your's and my sequence would be dependent to each dice, and difference to each others sequence.

If we swapped dice, our probabilities remain 1/6, and even if we bet each others spin , our probabilities remain 1/6.

If you rolled number 1, and I bet against you hitting another 1, my probabilities are 1/6, where to you repeat the number one, your probabilities are now 6^2, or 36-1.

In the second game we change the rules, stacked in a que is already predefined roles and results of each dependent dice, a sequence dependent to each of us based on one dice, for every time an even number comes out, we win, for every time an odd number comes out we lose.

We see 100 numbers each, I win only 33% of my numbers, where you win 88% of your numbers,

Your sequence was a wining sequence for you, with a 2-1 chance of getting even , ready predefined by the stored sequence,

Now if you were to alternate between dice, you would then have a bimodal distribution, working off two sequences rather than one sequence, both sequences then becoming differential to the original sequence.

So instead of just a predetermined random luck, you are changing the sequence luck by timing of choice,

sequence one- 246412431.......................
.........bimodal - 236511213
sequence two - 331564115.......................

would this be bimodal or multimodal distribution?

« Last Edit: 15/04/2015 08:18:30 by chris »

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #1 on: 20/04/2015 10:25:37 »

If I have two decks of cards pre-shuffled, and you have to pick one deck , how many opportunities do you have of aces been in your seat order in the deck sequence?

if you pick one of the two decks, no longer is the shuffle of the deck defining what cards you get, your choice defines what cards you get?

#### Colin2B

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##### Re: Is this sequence a binomial distribution?
« Reply #2 on: 20/04/2015 10:35:01 »
Hello, if I had  100 different sequences of sets of numbers

sequence one- 246412431.......................
.........bimodal - 236511213
sequence two - 331564115.......................

would this be bimodal or multimodal distribution?

So, 100=3 different sequences?

sequence one- 246412431=9
So do the others, so they are all the same, yes.

Good, thought I'd got it

==========

Seriously, are you looking for a genuine answer? I haven't even bothered to look at this one due to a suspicion that any answer will be met with a rebuttal (binomial is book learning and complacent thinking) followed by an invented theory of probability.

« Last Edit: 20/04/2015 10:45:26 by Colin2B »

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #3 on: 20/04/2015 12:51:02 »
Hello, if I had  100 different sequences of sets of numbers

sequence one- 246412431.......................
.........bimodal - 236511213
sequence two - 331564115.......................

would this be bimodal or multimodal distribution?

So, 100=3 different sequences?

sequence one- 246412431=9
So do the others, so they are all the same, yes.

Good, thought I'd got it

==========

Seriously, are you looking for a genuine answer? I haven't even bothered to look at this one due to a suspicion that any answer will be met with a rebuttal (binomial is book learning and complacent thinking) followed by an invented theory of probability.

I am genuine looking for an answer, this s were I started science looking for an answer and got lost in science.   Poker forums sent me border line insane, denying what I was saying when I know it is the truth.

If we have 10 decks of cards pre-shuffled, and had 2 tables playing, in which by timing of the hands, they get a new deck every hand,   then timing of the decks is picking the hand you receive and not the pre-shuffle?

#### alancalverd

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##### Re: Is this sequence a binomial distribution?
« Reply #4 on: 21/04/2015 14:42:01 »

2 dice would have a 6^2 chance of rolling the same number

No. If A rolls any number, B has a 1/6 probability of rolling the same number. So "any double" is 1/6 but a specific double is indeed 1/36.

And "binomial" is not the same as "bimodal".
« Last Edit: 21/04/2015 16:42:26 by alancalverd »

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #5 on: 21/04/2015 18:09:32 »

2 dice would have a 6^2 chance of rolling the same number

No. If A rolls any number, B has a 1/6 probability of rolling the same number. So "any double" is 1/6 but a specific double is indeed 1/36.

And "binomial" is not the same as "bimodal".

Sorry binomial was a typo in the title I did not know how to correct.

#### Colin2B

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##### Re: Is this sequence a binomial distribution?
« Reply #6 on: 21/04/2015 18:49:20 »
Just to back up what Alan is saying:

Let's take a simple example which is similar to your dice rolling. You have 2 coins which you toss one after the other. Clearly the chance of 2 heads is 1/4. But if you toss the 1st coin and get a head, what is the probability of getting 2 heads? Many people will say 1/4, but as you know the probability that the 2nd coin will fall heads is 50%, 1/2. So it depends when you make your decision or bet.
In the case of the rolling dice, once the first 1 has appeared the probability of the second is as Alan says 1/6, but that is the same for both players. Unless you are saying the 2nd player is only allowed to bet on the total outcome before any dice is rolled, whereas the 1st player has a bet after the first roll. In this case yes, the probabilities are different for each player.

Sorry binomial was a typo in the title I did not know how to correct.
It's a pity your spelling mistake wasn't the other way round!
The rolling of 2 dice can create a binomial distribution, with a mode of 7, if you add the face values of the 2 dice.
The individual dice have an equal chance for each face, so this is a random distribution and has no mode.

While I have your attention I will make a small point. Although we might discuss dice or coins, one of the strengths of maths and science is that we can take something specific and then remove the numbers describing the outcomes and develop a theory which is independent of specific objects, just using numbers and symbols on their own. So what is true for dice can be applied to children in a school, queues in a bank, or traffic approaching an intersection. But while considering the theorems, we do not need to specify particular objects. Just a small point

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #7 on: 21/04/2015 20:19:16 »
Just to back up what Alan is saying:

Let's take a simple example which is similar to your dice rolling. You have 2 coins which you toss one after the other. Clearly the chance of 2 heads is 1/4. But if you toss the 1st coin and get a head, what is the probability of getting 2 heads? Many people will say 1/4, but as you know the probability that the 2nd coin will fall heads is 50%, 1/2. So it depends when you make your decision or bet.
In the case of the rolling dice, once the first 1 has appeared the probability of the second is as Alan says 1/6, but that is the same for both players. Unless you are saying the 2nd player is only allowed to bet on the total outcome before any dice is rolled, whereas the 1st player has a bet after the first roll. In this case yes, the probabilities are different for each player.

Sorry binomial was a typo in the title I did not know how to correct.
It's a pity your spelling mistake wasn't the other way round!
The rolling of 2 dice can create a binomial distribution, with a mode of 7, if you add the face values of the 2 dice.
The individual dice have an equal chance for each face, so this is a random distribution and has no mode.

While I have your attention I will make a small point. Although we might discuss dice or coins, one of the strengths of maths and science is that we can take something specific and then remove the numbers describing the outcomes and develop a theory which is independent of specific objects, just using numbers and symbols on their own. So what is true for dice can be applied to children in a school, queues in a bank, or traffic approaching an intersection. But while considering the theorems, we do not need to specify particular objects. Just a small point

just to confirm

1/221 is not the same as 4524/1,000,000?

#### Colin2B

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##### Re: Is this sequence a binomial distribution?
« Reply #8 on: 22/04/2015 09:07:38 »
just to confirm

1/221 is not the same as 4524/1,000,000?
To use a technical term "it is near as dammit"

It all depends what you are doing.
If parking a car this difference is unlikely to trouble you, but if you are manoeuvring a sample under an electron microscope, different story.

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #9 on: 22/04/2015 12:13:12 »
just to confirm

1/221 is not the same as 4524/1,000,000?
To use a technical term "it is near as dammit"

It all depends what you are doing.
If parking a car this difference is unlikely to trouble you, but if you are manoeuvring a sample under an electron microscope, different story.

I have just had agreement elsewhere that I am correct, I will redefine for you

random x has 52 variants and the probabilities of receiving (a) from x is 1/221

random x + random x =2/442

random x = (1/221)/t

random x + random x = (2/442)/xt

The probabilities are not the same as such,

(1/221)/t= EV

(1/221)+(1/221)/xt=(2/442)/xt = X

where x is random and t is time

random probability distribution.

#### Colin2B

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##### Re: Is this sequence a binomial distribution?
« Reply #10 on: 22/04/2015 14:17:12 »
Flirts of all congratulations in spotting the f.point's new maths is in fact a reinvention of std vector maths (albeit in 1 dimension).
I've left a couple of questions there for you. One is easy, the other less so, have a think.

I will redefine for you
random x has 52 variants and the probabilities of receiving (a) from x is 1/221

random x + random x =2/442

In probability, definitions are vital, so good start. Some questions:

Are you saying that the set of items called x contains 52 items all different? Or are you saying x contains 221 items but only 52 different types of item?

I assume (a) is any single item selected from x.

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #11 on: 22/04/2015 16:40:31 »
Flirts of all congratulations in spotting the f.point's new maths is in fact a reinvention of std vector maths (albeit in 1 dimension).
I've left a couple of questions there for you. One is easy, the other less so, have a think.

I will redefine for you
random x has 52 variants and the probabilities of receiving (a) from x is 1/221

random x + random x =2/442

In probability, definitions are vital, so good start. Some questions:

Are you saying that the set of items called x contains 52 items all different? Or are you saying x contains 221 items but only 52 different types of item?

I assume (a) is any single item selected from x.

I am referring to texas holdem poker, and x contains 52 different variants, (a deck of cards).

(a) being pocket pair aces for example purposes.

you have 1/221 chance of pocket aces from a single deck or to clarify for every 221 hands you receive on average you will see pocket aces at least once.

For 221 hands you see 221 shuffles of the deck, so we can say that 221 hands is the same as 221 pre-shuffled decks.

#### Colin2B

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##### Re: Is this sequence a binomial distribution?
« Reply #12 on: 22/04/2015 17:45:10 »

I am referring to texas holdem poker, and x contains 52 different variants, (a deck of cards).

(a) being pocket pair aces for example purposes.
Thanks for the clarification.
Not sure what a pocket pair ace is, but as I'm unlikely to play it tells me this isn't info I need to know.
Sounds as though you should be an expert on probability!

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #13 on: 22/04/2015 20:28:42 »

I am referring to texas holdem poker, and x contains 52 different variants, (a deck of cards).

(a) being pocket pair aces for example purposes.
Thanks for the clarification.
Not sure what a pocket pair ace is, but as I'm unlikely to play it tells me this isn't info I need to know.
Sounds as though you should be an expert on probability!

pocket pair aces is 2 aces as your starting hand you are dealt , so you know.

#### Colin2B

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##### Re: Is this sequence a binomial distribution?
« Reply #14 on: 22/04/2015 23:02:06 »
pocket pair aces is 2 aces as your starting hand you are dealt , so you know.
Thanks
It all sounds much too complicated for me.

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #15 on: 23/04/2015 11:40:24 »
1/221+1/221=2/442 (/xt˛) = X

if the ratio for one deck is 1/221 is my calculation correct if we use multiple decks and multiple receivers?

(a)+(b)=(ab)

(ab)/xt˛=X

1,000,000 decks = 4,000,000 aces

4,000,000x/52,000,000x =0.07692307692% chance of receiving an ace , where x equals random position in a sequence.

compared to 1x/221x=0.00452488687% chance of receiving an ace, where x equals random position in a sequence.
« Last Edit: 23/04/2015 13:27:24 by Thebox »

#### Thebox

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##### Re: Is this sequence a binomial distribution?
« Reply #16 on: 25/04/2015 00:52:41 »
added - lets draw a picture of how set card sequences look from a pre-shuffled deck, and sequencing stacking.

Based using cards and the top five card sequences of each deck to further define, this we will call a virtual number variants position grid (vnvpg), the second part of on-line process after the RNG and before the deal.

X/(xt)
.deck 1 - ah jd 6c 7d ks
.deck 2 - 4s 2d kh jc 8d
.deck 3 - ts qd 8s 3h ah
.deck 4 - kc 9s ac 6d js
..................................y/(t)

X being the vertical columns and y being the horizontal rows. (t) being time and x being random

In example X1,y1=ah

The third part we call it the players variant position (pvp)

.X/(xt)
T1 - p1 p2 p3 p4 p5
T2 - p1 p2 p3 p4 p5
T3 - p1 p2 p3 p4 p5
T4 - p1 p2 p3 p4 p5
..................................y/(t)

If grid 1 was in random rotation of X, what are the probabilities of XT3, y4 of receiving X3,y4 ?
« Last Edit: 25/04/2015 01:02:53 by Thebox »

#### The Naked Scientists Forum

##### Re: Is this sequence a binomial distribution?
« Reply #16 on: 25/04/2015 00:52:41 »