Naked Science Forum
On the Lighter Side => New Theories => Topic started by: guest39538 on 18/11/2015 17:39:21
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Most of you know my poker theory , I have it down to this model,
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Is this understood? if not how can I improve it? what maths do I need?
i am trying to show this
https://en.wikipedia.org/wiki/Multivalued_function
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added diagram to show what I am trying to get at.
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added maths to represent the diagram
X^²=Y={σ²(x)/Y}/σ²(z)/t
P(x^²)/y/z=0_1/t
Y=∭(x}
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pi∫∫le
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pi∫∫le
Only because you do not recognise this as maths you already know, you don't know this maths because it does not exist yet. So you can't recognise it instantly.
It reads elegantly.
x is constant
y and z are variable
x=52
y=104
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It reads elegantly.
x is constant
y and z are variable
x=52
y=104
how is y a variable and 104?
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It reads elegantly.
x is constant
y and z are variable
x=52
y=104
how is y a variable and 104?
X^² is x to the power of 2
which is equal to Y.
X²=Y²
X only contains a specific 1 of 1/52 where Y contains 2 or more of a specific
If we have 100 decks of cards we have 100 chances that the top card is 1/52 to a specific card, all 100 top cards could be the same specific card, or all 100 top cards could contain no specific of the card.
It is a variable compared to 1/52
X is rows and Y is columns, X can only have 1 out of 52 where y can have several of 1 of 52
X^²=Y={σ²(x)/Y}/σ²(z)/t
P(x^²)/y/z=0_1/t
X to the power of 2 is equal to y which is equal to a variation of x divided by Y which is distributed to multiple Z that is ever changing over time
so it cant be 1/52 because the 1 could be several or none. so the P=0_1/t
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we have a 1/36 chance of intercepting a 4/52 chance
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1.6% chance there is no aces based on a 52 value wheel
4/52
{x}Δ{σ²X}={Y}=Δ4=σ²P[A]/{Y}
(ε, δ)=4/X
(ε, δ)=52/Y
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we have a 1/36 chance of intercepting a 4/52 chance
Wrong. It doesn't matter where the ball stops, or how many packs there are: the probability of any position being an ace is exactly the same, i.e. 1/13.
Take a simpler example. Toss a coin as many times as you like. The probability of the nth toss being a head is 1/2 whatever the value of n.
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we have a 1/36 chance of intercepting a 4/52 chance
Wrong. It doesn't matter where the ball stops, or how many packs there are: the probability of any position being an ace is exactly the same, i.e. 1/13.
Take a simpler example. Toss a coin as many times as you like. The probability of the nth toss being a head is 1/2 whatever the value of n.
No Alan, honestly mate
φ{X}/t=1/52
Z
≠
φ{Y}/t=0_52/52
^Z
∑{Y}≠∑{X}
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added - surely this is understandable?
φ{X}=1:52
φ{Y1.....Y52}=(<1.=1.>1):52
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added - surely this is understandable?
φ{X}=1:52
φ{Y1.....Y52}=(<1.=1.>1):52
nope. I might understand the first line, but I have no clue what the second line means...
typically = means "equal to" as in a = b means a is equal to b
and > means "greater than" as in c > d means c is greater than d
and < means "less than" as in d < c means d is less than c
(if c > d then d < c; if a = b then b = a)
by these definitions "<1.=1.>1" is completely nonsensical.
if you are not using standard, generally accepted notation, then you must explain what it is that you mean (at the very least define your operations)
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added - surely this is understandable?
φ{X}=1:52
φ{Y1.....Y52}=(<1.=1.>1):52
nope. I might understand the first line, but I have no clue what the second line means...
typically = means "equal to" as in a = b means a is equal to b
and > means "greater than" as in c > d means c is greater than d
and < means "less than" as in d < c means d is less than c
(if c > d then d < c; if a = b then b = a)
by these definitions "<1.=1.>1" is completely nonsensical.
if you are not using standard, generally accepted notation, then you must explain what it is that you mean (at the very least define your operations)
Yes greater than and less than and equal to, I have put . to represent and/or. And put it in brackets to rep a set
Help me please, I am trying to create something like this P(a ≤ x ≤ b) = ∫ f (x) dx to rep Y change.
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X²=52*52=X,Y
φ{X}=1:52
φ{Y1...............Y52}=?/52
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This
ppp123,
ppp231,
ppp231,
ppp213
123,
231,
231,
213
ppp
ppp
ppp
ppp
spot the difference competition.
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I have it simplified for you all.
draw card 1 from deck 1, a 1/52 to be any value
draw card 1 from deck 2, a 1/52 to be any value
1/52=1/52
Both cards have equal chance to being the same value.
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If you dont get this, its time give up .
https://theoristexplains.wordpress.com/2015/12/01/apparently-this-is-wrong/
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Thank you A.
''In statistics, the autocorrelation of a random process describes the correlation between values of the process at different times, as a function of the two times or of the time lag. Let X be some repeatable process, and i be some point in time after the start of that process. (i may be an integer for a discrete-time process or a real number for a continuous-time process.) Then Xi is the value (or realization) produced by a given run of the process at time i. Suppose that the process is further known to have defined values for mean μi and variance σi2 for all times i. Then the definition of the autocorrelation between times s and t is''
I am trying to explain this .
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This must be close?
λ(X)=φt
λ(Y)=σ/
λ(X)=φt1=
λ(Y)=σ=t2
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==
===