(Agatha, Beniko)

A and B have 1 coin each.
They throw all of their coins and count the number of heads.
prob. of a tie must be ... 1/2

A and B have 2 coins each. (...)
prob. of a tie must be ... (smaller)

A and B have 3 coins each. (...)
prob. of a tie must be ... (even smaller)

(as N increases)

(1) Is this value going down to zero ?

(2) The same question, where A has an extra coin.

In article <2cfbb9cb-bf93-4455-b820-f4c521c5975bn@googlegroups.com>,
henh...@gmail.com <henhanna@gmail.com> wrote:

> A and B have 1 coin each.
> They throw all of their coins and count the number of heads.
> prob. of a tie must be ... 1/2
>
> A and B have 2 coins each. (...)
> prob. of a tie must be ... (smaller)
>
> A and B have 3 coins each. (...)
> prob. of a tie must be ... (even smaller)
>
> (1) Is this value going down to zero ?

I am terrible at probability but here are my immediate thoughts...

In n flips, the probability of getting k heads is (nCk)/2^n, so the
probability that they both get k heads is its square.

Summing over all k (which we can do as the events are disjoint) gives

[ (nC0)^2 + (nC1)^2 + ... + (nCn)^2 ] / 4^n

The sum in brackets here equals (2n)Cn.

(Proof? Use nCk = nC(n-k) and the sum is a convolution.)

One way to show that (2n)Cn / 4^n tends to 0 is via logs.

I can fill in the details and steps if needed, but since I've had no
replies to any of my posts to your questions, I'm unsure that they're
getting through. So let me know if you'd like steps.

On the other hand, maybe someone has a _nice_ answer. I'm rubbish at
probability.

Gareth

On Tuesday, October 4, 2022 at 2:16:15 PM UTC-7, Gareth Taylor wrote:
> In article <2cfbb9cb-bf93-4455...@googlegroups.com>,
> henh...@gmail.com <henh...@gmail.com> wrote:
>
> > A and B have 1 coin each.
> > They throw all of their coins and count the number of heads.
> > prob. of a tie must be ... 1/2
> >
> > A and B have 2 coins each. (...)
> > prob. of a tie must be ... (smaller)
> >
> > A and B have 3 coins each. (...)
> > prob. of a tie must be ... (even smaller)
> >
> > (1) Is this value going down to zero ?

> I am terrible at probability but here are my immediate thoughts...
>
> In n flips, the probability of getting k heads is (nCk)/2^n, so the
> probability that they both get k heads is its square.
>
> Summing over all k (which we can do as the events are disjoint) gives
>
> [ (nC0)^2 + (nC1)^2 + ... + (nCn)^2 ] / 4^n
>
> The sum in brackets here equals (2n)Cn.
>
> (Proof? Use nCk = nC(n-k) and the sum is a convolution.)
>
> One way to show that (2n)Cn / 4^n tends to 0 is via logs.
>
> I can fill in the details and steps if needed, but since I've had no
> replies to any of my posts to your questions, I'm unsure that they're
> getting through. So let me know if you'd like steps.
>
> On the other hand, maybe someone has a _nice_ answer. I'm rubbish at
> probability.
>
> Gareth

> I am terrible at probability but here are my immediate thoughts...

waht are you good (better) at? do you write Limericks or Sonnets ?

> The sum in brackets here equals (2n)Cn.

that's a nice one....

i recently learned of the name [Hockey-stick identity]
( but i thik i was vaguely aware of it....)

In article <828fc0db-e4ce-45e9-9600-bf93594ef524n@googlegroups.com>,
henh...@gmail.com <henhanna@gmail.com> wrote:

>> I am terrible at probability but here are my immediate thoughts...
> waht are you good (better) at? do you write Limericks or Sonnets ?

Number theory or algebra is more fun. I teach maths, but I have a bit
of a mental block for probability. You'll notice that I turned your
question into combinatorics rather than probability!

Oddly, I have written silly poems in the past!

> i recently learned of the name [Hockey-stick identity]

I had to look that up. I know the result but not that name. The result
I invoked above was an example of Vandermonde's Convolution.

Gareth

On Tuesday, October 4, 2022 at 2:16:15 PM UTC-7, Gareth Taylor wrote:
> In article <2cfbb9cb-bf93-4455...@googlegroups.com>,
> henh...@gmail.com <henh...@gmail.com> wrote:
>
> > A and B have 1 coin each.
> > They throw all of their coins and count the number of heads.
> > prob. of a tie must be ... 1/2
> >
> > A and B have 2 coins each. (...)
> > prob. of a tie must be ... (smaller)
> >
> > A and B have 3 coins each. (...)
> > prob. of a tie must be ... (even smaller)
> >
> > (1) Is this value going down to zero ?
> I am terrible at probability but here are my immediate thoughts...
>
> In n flips, the probability of getting k heads is (nCk)/2^n, so the
> probability that they both get k heads is its square.
>
> Summing over all k (which we can do as the events are disjoint) gives
>
> [ (nC0)^2 + (nC1)^2 + ... + (nCn)^2 ] / 4^n
>
> The sum in brackets here equals (2n)Cn.
>
> (Proof? Use nCk = nC(n-k) and the sum is a convolution.)
>
> One way to show that (2n)Cn / 4^n tends to 0 is via logs.
>

not too surprised that it goes to Zero...

( What 's a probability value that (interestingly) Doesn't go to Zero as N increases ? )

simulation:

(2-sided) 1 : 0.50003
(2-sided) 2 : 0.37273
(2-sided) 3 : 0.31224
(2-sided) 4 : 0.27469
(2-sided) 5 : 0.24521
(2-sided) 6 : 0.22711
(2-sided) 7 : 0.20843
(2-sided) 8 : 0.19551
(2-sided) 9 : 0.18549
(2-sided) 10 : 0.17606

(2-sided) 100 : 0.0559
(2-sided) 1000 : 0.0172

(2-sided) 10000 : 0.006

In article <bd470200-70b0-46e7-af4d-cf46bd66f877n@googlegroups.com>,
henh...@gmail.com <henhanna@gmail.com> wrote:

> ( What's a probability value that (interestingly) Doesn't go to Zero
> as N increases ? )

Suppose you have 26 cards labelled from A to Z. You shuffle them and
then deal them out in a row. What is the probability that none of the
letters is in the correct position?

With just two cards, it's obviously 1/2.

How about three? The orders BCA and CAB have none in the correct
position, while the orders ACB, CBA, BAC, ACB have at least one in the
correct position. So it's 1/3.

What happens for larger alphabets?

Gareth

As Gareth Taylor said, the probability of both players getting the
same number of heads is (2n)Cn / 4^n.

But (2n)Cn / 4^n is *also* the probability of the two players getting
exactly n heads between them.

This is unsurprising as one is

nC0 . nC0 + nC1 . nC1 + ... + nCn . nCn

and the other is

nC0 . nCn + nC1 . nC(n-1) + ... + nCn . nC0

and nCk = nC(n-k)

Does anyone have a more intuitive explanation as to why these two
probabilities are the same?

-- Richard

On 05/10/2022 00:35, Richard Tobin wrote:
> As Gareth Taylor said, the probability of both players getting the
> same number of heads is (2n)Cn / 4^n.
>
> But (2n)Cn / 4^n is *also* the probability of the two players getting
> exactly n heads between them.
>
> This is unsurprising as one is
>
> nC0 . nC0 + nC1 . nC1 + ... + nCn . nCn
>
> and the other is
>
> nC0 . nCn + nC1 . nC(n-1) + ... + nCn . nC0
>
> and nCk = nC(n-k)
>
> Does anyone have a more intuitive explanation as to why these two
> probabilities are the same?
>

I'm not sure if this is any more intuitive...

If A and B both throw n coins and get a tie for heads, then by reversing all of B's coins we would
get a total of n heads between them. And vice-versa. So if we look at the space of all 2n coin
tosses, there is a 1-1 correspondence [viz. through reversing B's tosses] which associates each
"equal heads" outcome to a "n heads total" outcome and vice versa.

So the count of outcomes for each condition is the same, hence the probabilities are the same
(dividing the common count by 2^(2n)).

Mike.

Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:
> On 05/10/2022 00:35, Richard Tobin wrote:
>> As Gareth Taylor said, the probability of both players getting the
>> same number of heads is (2n)Cn / 4^n.
>>
>> But (2n)Cn / 4^n is *also* the probability of the two players getting
>> exactly n heads between them.
>>
>> This is unsurprising as one is
>>
>> nC0 . nC0 + nC1 . nC1 + ... + nCn . nCn
>>
>> and the other is
>>
>> nC0 . nCn + nC1 . nC(n-1) + ... + nCn . nC0
>>
>> and nCk = nC(n-k)
>>
>> Does anyone have a more intuitive explanation as to why these two
>> probabilities are the same?
>
> I'm not sure if this is any more intuitive...
>
> If A and B both throw n coins and get a tie for heads, then by
> reversing all of B's coins we would get a total of n heads between
> them. And vice-versa. So if we look at the space of all 2n coin
> tosses, there is a 1-1 correspondence [viz. through reversing B's
> tosses] which associates each "equal heads" outcome to a "n heads
> total" outcome and vice versa.
>
> So the count of outcomes for each condition is the same, hence the
> probabilities are the same (dividing the common count by 2^(2n)).

Perfect!

Phil
--
We are no longer hunters and nomads. No longer awed and frightened, as we have
gained some understanding of the world in which we live. As such, we can cast
aside childish remnants from the dawn of our civilization.
-- NotSanguine on SoylentNews, after Eugen Weber in /The Western Tradition/

On Thursday, October 6, 2022 at 5:57:29 AM UTC-7, Phil Carmody wrote:
> Mike Terry <news.dead.p...@darjeeling.plus.com> writes:
> > On 05/10/2022 00:35, Richard Tobin wrote:
> >> As Gareth Taylor said, the probability of both players getting the
> >> same number of heads is (2n)Cn / 4^n.
> >>
> >> But (2n)Cn / 4^n is *also* the probability of the two players getting
> >> exactly n heads between them.
> >>
> >> This is unsurprising as one is
> >>
> >> nC0 . nC0 + nC1 . nC1 + ... + nCn . nCn
> >>
> >> and the other is
> >>
> >> nC0 . nCn + nC1 . nC(n-1) + ... + nCn . nC0
> >>
> >> and nCk = nC(n-k)
> >>
> >> Does anyone have a more intuitive explanation as to why these two
> >> probabilities are the same?
> >

> > I'm not sure if this is any more intuitive...
> >
> > If A and B both throw n coins and get a tie for heads, then by
> > reversing all of B's coins we would get a total of n heads between
> > them. And vice-versa. So if we look at the space of all 2n coin
> > tosses, there is a 1-1 correspondence [viz. through reversing B's
> > tosses] which associates each "equal heads" outcome to a "n heads
> > total" outcome and vice versa.
> >
> > So the count of outcomes for each condition is the same, hence the
> > probabilities are the same (dividing the common count by 2^(2n)).

> Perfect!
>
> Phil
> --

1. i wonder what is another example of ... looking in a novel way to show that two problems are actually the same

2. What else is (2n C n) the answer for ?