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 ?
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...
The sum in brackets here equals (2n)Cn.
I am terrible at probability but here are my immediate thoughts...waht are you good (better) at? do you write Limericks or Sonnets ?
i recently learned of the name [Hockey-stick identity]
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.
( What's a probability value that (interestingly) Doesn't go to Zero
as N increases ? )
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?
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 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
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