Re: A card game probability



On Jan 27, 1:16 pm, "Faton Berisha" <fberi...@xxxxxxxxxx> wrote:
On Jan 27, 2:09 am, matt271829-n...@xxxxxxxxxxx wrote:



I don't quite follow this. I'm guessing that "bar" signifies the
complement, so "bar p_n" means the probability of *not* getting the
first match at card n. Is that right?Not exactly. Here bar p_n is just a notation. I explained what
precisely it means.
If it troubles you, just denote it by another symbol, say q_n.

So if n = 1 should we have bar
p_n = 12/13? I don't see how to get to that.If you substitute n=1, that's exactly what you'll get: \bar p_1=12/13.
(See the computation in my previous message.)

What does binom{n-1}i
mean? Initially I thought it might mean "n-1 choose i" but that seems
not to work.Yes, it means {n - 1} chose i, and yes it does work.

And what does "A" signify?Denote by A an event. Then bar A is the complement of A.
Denote by Pr(A | B) the conditional probability of an event A given
the occurrence of event B.

Denote by A_n: The card flipped in the n-th trial matches its pair.
Then,

Pr(bar A_n | bar A_1 bar A_2 ... bar A_{n-1})

is the probability of not having a match in the n-th trial, given that
no match occurred in previous trials; i.e., the probability of not
having a match after n trials given the fact that no match occurred in
n-1 first trials.



p_n = 1 - prod_{n=1}^n bar p_n.This equation does not make a whole lot of sense as written. I'm
guessing that you mean something like q_n = 1 - prod_{k=1}^n bar p_k,You are, of course, right here. I mistakenly used the same symbol for
the iterator and the upper limit.
It should be

p_i = prod_{i=1}^n \bar p_i

for the limit of having at least one match in n trials.

But then why are you multiplying the complementary probabilities and
subtracting from one? If p_k is the probability of getting the first
match at card k, then the probability of getting at least one match in
the first n cards is sum_k=1^n p_k, isn't it?
...Be careful here. In my notation \bar p and p are not complementary
probabilities (although A and \bar A are complementary events).



On Jan 27, 1:16 pm, "Faton Berisha" <fberi...@xxxxxxxxxx> wrote:
On Jan 27, 2:09 am, matt271829-n...@xxxxxxxxxxx wrote:



I don't quite follow this. I'm guessing that "bar" signifies the
complement, so "bar p_n" means the probability of *not* getting the
first match at card n. Is that right?Not exactly. Here bar p_n is just a notation. I explained what
precisely it means.
If it troubles you, just denote it by another symbol, say q_n.

So if n = 1 should we have bar
p_n = 12/13? I don't see how to get to that.If you substitute n=1, that's exactly what you'll get: \bar p_1=12/13.
(See the computation in my previous message.)

What does binom{n-1}i
mean? Initially I thought it might mean "n-1 choose i" but that seems
not to work.Yes, it means {n - 1} chose i, and yes it does work.

And what does "A" signify?Denote by A an event. Then bar A is the complement of A.
Denote by Pr(A | B) the conditional probability of an event A given
the occurrence of event B.

Denote by A_n: The card flipped in the n-th trial matches its pair.
Then,

Pr(bar A_n | bar A_1 bar A_2 ... bar A_{n-1})

is the probability of not having a match in the n-th trial, given that
no match occurred in previous trials; i.e., the probability of not
having a match after n trials given the fact that no match occurred in
n-1 first trials.



p_n = 1 - prod_{n=1}^n bar p_n.This equation does not make a whole lot of sense as written. I'm
guessing that you mean something like q_n = 1 - prod_{k=1}^n bar p_k,You are, of course, right here. I mistakenly used the same symbol for
the iterator and the upper limit.
It should be

p_i = prod_{i=1}^n \bar p_i

for the limit of having at least one match in n trials.

But then why are you multiplying the complementary probabilities and
subtracting from one? If p_k is the probability of getting the first
match at card k, then the probability of getting at least one match in
the first n cards is sum_k=1^n p_k, isn't it?
...Be careful here. In my notation \bar p and p are not complementary
probabilities (although A and \bar A are complementary events).


[More Google Grief. Sorry if this appears more than once.]

Right, I figured out how you get the value of the expression to be
12/13 for n = 1. I didn't realise that you were assuming binom{n-1}i =
0 when i > n-1. I just looked at it and thought "that must be wrong".

There are several other things I don't understand in your reply, but
rather than labouring over those, let's cut to the chase and look at
your final result, which I quote below:

"Hence, the probability of flipping at least one such card in n
trials
is

p_n = 1 - prod_{n=1}^n bar p_n.

As it is, the formula is not easily evaluated for large values of n.
I
made a quick test in my computer, and these are the corresponding
values of p_n that I obtained:

( 0.07692, 0.14770, 0.21282, ..."

I agree with the first result (n = 1). For n = 2 I calculate the
probability as the sum of the following cases:

First card = 1 (prob 4/52); second card = anything (prob 1)
First card = 2 (prob 4/52); second card = 2 (prob 3/51)
First card not 1 or 2 (prob 44/52); second card = 2 (prob 4/51)

Total probability = 4/52*1 + 4/52*3/51 + 44/52*4/51 = 98/663 =
0.14781...

And for n = 3, using an analogous method, I get a probability of
3533/16575 = 0.21315...

These are rather different from your results.

.

Relevant Pages

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