# Zero Probability is hard to obtain in a finite sense. Right?

Probability theory and statistics

### Zero Probability is hard to obtain in a finite sense. Right?

Let's say we have a deck of 52 index cards each labeled with a distinct number from one to fifty-two on one side and left blank on the other side. We shuffle the deck at least eight times. What is the probability we do not select an index card that has been labeled one without replacement by the 52nd selection?

Of course, we expect the answer to be zero or have a zero probability. In practice, the answer is $$(\frac{1}{52})^{52} \approx 5.859 * 10^{-90}$$ which is clearly not zero.

However, if we have an infinite deck of index cards labeled from one to infinity, then we approach a zero probability for selecting that index card labeled one by the infinite selection... Right?
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Oops! We corrected our error.

Guest wrote:Let's say we have a deck of 52 index cards each labeled with a distinct number from one to fifty-two on one side and left blank on the other side. We shuffle the deck at least eight times. What is the probability we do not select an index card that has been labeled one without replacement by the 52nd selection?

Of course, we expect the answer to be zero or have a zero probability. In practice, the answer is $$(\frac{1}{52!}) \approx 7.2107 * 10^{-81}$$ which is clearly not zero.

However, if we have an infinite deck of index cards labeled from one to infinity, then we approach a zero probability for selecting that index card labeled one by the infinite selection... Right?
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

... $$(\frac{1}{52!}) \approx 1.2398 * 10^{-68}$$ ...
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Okay! However, in a deck of one index card which is labeled as one, the probability of not selecting it is zero! Right! Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Guest wrote:... $$(\frac{1}{52!}) \approx 1.2398 * 10^{-68}$$ ... is clearly not zero, but it is almost zero which is good enough! Right?

We will certainly select eventually (without replacement) the index card labeled one from a deck of 52 index cards each labeled uniquely with a number from 1 to 52... Right! Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Collatz Conjecture states for any positive odd integer, $$N_{m }$$, we will generate a sequence of odd integers that will converge to one.

Now, please consider the following proposition:

Probability(Collatz sequence does not converge to 1)

= Prob($$N_{m }$$ as $$m \rightarrow \infty$$)

$$=\prod_{j=1}^{\infty} \sum_{i=0}^{l_{j } - 1 }(1/2)^{i+1} \rightarrow 0.$$

'Proof of Collatz Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1485.
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Guest wrote:Collatz Conjecture states for any positive odd integer, $$N_{m }$$, we will generate a sequence of odd integers that will converge to one.

Now, please consider the following proposition:

Probability(Collatz sequence does not converge to 1)

= Prob($$N_{m } \ne 1$$ as $$m \rightarrow \infty$$)

$$=\prod_{j=1}^{\infty} \sum_{i=0}^{l_{j } - 1 }(1/2)^{i+1} \rightarrow 0.$$

'Proof of Collatz Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1485.
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Goldbach conjecture states that all positive even integers, e, greater than four is the sum of two odd primes.

Now, please consider the following proposition:

Probability( $$e \ne p + q$$ ) $$= \prod_{j=1}^{k \to \infty }\frac{\pi (\sqrt{e - p_{j}})}{\pi (\sqrt{e - p_{j}}) + 1} \rightarrow 0$$

where p and q are any odd primes less than e.

Moreover, $$p_{j}$$ is any distinct odd prime less than any even integer, e, greater than 100.

We let k be the number of distinct odd primes less than e.

And we let π() be the odd prime-counting function.

Note: One counts, and it is counted.

'Proof of Goldbach Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1548.
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Guest wrote:Goldbach conjecture states that all positive even integers, e, greater than four is the sum of two odd primes.

Now, please consider the following proposition:

Probability( $$e \ne p + q$$ ) $$= \prod_{j=1}^{k \to \infty }\frac{\pi (\sqrt{e - p_{j}})}{\pi (\sqrt{e - p_{j}}) + 1} \rightarrow 0$$

where p and q are any odd primes less than e.

Moreover, $$p_{j}$$ is any distinct odd prime less than any even integer, e, greater than 100.

We let k be the number of distinct odd primes less than e.

And we let π() be the odd prime-counting function.

Note: One counts, and it is counted.

'Proof of Goldbach Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1548.

Example:

Probability( $$110,800,888 \ne p + q$$) $$= 4.94066*10^{-324}$$.

Therefore, we are extremely confident that 10,800,888 is a sum of two odd primes, p and q.

What are p and q?
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Example:

Probability( $$110,800,888 \ne p + q$$) $$= 4.94066*10^{-324}$$.

Therefore, we are extremely confident that 10,800,888 is a sum of two odd primes, p and q.

What are p and q?

10,800,888 = 179 + 10800709 . The prime pair, 179 and 10800709, is called a Goldbach partition for the even integer, 10,800,888. And there are many more Goldbach partitions for that number...
Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Guest wrote:Goldbach conjecture states that all positive even integers, e, greater than four is the sum of two odd primes.

Now, please consider the following proposition:

Probability( $$e \ne p + q$$ ) $$= \prod_{j=1}^{k \to \infty }\frac{\pi (\sqrt{e - p_{j}})}{\pi (\sqrt{e - p_{j}}) + 1} \rightarrow 0$$

where p and q are any odd primes less than e.

Moreover, $$p_{j}$$ is any distinct odd prime less than any even integer, e, greater than 100.

We let k be the number of distinct odd primes less than e.

And we let π() be the odd prime-counting function.

Note: One counts, and it is counted.

'Proof of Goldbach Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1548.

Hmm. That square root power (1/2) inside the important expression, $$\pi (\sqrt{e - p_{j}})$$, is a big indicator of the truth of the Riemann Hypothesis.
Go figure! And Go Blue! Guest

### Re: Zero Probability is hard to obtain in a finite sense. Ri

Guest wrote:
Example:

Probability( $$10,800,888 \ne p + q$$) $$= 4.94066*10^{-324}$$.

Therefore, we are extremely confident that 10,800,888 is a sum of two odd primes, p and q.

What are p and q?

10,800,888 = 179 + 10800709 . The prime pair, 179 and 10800709, is called a Goldbach partition for the even integer, 10,800,888. And there are many more Goldbach partitions for that number...

Correction! Probability( $$10,800,888 \ne p + q$$) $$< 5*10^{-324}$$. And 110, 800,888 is a mistake.
Guest