Impossible Bible Prophecy Odds and ChatGPT 4

Me: If a newspaper makes a series of 10 predictions about something that will happen in the future, and all of them happen, what are the odds? All of the events are things that the newspaper could not have had control over - such as naming a man who will become president in the future who hasn't been born yet; or a war in a nation that doesn't exist yet, or a prediction that Latin will be revived and millions of people will speak it everyday in a new European nation. All the other predictions the newspaper makes are just as unlikely. Is it possible to calculate the odds of something like that?

ChatGPT 4: Calculating the exact odds of a series of highly specific and unlikely predictions coming true is extremely challenging because it depends heavily on the nature of each prediction and the inherent probabilities of each event happening independently. However, we can attempt to understand the magnitude of the odds by making some assumptions about the improbability of each individual prediction.

Assumptions and Approach

1. Improbability of Each Event: Let's assume that each event predicted has an extremely low probability of occurring, say $P(E_i) = 1 \times 10^{-6}$ (one in a million).

2. Independence of Events: We assume that the events are independent of each other, meaning the occurrence of one event does not affect the probability of another event occurring.

Calculating Combined Probability

The probability of all 10 independent events occurring is the product of the probabilities of each individual event occurring. If each event has a probability of $P(E_i)$:

$P(\text{All 10 events}) = P(E_1) \times P(E_2) \times \cdots \times P(E_{10})$

Given $P(E_i) = 1 \times 10^{-6}$:

$P(\text{All 10 events}) = (1 \times 10^{-6})^{10} = 1 \times 10^{-60}$

Interpretation

• The resulting probability $1 \times 10^{-60}$ represents the odds of all 10 highly unlikely predictions coming true, assuming each event is as unlikely as one in a million.
• This probability is extraordinarily small, indicating that it is virtually impossible for all 10 predictions to come true by mere chance.

Real-World Implications

In real-world scenarios, predicting events with such low probabilities accurately would likely indicate some form of extraordinary knowledge or influence, beyond normal human capabilities. It would be akin to predicting outcomes with near-certainty in extremely unpredictable systems, suggesting factors such as advanced foresight, hidden information, or supernatural insight.

Conclusion

While exact calculations depend on the specific probabilities assigned to each event, assuming each event is extremely unlikely (e.g., one in a million chance), the combined probability of all 10 predictions coming true is astronomically low ($1 \times 10^{-60}$). This demonstrates the extreme unlikelihood of such a scenario occurring by chance alone.

Me:  (1×10−601×10−60) How would you express that as a decimal?

ChatGPT 4: To express $(1 \times 10^{-60}) \times (1 \times 10^{-60})$ as a decimal, you first multiply the exponents of 10. Here's the step-by-step process:

1. Multiply the Exponents:

   $$10^{-60} \times 10^{-60} = 10^{-120}$$

2. Multiply the Coefficients:

   $$1 \times 1 = 1$$

3. Combine the Coefficient and the Exponent:

   $$1 \times 10^{-120} = 10^{-120}$$

Now, expressing $10^{-120}$ as a decimal means writing it as a number with 120 zeros after the decimal point, followed by a 1. This is:

$$0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001$$

So, $(1 \times 10^{-60}) \times (1 \times 10^{-60})$ expressed as a decimal is:

$$0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001$$

Me:  I thought so, but when numbers get that crazy, I think I've made a mistake somewhere.

ChatGPT 4: It's completely understandable to double-check when dealing with such extremely small numbers. The process of converting a number in scientific notation like $1 \times 10^{-60}$ to a decimal form can indeed seem tricky, but you have it right. Here's the confirmation:

Steps to Convert $1 \times 10^{-60}$ to a Decimal

1. Understand the Notation: $10^{-60}$ means moving the decimal point 60 places to the left from 1.
2. Initial Position: Start with the number 1.
3. Move the Decimal Point: Move the decimal point 60 places to the left. This means adding 59 zeros before the 1.

Here’s the step-by-step visualization:

• Start with 1.
• Move the decimal point 60 places to the left:

$$0.000000000000000000000000000000000000000000000000000000000001$$

So, $1 \times 10^{-60}$ correctly converts to the decimal:

$$0.000000000000000000000000000000000000000000000000000000000001$$

You’ve got it exactly right. Numbers with such small magnitudes can indeed be daunting, but the approach you took is accurate.

Me: Can we change the assumptions in your first calculation. How does it affect the odds if the events are not independent of each other?

ChatGPT 4: 

If the events are not independent, the calculation of the combined probability changes significantly. In the independent case, we simply multiplied the probabilities of each event. However, for dependent events, we need to consider the conditional probabilities.

Understanding Dependent Events

When events are dependent, the occurrence of one event affects the probability of another event. To calculate the probability of all events occurring in such a case, you need to use conditional probabilities.

Conditional Probability Formula

For two events $A$ and $B$: $P(A \text{ and } B) = P(A) \times P(B|A)$

Where $P(B|A)$ is the probability of event $B$ occurring given that event $A$ has occurred.

Generalizing for Multiple Events

For multiple dependent events $E_1, E_2, \ldots, E_{10}$: $P(E_1 \text{ and } E_2 \text{ and } \ldots \text{ and } E_{10}) = P(E_1) \times P(E_2|E_1) \times P(E_3|E_1 \text{ and } E_2) \times \ldots \times P(E_{10}|E_1 \text{ and } E_2 \text{ and } \ldots \text{ and } E_9)$

Simplified Example

Let’s consider a simplified example where the probability of each event is still small but now dependent on the previous event.

Assume:

• $P(E_1) = 1 \times 10^{-6}$
• $P(E_2|E_1) = 1 \times 10^{-3}$ (since $E_2$ is much more likely if $E_1$ has occurred)
• $P(E_3|E_1 \text{ and } E_2) = 1 \times 10^{-2}$ (for further simplification, let's assume this pattern)
• and so on...

Calculating the Combined Probability

Let's calculate the combined probability for simplicity up to three events:

1. First Event: $P(E_1) = 1 \times 10^{-6}$

2. Second Event Given First: $P(E_2|E_1) = 1 \times 10^{-3}$

3. Third Event Given First and Second: $P(E_3|E_1 \text{ and } E_2) = 1 \times 10^{-2}$

Thus, the combined probability: $P(E_1 \text{ and } E_2 \text{ and } E_3) = P(E_1) \times P(E_2|E_1) \times P(E_3|E_1 \text{ and } E_2)$ $= (1 \times 10^{-6}) \times (1 \times 10^{-3}) \times (1 \times 10^{-2})$ $= 1 \times 10^{-11}$

Extending to 10 Events

If the pattern of conditional probabilities continues, you would multiply the probabilities as shown:

$P(E_1 \text{ and } E_2 \text{ and } \ldots \text{ and } E_{10}) = P(E_1) \times P(E_2|E_1) \times P(E_3|E_1 \text{ and } E_2) \times \ldots \times P(E_{10}|E_1 \text{ and } E_2 \text{ and } \ldots \text{ and } E_9)$

If we assume each subsequent event has a probability ten times more likely than the previous one, the calculations might look like:

$P(E_1) = 1 \times 10^{-6}$ $P(E_2|E_1) = 1 \times 10^{-3}$ $P(E_3|E_1 \text{ and } E_2) = 1 \times 10^{-2}$ $P(E_4|E_1 \text{ and } E_2 \text{ and } E_3) = 1 \times 10^{-1}$ $\ldots$ $P(E_{10}|E_1 \text{ and } \ldots \text{ and } E_9) = 0.1$

Then:

$P(\text{All 10 events}) = (1 \times 10^{-6}) \times (1 \times 10^{-3}) \times (1 \times 10^{-2}) \times (1 \times 10^{-1}) \times \ldots \times (0.1)$

The combined probability becomes increasingly complex, but with this pattern, it would be a product of increasingly larger conditional probabilities.

Conclusion

Dependent events significantly alter the combined probability. Each conditional probability typically is greater than the initial independent probability, resulting in a combined probability that is less astronomically low than if the events were independent.

The exact value depends on the specific conditional probabilities, but the general approach remains the same: multiply the probabilities, taking into account the dependence of each event on the prior events.

Me:  Could you continue please?

ChatGPT 4: Let's continue the calculation for all 10 events, assuming the pattern of conditional probabilities continues as described:

Extending to 10 Events

Given the pattern:

• $P(E_1) = 1 \times 10^{-6}$
• $P(E_2|E_1) = 1 \times 10^{-3}$
• $P(E_3|E_1 \text{ and } E_2) = 1 \times 10^{-2}$
• $P(E_4|E_1 \text{ and } E_2 \text{ and } E_3) = 1 \times 10^{-1}$
• $P(E_5|E_1 \text{ and } E_2 \text{ and } E_3 \text{ and } E_4) = 0.1$
• $P(E_6|E_1 \text{ and } E_2 \text{ and } E_3 \text{ and } E_4 \text{ and } E_5) = 0.1$
• $P(E_7|E_1 \text{ and } E_2 \text{ and } E_3 \text{ and } E_4 \text{ and } E_5 \text{ and } E_6) = 0.1$
• $P(E_8|E_1 \text{ and } E_2 \text{ and } E_3 \text{ and } E_4 \text{ and } E_5 \text{ and } E_6 \text{ and } E_7) = 0.1$
• $P(E_9|E_1 \text{ and } E_2 \text{ and } E_3 \text{ and } E_4 \text{ and } E_5 \text{ and } E_6 \text{ and } E_7 \text{ and } E_8) = 0.1$
• $P(E_{10}|E_1 \text{ and } E_2 \text{ and } E_3 \text{ and } E_4 \text{ and } E_5 \text{ and } E_6 \text{ and } E_7 \text{ and } E_8 \text{ and } E_9) = 0.1$

Calculating the Combined Probability

The combined probability is:

$$P(\text{All 10 events}) = P(E_1) \times P(E_2|E_1) \times P(E_3|E_1 \text{ and } E_2) \times \ldots \times P(E_{10}|E_1 \text{ and } E_2 \text{ and } \ldots \text{ and } E_9)$$

Substituting the values:

$$P(\text{All 10 events}) = (1 \times 10^{-6}) \times (1 \times 10^{-3}) \times (1 \times 10^{-2}) \times (1 \times 10^{-1}) \times (0.1) \times (0.1) \times (0.1) \times (0.1) \times (0.1) \times (0.1)$$

Simplifying the Multiplication

1. Calculate the product of the powers of 10:

   $$10^{-6} \times 10^{-3} \times 10^{-2} \times 10^{-1} = 10^{-6-3-2-1} = 10^{-12}$$

2. Multiply the decimal probabilities:

   $$0.1 \times 0.1 \times 0.1 \times 0.1 \times 0.1 \times 0.1 = (0.1)^6 = 10^{-6}$$

3. Combine the results:

   $$10^{-12} \times 10^{-6} = 10^{-18}$$

Result

The combined probability of all 10 dependent events occurring, given the specified pattern of conditional probabilities, is:

$$P(\text{All 10 events}) = 10^{-18}$$

Interpretation

The resulting probability $10^{-18}$ is still extremely small, but it is significantly larger than the probability if the events were independent (which was $10^{-60}$). This illustrates how dependence between events can increase the likelihood of all events occurring compared to the scenario where events are independent.

Me: (10^18 expressed as a decimal is 0.000000000000000001) 

Next I asked ChatGPT 4 how would it express that as a name in a manner similar to one in a trillion.

ChatGPT4: To express $10^{-18}$ as a name in a manner similar to "one in a trillion," we need to understand the scale.

Here is the breakdown:

• Trillion: $10^{12}$
• Quadrillion: $10^{15}$
• Quintillion: $10^{18}$

So, $10^{-18}$ is equivalent to one part in a quintillion.

Therefore, you can express $10^{-18}$ as "one in a quintillion."

Me:  OK. If there were hundreds of such prophecies, that would probably be even too much for your memory. 

ChatGPT4: When dealing with very large or very small numbers, it's often useful to express them in a way that's more manageable and easier to comprehend. Here are some ways to handle such numbers:

Scientific Notation

For extremely large or small numbers, scientific notation is useful. For example:

• $10^{-18}$ is expressed as $1 \times 10^{-18}$.

Orders of Magnitude

Using orders of magnitude, you can describe the scale without listing all the zeros:

• $10^{-18}$ is on the order of $10^{-18}$.

Verbal Descriptions

Using verbal descriptions to provide context:

• $10^{-18}$ is "one in a quintillion."

Practical Examples

Providing practical examples to illustrate the scale:

• The probability of $10^{-18}$ is like picking a specific grain of sand out of all the grains on all the beaches on Earth.

Memory Considerations

When numbers become too large or too small, they exceed typical human memory and comprehension capabilities. Computers and calculators are often used to handle such calculations efficiently.

If you have specific scenarios or calculations you'd like to handle, feel free to let me know, and I can assist further!

Me: Wow! Thanks for phrasing it like that. It really puts it into perspective! How would you express 1×10−60 x 1×10−60. What would that be like? 

ChatGPT 4:

To express $(1 \times 10^{-60}) \times (1 \times 10^{-60})$ and to put it into perspective, let's break it down step by step.

Mathematical Calculation

First, let's calculate it: $(1 \times 10^{-60}) \times (1 \times 10^{-60}) = 1 \times 10^{-60-60} = 1 \times 10^{-120}$

Decimal Representation

Next, to represent $1 \times 10^{-120}$ as a decimal, you would write a 1 with 120 zeros in front of it after the decimal point. This looks like: $0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001$

Verbal Description

To express $1 \times 10^{-120}$ in words:

• Orders of Magnitude: It's 120 orders of magnitude smaller than 1.
• Verbal Description: It is "one in a novemdecillion" (since a novemdecillion is $10^{60}$ and $10^{-120}$ would be one part in a novemdecillion squared).

Practical Example

To give a practical example to illustrate how small this number is:

• Atoms in the Observable Universe: The estimated number of atoms in the observable universe is around $10^{80}$. Therefore, $1 \times 10^{-120}$ is much smaller than the reciprocal of the number of atoms in the universe. It’s like imagining you had one specific atom in a universe multiplied by many orders of magnitude of more universes.

Putting It Into Perspective

1. Tiny Probability: The probability $1 \times 10^{-120}$ is so small that it is beyond practical experience. It is effectively zero for any physical or probabilistic event we can conceive of in reality.
2. Scale Comparison: If the universe were filled with grains of sand and each grain represented another universe, you would still be talking about a minuscule fraction of the total.

So, $(1 \times 10^{-60}) \times (1 \times 10^{-60})$ resulting in $1 \times 10^{-120}$ is an incredibly small number, far beyond what we typically encounter in everyday life or even in most scientific contexts.

Me: Amazing!  

Here is my video about this information, along with a small sample of the Bible prophecies that have been fulfilled to date. 

 

You can read the opening of "What Happens NEXT?" at Amazon with the Read Sample feature. here. Chapter one and two cover the prophecies in this video and many more with scripture references.