Channeling my inner Bendable here...

Lets say you have a 6 digit number.

How does one calculate the probability that any number will repeat in that 6 digit number, ie "1233.45" vs. "1234.56"

Is it as simple as 1/10 chance? Every digit has an equal chance of repeating.

Similar to the pick 4 lottery. The number 8 has the same chance every time the set of numbers is drawn.

42

The answer to the Ultimate Question of Life, the Universe, and Everything.

Each number has a 1 in 10 probability so for two in a row it's 1 in 100.

quote:

Originally posted by tatortodd:
Each number has a 1 in 10 probability so for two in a row it's 1 in 100.

Yep

Depends on how well you factor decimalization and define each term.

If the same number is factored in as a placeholder, it can't.

If the same digit is factored, it would be one in 10.

The odds of 100 would only apply if the prespecified number was factored

The definition of digit and number is not interchangeable.

I have a special set of 6-sided dice that always come up 3,4.

quote:

Originally posted by .38supersig:
The odds of 100 would only apply if the prespecified number was factored

Yes.
Once a number has been selected by whatever means, the probability of the next number being the same is 1/10—just as it would be if no number had been selected before.

I would look at it this way

The prob that the second number is not the same as the first is .9

prob third is not same as second is .9

prob fourth is not same as third is .9

prob fifth is not same as fourth is .9

prob sixth is not same as fifth is .9

So the prob that no two consecutive numbers of the group of 6 are the same is .9^5 = .59

then the prob that there is at least one repeating pair is .41

Better yet, what are the odds of 2 people having the same birthday in a random group of 23 people?

Of 30 people?

Of 70 people?

Disregard Leap Year, for simplicity use 365 days per year.

(Same birthday defined as Month & Day, not Month Day and Year)

No Googling!

This is fun, a great conversation starter, and wins bets. IIRC Amarillo Slim won a lot of bets on this.
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Don't forget Benford's law:

"Benford’s Law, also known as the Law of First Digits or the Phenomenon of Significant Digits, is the finding that the first digits (or numerals to be exact) of the numbers found in series of records of the most varied sources do not display a uniform distribution, but rather are arranged in such a way that the digit “1” is the most frequent, followed by “2”, “3”, and so in a successively decreasing manner down to “9”."

The odds are fifty-fifty. Either it will happen, or it will not happen. One or the other.

So then, If there are apples on a tree and none on the ground.

A strong gust of wind blows, then there are no apples on the tree and there are no apples on the ground.

Nothing has eaten or ran off with any apples. How is this possible?

I'm no math or probability expert, but is the answer 1 in 6 factorial (1 in 720) ?

It's Yatzy.

quote:

then the prob that there is at least one repeating pair is .41

You can easily prove it.

Go to Excel and use RAND to generate a slew of random numbers.

I generated 64 sets of 6 random digits (takes less than a minute)


38/64 = .594 had no repeating pair

26/64 = .406 had at least one repeating pair

quote:

Originally posted by .38supersig:
So then, If there are apples on a tree and none on the ground.

A strong gust of wind blows, then there are no apples on the tree and there are no apples on the ground.

Nothing has eaten or ran off with any apples. How is this possible?

Easy -- this is a riddle in the singular and plural!

It's a Bayesian analysis problem, coming down to a yes/no probability of the next number, like saying yes/no for what the next digit will be. Each number would be a 10% probability, with a 90% probability that it won't be that number. Similar to determining the odds on coin flips or winning the lottery, maybe a bit more difficult with ten digits involved. The CIA uses this methodology for predictive analysis, except they get to do it on a Cray super computer instead of a legal pad.