Forgive me, but high school math was a long time ago: how would one determine the odds of randomly selecting a telephone number (10 digits) from out of the air (so, no forces)? Sort of like the Human Phone Number trick (which is no longer being made) but with an explanation of how hard it would truly be.

If I recall the lottery (6/49) math correctly, it is 49x48x47...divide by 6x5x4...yes? So, how to apply that to a telephone number (if it is correct)?

Lee

Phone numbers are a little different from the lottery because the numbers in a phone number may repeat, and they have a distinct order.

If you assume that any digit can be used for any place in the phone number, and that digits may repeat, then for each place you have 10 choices, so the number of 10-digit phone numbers you're looking at is 10x10x10x10...x 10= 10^10 = 10,000,000,000 possible phone numbers. The probability of randomly matching such a number would be 1 out of 10,000,000,000.

In actual practice, the first number is somewhat restricted--I don't think (though I may be wrong now) any area code can begin with a zero or one. The same for the first number of the seven digit local number. There are also certain restricted exchanges such as 555, but this doesn't affect the total that much. If we eliminate the zero or one in the first and fourth places then we have 8x 10 x 10 x 8 x 10 . . . x 10 = 64x10^8 or 6,400,000,000 possible phone numbers.

Which is about the same as the number of people on the planet.

Mike

I guess those new born babies are just going to have to use email.

Fabulous! Thank you so much!

Lee

Hey Lee. Sorry I couldn't make it out on Saturday.

For the Lotto 6/49, I have used the following. If you started in downtown Edmonton and laid loonies edge to edge all the way down Highway 2 to Downtown Calgary, it would take the same number of coins as there are possible Lotto 6/49 tickets. Now if one coin were Tails up, with all the others Heads up, the Tail would be the winning ticket. So here's the game. You pay me to walk down Highway 2 blindfolded until you feel like picking up a coin. Get the tail and you win.

As for the phone numbers, Landmark is correct.

John

John - can I use my own blindfold?

Grinning,

Lee

This is similar to a question I have, I must be bad at maths as I still cant answer it, so could anyone tell me the odds on this..

free choice of 1 of 52 cards, free choice of 1 of 5 evelopes containing a different prediction, what are the odds of a match?
sorry if this is classed as hijacking, its not my intention -just pretty related.

thank you in advance.

1 in 52 - the number of envelopes is irrelevant.

Mike

Thanks Mike, not the answer I expected though, disappointing really.

Catweazle -

Then start your own thread and ask the question. Mike is correct.

Lee

Funny thing about apparent odds vs. true odds: When I used to do the invisible deck presentation where 2 people would each shuffle a deck, then deal out cards face-up until there was a matching pair (which was revealed to match the card facing the other way in the invisible deck), I would often mention the odds.

Whenever I said the odds were 1 in 52 of a match (the true odds), people would inevitably correct me and mention that odds should be far greater, like 1 in 2500 (50 times 50 - an estimate often used when they couldn't square 52 in their head). Whenever I claimed the odds of a match were 1 in over 2,700 (1 in 2,704 - the wrong odds), I'd only get corrected on this by people who knew and understood probability, which didn't happen that often.

In that case, I unfortunately had to lean towards the wrong odds for a presentation with a smoother flow.

I remember one presentation where I specifically KNEW there was a statistics teacher in my audience, and I still remember the patter line:

[speaking to everyone at the table]

"...and the odds of any card matching are over 1 in 2,700!"

[leans toward statistics professor and whispers without missing a beat]

"Yes, I know. Shhh..."

(He smiled.)

Prepared for your audience? A true magician!

Grinning,

Lee