I'm thinking of a number between 1 and 1000...

Miscellaneous and Personal Stuff I Must Share
41

318?

42

Well, Bill H did get it, but no one else has.

As an aside for an statisticians out there, what is the expected number of guesses for a randomly picked number?

43

the number Q?

how about seksan bir? (81 in Turkish)

44

A life. Which, after typing all that out, you obviously are in desperate need of. :wink:

45

138!

46

Tretiak:

Let the number be x

So there! Ha!

47

Now, where’d I put that bottle of Rolling Rock?

48

Is that all? Wow I can think of at least 998 integers between 1 and 1000. But then I love math.

49

The number, of course, is 243.

I think.

50

I’m going to assume it’s in binary, so it must be one of these:

1 10 11 100 101 110 111 1000

51

I’ll just go with my favorite number, 13.

52

333

53

I’m not sure if ‘the expected number of guesses’ has a precise definition. Let’s try for the nth guess, where the probability hits 50%. Here’s what I mean.

The chance of the first guess being wrong is 999/1000.
So the chance of the first guess being correct is 1 - (999/1000), or 0.1%.

The chance of the first two guesses being wrong is 1 - (999/1000 * 998 / 1000), or approx 99.7%.
So the chance of the second guess being correct is approx 0.3%.

The chance of the first three guesses being wrong is 1 - (999/1000 * 998 / 1000 * 997/1000), or approx 99.4%.
So the chance of the third guess being correct is approx 0.6%.

Keep that up, and you’ll reach some guess where the chance goes over 50%. I think that’s the best approximation of what you’re looking for.
(But I’m not doing the calculations!)

54

Assuming that we are talking about 1-1000 inclusive, if the person guessing is told, “higher” or “lower”, then the number can always be guessed with absolute certainty in 10 guesses. The probability doesn’t become greater than 50% until guess 9, at which point the probability is 75% that the number will have been guessed. If the guesser guesses perfectly, it will take 9 guesses about 1/2 the time, less than nine about 1/4 of the time, and 10 guesses about 1/4 of the time.

I could figure out the number of guesses for no hints, but don’t want to spend the time. It’s going to be in the hundreds.

55

Without the higher/lower hints, it is a lot more complex.

glee is on the right track, but has the numbers a little off.

The chance of the first guess being wrong is:

1-(1/1000)=99.9%

The chance of the first two guesses being wrong is:

(1-(1/1000))*(1-(1/999))=99.8%

The chance of the first three guesses being wrong is:

(1-(1/1000))(1-(1/999))(1-(1/998))=99.7%

The chance of the first four guesses being wroing is:

(1-(1/1000))(1-(1/999))(1-(1/998))*(1-(1/997))=99.6%

and so on. Keep this up until you get the cumulative odds at less than 50%, and you have the expected number of guesses. It’s going to take hundreds of guesses to reach 50% probability.

56

Arken wrote

Well, it took under two minutes using Excel. I’m desparate for life, but not that desparate.

57

Using this method, it’ll take 501 guesses to go over 50%.

(1-(1/1000))= 999/1000

(1-(1/999))= 998/999

(1-(1/998))= 997/998

Watch what happens when you multiply them:

999 * 998 * 997
1000 * 999 * 998

Cancelling on the top and the bottom gives you 997/1000.

When you have guessed 501 times, the probability will be 501/1000, slightly tipping the scales in your favor.

58

Ahem. I beg to differ.

59

Yes, thanks for the correction. I know why I was confused!
I was doing the ‘how many people in a room before there’s a 50% chance that 2 share a birthday?’ maths.
So my calculations would work for a different problem:
If a line of n people all pick a random number between 1 and 1000, how likely is it that any 2 of them get the same one?

60

OK, no one has gotten it yet. And my feeble mind has relaized that it may take a whiel without another hint.

Two of the digits in the number are the same.