Rolling a die

You start rolling a die. What is the expected (average) number of rolls you would have to make before you get three consecutive 6s?

Bill Ryan

about 260

there are 216 possible 3-roll combinations. you would expect to (on average) get any specific order (such as three 6's) by set 108. the first two rolls don't give a combination, and every roll after gives one. Thus 110 rolls should be the average.

SPOILER

Let A = the expected number of further rolls, if the last roll was not a 6 (or it is the first roll); B = the expected number of further rolls, if the last roll was a 6, but the next-to-last roll was not (or it is the second roll and the first roll was a 6); C = the expected number of further rolls, if the last two rolls were 6s.

Then:

A = 5A/6 + B/6 + 1 B = 5A/6 + C/6 + 1 C = 5A/6 + 1.

Solving the equations gives A = 258, B = 252, C = 216, so the answer is 258. I confess I would have guessed that the answer was smaller.

: Bill Ryan

The answer is

6^3 + 6^2 + 6 = 258.

And the natural extension to strings of length k for an n-sided die is also true.

As you may know, there is a relatively simple expression for the expected waiting time to see an arbitary string from an arbitrary alphabet.

The concept of probablity fits well in theory , but practically it's results are questinable.

Probablity of getting a Head from a toss of a coin is 0.5 , but i once tried this using a program , where i observed the worst case as 0.18.

Whatever the probablity , an event can happen at any time(possibly next).

I still rememeber one scholar saying ' every incidence in this world is a mere coincidence' (ie. 1 in infinity)

Regards, Tiger.

Do you have some evidence to support this assertion?

And?

1) it's trivial to write a broken program which calculates or displays wrong results. 2) probability theory explains what will happen on average. It does not attempt to predict with vast accuracy the outcome of a short run.

This is not true. Certain events cannot happen. Predicting how likely an event is to occur is the proper province of probability theory, but predicting *whether* that event will occur is possible in only two cases: p = 0, and p = 1.

If he's speaking mathematically, he's wrong. If he's speaking philosophically, whether he's right or wrong is open to debate. If he's speaking practically, he's wrong.

A quick Monte Carlo program produced the following results for a mere 10 consecutive runs:

113 19 135 326 287 37 198 474 772 188

The range here is quite large: 19 through 772. And yet, if you find the arithmetic mean of these numbers, it comes to 254.9 (if I added them up right). So *already* (and much sooner than I'd have expected, to be honest) we see that the theoretical probability (as calculated by two respondents, presumably independently) matches the experimental probability startlingly closely (purists please note: I used fread to grab data from /dev/random, and divided by INT_MAX (as a double), then multiplied by 6 and added 1. No rand() calls in sight).

254.9 is quite astonishingly close, in fact, to the calculated answer of 258.

A naive understanding of probability and infinity will do you no

Richard,

The probability of you replying to this message was 0.999 recurring (not equal to 1), although there was an infinitely small chance that I will be wrong.

Michael

But 0.999 recurring *is* equal to 1.

This is easily proved. Since I recently proved, in this newsgroup, that the sum of the infinite series 1/2 + 1/4 + 1/8 + 1/16 + ... is 1, I hope you'll accept that the sum of the infinite series 9/10 + 9/100 + 9/1000 + 9/10000 + ... is also 1. If not, please take it up with sci.math.

Incidentally, events which are known to have happened have probability 1 of having occurred, irrespective of the original probability they had of

By which, of course, he actually means 'please read the sci.math FAQ'. Please, please don't take this up with sci.math.

Richard,

You may be very good at proving to non-mathematicians the error of their ways, but you aren't very good at spotting a troll when it comes along

ObPuzzle:

D R O I S A N B T R E P A O N L L

From the above set of letters, find the start and make a set of three sentences (punctuation excluded), by moving from letter to letter horizontally, vertically or diagonally.

How much better can this sentence be packed? Can it be made to look more like a dwarfish troglodyte?

Cheers,

Michael