What do you think?

bobbym · 2011-08-30 20:14:28

New problem:

Difficulty: Incredibly hard if you do it my way.

E poses another one:
From the set, S = {2,3,5,7,11,13}, numbers are picked with replacement until a run of 6 different numbers appears for the first time. What is the average number of picks?

Here is an example:

2,2,3,5,7,13,2,5,2,2,3,7,5,11,11,3,2,3,3,7,7,2,5,2,5,5,13,13,11,13,2,5,3,7,11,13

Here the answer is 36, because the 36th one completes the first run of 6 distinct numbers.

A says) How boring, why does she keep coming up with such easy ones? The answer is obviously 36.
B says) Maybe because you never get them right. It is a little more than 83.
C says) 36 is correct by simulation.
D says) 216 is the obvious answer.
E says) You are all wrong!

Who is right?

gAr · 2011-08-31 04:36:07

Hi bobbym,

bobbym · 2011-08-31 04:51:46

Hi gAr;

gAr · 2011-08-31 05:07:00

Hi bobbym,

Thanks!
Did you pick the question from that?

bobbym · 2011-08-31 05:12:46

Hi gAr;

Yes, I was playing with it and wanted to see how much of the
method in the PDF I could remember. Took a couple of hours
before I could get his idea to work.

I had not looked at the markov chain idea so thanks for putting it
down. Thanks for looking at the problem.

gAr · 2011-08-31 05:50:28

Hi bobbym,

Okay.
Actually, I misunderstood it as a CCP and was about to answer that it's 14.7

What would be the probability that all different numbers have been picked consecutively after 83 guesses?
Is that P^83?

Last edited by gAr (2011-08-31 06:29:22)

bobbym · 2011-08-31 07:23:43

Hi gAr;

that all different numbers

How could we get 83 in a row different? Do you mean 83 with no
run of 6?

gAr · 2011-08-31 15:22:24

Hi bobbym,

I meant to ask what is the probability that there's atleast one run of 6 different numbers in 83 guesses.

bobbym · 2011-08-31 16:05:39

Hi gAr;

I think I can answer that. Will post it as soon as I get it.

I start with a simulation which indicates a probability of about .632 that there will be 1 or more runs of 6 distinct in 83 picks.

Looks like to get the probability you just raise your matrix up to the 83 power and check the first row, last column.

gAr · 2011-09-01 00:23:52

Hi bobbym,

Thanks for verifying!
But feels a little strange that we expect a run of 6 different numbers in 83.2 picks, whereas the probability is only about 63%.

bobbym · 2011-09-01 00:28:29

Hi gAr;

Yes, that always bothered me too. One would think it
would be close to 1.

gAr · 2011-09-01 00:51:27

Hi,

Yes, I too thought like that!

bobbym · 2011-09-01 00:55:28

At least as far as I can remember. The prob. has nver been less than .5

gAr · 2011-09-01 01:03:13

The prob. has nver been less than .5

In other problems?

bobbym · 2011-09-01 01:09:12

That is what I meant. Have you ever seen a case where the prob was less than 1 / 2
for the expected number?

gAr · 2011-09-01 01:16:25

Okay.
I haven't done many problems on expectation. I'll observe from now on.

bobbym · 2011-09-01 01:23:34

I have not done enough of them to answer that question. Now I think
that if we watch we will find some with a prob. less than .5

gAr · 2011-09-01 01:30:56

Okay.
I'll look at some more expectation problems.

bobbym · 2011-09-01 01:49:49

Hi gAr;

And good work with the matrix you came up with!

gAr · 2011-09-01 01:57:07

Hi bobbym,

I started with a wrong matrix yesterday. After wasting a couple of hours with that, I carefully wrote it down with greater detail instead of guessing. Finally I was glad that it was correct!

bobbym · 2011-09-01 02:11:34

Hi;

It worked real nice. Fit the simulation to almost 3 decimal
places.

gAr · 2011-09-01 02:24:42

I was trying a problem from chapter 4 in that file

3. If a die is rolled 100 times (say), what is the probability that all six sides have appeared at least one?
If a die is rolled 100 times (say), what is the probability that there is a run of 6 that has all six possible
faces? What about a run of 10 with all six faces at least once?

Isn't this the matrix for that

and one more:

4. A die is rolled repeatedly and summed. What is the expected number of rolls until the sum is:
(a) a prime? Experimentally it appears to be around 2.432211 (one million trials).
(b) a power of 2? Hmmmm...
(c) a multiple of n? The answer appears to be n.

For the part (c), I used an absorbing markov chain for a particular case of n=10.

First row is the initial state.
Other rows are {sum of the rolls} (mod 10)
The expected number of rolls for this is indeed 10.
Perhaps there is a recurrence relation for this pattern which we can generalize to 'n'?

Last edited by gAr (2011-09-01 05:33:23)

bobbym · 2011-09-01 08:04:27

Hi gAr;

I will start to work on them.

3. If a die is rolled 100 times (say), what is the probability that all six sides have appeared at least one?

This one has such a small probability that in 10 000 000 trials it did not happen once.

If a die is rolled 100 times (say), what is the probability that there is a run of 6 that has all six possible
faces? What about a run of 10 with all six faces at least once?

These two can be answered with the matrix in post #1303

gAr · 2011-09-02 19:05:59

Hi bobbym,

This one has such a small probability that in 10 000 000 trials it did not happen once.

Using the markov chain in my previous post,
the probability ≈ 0.999999927551959

These two can be answered with the matrix in post #1303

Oh, yes. I didn't see the word "run" there in the question.
So, question #3 was easy enough!

Did you try #4, (c)?

bobbym · 2011-09-02 19:13:10

Hi gAr;

Using the markov chain in my previous post,
the probability ≈ 0.999999927551959

That might be right! The probability is so small that you need a large simulation to
get any at all. We can sort of check in another way.

I have not looked at the next one yet because I am still stuck on 3a.

Math Is Fun Forum

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