Asking a girl on a second date.

I would say the same time.

Same time what????

the answer is greater I think based on that goat thing from earlier.

This thread is now about probability problems.

Here's the next one:

Let's say you repeatedly flip a coin until you see the sequence "heads, tails, heads." When you see this sequence, you stop, write down the total number of flips you made, and start again. You do this millions of times, and calculate the average number of flips it takes until you first see this sequence.

Now let's say you repeat this experiment except this time you look for the sequence "heads, tails, tails." You calculate the average number of flips it takes until you first see this sequence.

The question is, would you expect the average number of flips for "heads, tails, heads" to be less than, greater than, or the same as the average number of flips for "heads, tails, tails?"
Since the first two events of the sequence are both 'heads, tails' we can ignore them.

After each one there's a 50% chance of sequence A or sequence B occuring.

So, given a large enough sample size they'll be the same.

I thought it was the same the first time I heard the problem, too. But it isn't.

I actually wrote a quick Java program to simulate this and I can assure you that the average number of flips is not the same.

So with that answer eliminated, try to reason out the correct one.

Same time what????

the answer is greater I think based on that goat thing from earlier.

The same amount of flips.

I actually wrote a quick Java program to simulate this and I can assure you that the average number of flips is not the same.

So with that answer eliminated, try to reason out the correct one.

Java is not accurate with math. And that also assumes you can write an accurate program in java.

Java is not accurate with math. And that also assumes you can write an accurate program in java.

...

mormons.

Given that the first two events of the sequence are the same, I don't understand how it could not be 50%...

greater

am i right or wrong

Given that the first two events of the sequence are the same, I don't understand how it could not be 50%...

It is another problem where it is simple to understand the problem itself, but the answer is counterintuitive.

Don't think about the probability of just flipping any desired three result sequence with three coin tosses. Think about the longer sequence.

am i right or wrong

Just google it if you want to know for sure.

its because the sequence can more easily repeat with hth then with htt

the last h can start the next sequence in hth

It is another problem where it is simple to understand the problem itself, but the answer is counterintuitive.

Don't think about the probability of just flipping any desired three result sequence with three coin tosses. Think about the longer sequence.
<?
$seqA = 0;
$seqB = 0;

while ($seqA < 100000 && $seqB < 100000) {
$i = rand(0,1);
$j = rand(0,1);
$k = rand(0,1);

if ($i == 0 && $j == 1 && $k == 0) {
$seqA++;
} elseif ($i == 0 && $j == 1 && $k == 1) {
$seqB++;
}
}

echo "Sequence A results:".$seqA;
echo "\n";
echo "Sequence B results:".$seqB;
echo "\n";
?>
kate > php ./probability.php
Sequence A results:99643
Sequence B results:100000
kate > php ./probability.php
Sequence A results:99719
Sequence B results:100000
kate > php ./probability.php
Sequence A results:100000
Sequence B results:99739
kate > php ./probability.php
Sequence A results:99730
Sequence B results:100000

I am right.

actually im right

Just google it if you want to know for sure.

or why dont u tell us

<?
$seqA = 0;
$seqB = 0;

while ($seqA < 100000 && $seqB < 100000) {
$i = rand(0,1);
$j = rand(0,1);
$k = rand(0,1);

if ($i == 0 && $j == 1 && $k == 0) {
$seqA++;
} elseif ($i == 0 && $j == 1 && $k == 1) {
$seqB++;
}
}

echo "Sequence A results:".$seqA;
echo "\n";
echo "Sequence B results:".$seqB;
echo "\n";
?>
kate > php ./probability.php
Sequence A results:99643
Sequence B results:100000
kate > php ./probability.php
Sequence A results:99719
Sequence B results:100000
kate > php ./probability.php
Sequence A results:100000
Sequence B results:99739
kate > php ./probability.php
Sequence A results:99730
Sequence B results:100000

I am right.

That isn't the problem.

its because the sequence can more easily repeat with hth then with htt

the last h can start the next sequence in hth

I thought it was the same the first time I heard the problem, too. But it isn't.

I actually wrote a quick Java program to simulate this and I can assure you that the average number of flips is not the same.

So with that answer eliminated, try to reason out the correct one.

oh ok i get it. heads tails tails comes up sooner.

heads tails tails, if you fuck up either of the two last flips you are one flip already into a new sequence (you got heads already)

vs heads tails heads, if you fuck up the last flip you have to start over

thats what i was trying to say but i said it backwards
you are right

oh ok i get it. heads tails tails comes up sooner.

heads tails tails, if you fuck up the last flip you are one flip already into a new sequence.

vs heads tails heads, if you fuck up the last flip you have to start over

This is the reasoning. HTT comes up after 8 flips on average and HTH coes up after 10 flips on average.

Scurvy With Fries didn't give an answer and his reasoning doesn't apply to THIS problem. If we were asking "Given a really long sequence, how many instances of each pattern would we expect to see?" the answer there is the same number of times, and despite HTH coming up later on average it tends to occur in bunches to even things out.

Here is a math problem.

Find a pythagorean triple whose area is 666,666.