Guessing Random Sequences

This is a neat applet that lets you see how good you
are at coming up with true random sequences. It's
almost impossible, even though to our eye they look
random when we're done with them.

Just enter a series of zeros and 1's in the boxes and
hit calculate. One integer to each box.

http://www.gametheory.net/mike/applets/Random/random.html

0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0

.9650, .6386, .6550

RANDOM!!!

http://www.gametheory.net/mike/applets/Random/random.html

The main mistake people make in trying to generate their own, "Random," numbers is that when people think, "Random," they automatically assume that the sample need be choppy, so they don't put any runs of more than two or three which would often naturally occur in a sample such as this one.

For example, the probability of four zeroes in a row is (.5)^4 = .0625

Given this probability, the probability of not having four zeroes in a row is .9375, however, we have seventeen total numbers (in a series of 20) that could potentially be the beginning of four consecutive zeroes. The probability of all seventeen of these numbers NOT starting a run of seventeen consecutive zeroes at any point is:

(.9375)^17 = 0.33381949729

Thus, in a sample size of twenty, you are more likely than not to have four zeroes occur at least once. It will happen at least once in about two out of three samples.

I got random on the first try and that was the
last time, the rest were not random.

Quote: Mission146

Thus, in a sample size of twenty,
you are more likely than not to have four zeroes occur at least once.
It will happen at least once in about two out of three samples.

not even.
your (.5)^4 = .0625 is just for the next 4 trials not a series of 20 trials.
I do not follow what you are calculating, but I do disagree with your statement
"in a sample size of twenty,
you are more likely than not to have four zeroes occur at least once"

0000 or at least 0000 (4 or more in a row - I use this one)

I say your math is just wrong for streak probability.
Time to learn how to do the math correctly.
(craps - if I can learn it, easy for the rest)

IF you are trying to figure out the expected number of steaks of length 4 in 20 trials
that is something different, (expected # of runs = 0.5625, an average and not a probability)

one should really learn about streak probability at Sally's page
http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

it links to this also here at WoV
https://wizardofvegas.com/forum/questions-and-answers/math/4855-ask-the-wizard-correction

But any one can believe what ever they want with streak probabilities.

Good Luck

No, Sally's number is definitely correct.

I'm not sure what was wrong with my approach, to be honest.

If the probability of a streak of (at least) four is .0625, then the probability of the next number NOT starting such a streak is .9375.

(.5) + (.5 * .5) + (.5 * .5 * .5) + (.5 * .5 * .5 * .5) = 0.9375

I want that .9375 to fail to happen seventeen consecutive times, so:

.9375^17 = 0.33381949729

Never mind, I see what's wrong now. It's because seventeen consecutive numbers do not have to fail to start the streak, you're already into the twenty if you go 0, 0, 0, 1...then the next number is the fifth overall number, so you don't always get seventeen chances. In fact, you'd almost never get seventeen chances in a sample of 20 decisions.

First try

1.0000 .9630 0.0088

Random!!!

You're not cool if you don't get a 1 on your first try. :P

Quote: Mission146

Never mind, I see what's wrong now. It's because seventeen consecutive numbers do not have to fail to start the streak, you're already into the twenty if you go 0, 0, 0, 1...then the next number is the fifth overall number, so you don't always get seventeen chances. In fact, you'd almost never get seventeen chances in a sample of 20 decisions.

You should consider that the seventeen possibilities are not independent, so you cannot simply multiply probabilities.

Quote: Mission146

No, Sally's number is definitely correct.

Never mind, I see what's wrong now.

how about the *first time* it occurs approach
many use this method as Sally did in her Excel sheet

for a run (at least 4) on the 4th trial = 0.5^4 = 0.0625 (p^run)
we check: 16 outcomes (2^4) and only 0000 is a success for 1/16
I think no one would disagree with this

for a run on the 5th trial but not the 4th = 0.5^5 = 0.03125 (q*p^run)
we sum the 4th and 5th trial prob to get 0.03125+0.0625=0.09375
=4thTrialProb+(q*p^run)

we check: now 32 possible outcomes (2^5) and only 00001,10000 or 00000 is a success for 3/32
the Wizard's method is the 7th (n+2) 4step Fibo# = number of sequences without a run of 4 or more
1,1,2,4,8,15,29 so 29/32
1- 29/32 = 3/32 another check

for a run on the 6th trial but not the 4th or 5th = this gets different as we can now consider it did not happen earlier
=5thTP+(1-3rdTP)*(q*p^run)
0.09375+(1*0.03125)=0.125
of course it could not happen on the 3rd trial here but we look back 4 trials including the 6th one (6,5,4,3)
and so on down the 7th trial to the 20th. keep adding
7thTP=6thTP+(1-4thTP)*(q*p^run)
8thTP=7thTP+(1-5thTP)*(q*p^run) etc

using the Wizard's method to check the 6th trial probability
8th (n+2) 4step Fibo#
29+15+8+4=56 (1,1,2,4,8,15,29,56)
2^6=64
1- 56/64 = 8/64
it does work

all this good stuff (?) for someone in the future to be a streak expert
for whatever reason

a chart of runs from length 2 to 10
prob of success = 50%