Repeated experiments

Anders Christensen · science forum beginner Joined: 19 Sep 2005 Posts: 2

Hi NG,

An event happens with probability p (say p=0.3). What is the probability
that the event happens n times in a row (say n=5)?
My own suggestion would be that it happened with probability 0.3^{5}. But
since the experiment can be done over and over again the sum of
probabilities (0.3^{1}+0.3^{2}+0.3^{3}+... ) won't converge to 1.

What is the "expected" number of times the event will happen in a row if I
repeat the experiment?

I'd be glad if some one could tip me how to calculate the probabilities
when repeating a randomized experiment :)

../Anders

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Anders wrote;
"An event happens with probability p (say p=0.3). What is the
probability that the event happens n times in a row (say n=5)? My own
suggestion would be that it happened with probability 0.3^{5}. But since
the experiment can be done over and over again the sum of probabilities
(0.3^{1}+0.3^{2}+0.3^{3}+... ) won't converge to 1.
What is the "expected" number of times the event will happen in a row if
I repeat the experiment?
I'd be glad if some one could tip me how to calculate the probabilities
when repeating a randomized experiment Smile

"
__________________________________
Re;
You're right, the probability of the event occurring 5 times in a row,
if the events are independent of each other, is indeed 0.3^5 or 0.00243.
The expectation of this is simply 1/0.00243=412, or 1 occurrence of 5
times in a row every 412 sets of 5 occurrences; ie, 412 x 5=2060
individual events consisting of both P=0.3 and all contrary events of
P=0.7.
For the occurrence of two events in a row P=0.09. Thus the expectation
is that 1 out of every 11 pairs of events, results in two events in a
row. Therefore, the expectation in terms of the individual event is
that the event (P=0.3) will occur two times in a row, once in every 2 x
11=44 events of both P=0.3 and contrary events of P=0.7.

-Dan Akers

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Anders wrote:
"An event happens with probability p (say p=0.3). What is the
probability that the event happens n times in a row (say n=5)? My own
suggestion would be that it happened with probability 0.3^{5}. But since
the experiment can be done over and over again the sum of probabilities
(0.3^{1}+0.3^{2}+0.3^{3}+... ) won't converge to 1.
What is the "expected" number of times the event will happen in a row if
I repeat the experiment?
I'd be glad if some one could tip me how to calculate the probabilities
when repeating a randomized experiment Smile

"
___________________________________
Re;
Sorry that my previous response only dealt with the expectation of the
event occurring X times in row. As for the probability of the event
(P=0.3) occurring X times in row in a run of Y experiments, you must
consider the contrary event with P=0.7. In a set of 5 experiments, the
probability of 5 "events" in a row is 0.3^5=0.00243. Therefore, the
probability of NOT having 5 in a row events from 5 experiments is
1-0.00243=0.99757. Thus, in a run of 20 sets of 5 experiments (100
experiments in all), the probability for 5 in a row events of P=0.3 is
1-0.99757^20=0.047 or about 5%.

For 1000 sets of 5 experiments (5000 experiments in all), the
probability of 5 in row events with P=0.3 rises to 1-0.99757^1000=0.921
or 92.1%. Thus as the number of experiments increases, the probability
of 5 in a row occurring at least once approaches unity (1).

Please note that the above is only valid when each set of 5 experiments
is taken as an occurence unto itself with no overlap with the next set
of 5 allowed.

-Dan Akers

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Anders wrote;
"An event happens with probability p (say p=0.3). What is the
probability that the event happens n times in a row (say n=5)? My own
suggestion would be that it happened with probability 0.3^{5}. But since
the experiment can be done over and over again the sum of probabilities
(0.3^{1}+0.3^{2}+0.3^{3}+... ) won't converge to 1."
__________________________________
Re;
I'm still thinking about the probability of 'n' in a row if the sets of
5 experiments are allowed to overlap (occur "anywhere" within the N
experiments). My thinking is that for say, 100 experiments, there are
N-n+1 or 100-5+1=96 opportunities to "make" 5 in row. Therefore, the
probability of 5 in a row events of P=0.3, overlap allowed, in 100
experiments is 1-0.99757^96=0.208 or 20.8%. For 5000 total experiments
there are 5000-5+1=4996 opportunities to make 5 in a row with overlap
allowed. Thus, P=1-0.99757^4996=0.9999 or very nearly 100%.

-Dan Akers

Anders Christensen · science forum beginner Joined: 19 Sep 2005 Posts: 2

Thank you for your detailed replies:)
../Anders.

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