A Drunk Hypothesis!

[DoomSlice]

8 deep, but it seems like a lot of my ideas come when I'm drunk.

My hypothesis is that, given any Bernoulli (spelling) trial, if the first trial is a success (X=1) then the expected value of the mean of the continuing series (finite) will be (ever so slightly) more than .5!

Prove/disprove.

[SheetWise]

The law of the conservation of luck.

[DoomSlice]

PPPPPPPPPPPPPPPPPPPPP-rove it.

Mean of any even chance series = .5

So the mean of any finite series with 50% chance of success = .5

So the mean of any finite series with 50% chance of success + 1 success >.5

Is that a good enough proof, or should I go more in depth?

[Siegmund]

Drunk hypotheses have a way of being somewhat ill-posed questions.

Given a series of N Bernoulli tries each with p=1/2,

E[(# succesful trials / # total trials) | 1st trial is a success] = (n+2)/(2n+2).

So, yes, if you pick a number of trials in advance, this is true. Similarly, if your poker room closes at a fixed time each day, winning your first hand does slightly increase the chance youll be ahead at closing time.

However, for any epsilon, there exists some K for which E < epsilon + 1/2 for all N>K.
Usually when a mathematician says something is true about the long-term behavior of a series, he means he can make the statement at the beginning, FIX epsilon, and show the statement to be *true* for all sufficiently large Ns.

Generally, results of the "I pick my constant and then let you pick yours" type are stronger results than results of the "pick any constant you like, and I can find..." type.

So, yes, your statement is true for finite sequences, but it's not exactly an earth-shattering result: in particular, it doesn't change the fact that lim (n+2)/(2n+2) as n->infinity is (exactly) 1/2, and it doesn't change your strategy playing a game based on Bernoulli trials, since knowing the outcome of the first trial doesn't help you decide whether calling heads or tails on the next trial is better.

[BruceZ]

[ QUOTE ]
Given a series of N Bernoulli tries each with p=1/2,

E[(# succesful trials / # total trials) | 1st trial is a success] = (n+2)/(2n+2).

[/ QUOTE ]

E = [1 + (N-1)/2]/N = (N+1)/(2N).

So your n must be the N-1 trials which exclude the first trial, n = N-1.

[pzhon]

I don't know why people are trying to help you with proofs. This is simply a tired old gambling fallacy.

If a red comes up on a roulette wheel, this neither means reds are hot, nor that blacks are due. Make your own proof.

[Cyrus]

If the Roulette wheel is going to be spun k times without a zero coming up, then it is expected that red will come up k/2 times and so will black. However, if the first trial comes up red, then the "expectation" has tilted slightly towards red, since the rest of the trials, i.e. k-1 trials, are still expected to come up evenly split between red and black. But, after those k trials, red is "expected" to have come up [(k-1)/2]+1 times.

Of course, that "expectation" includes trials already performed! The truth is that for the k-1 remaining trials, the result of the 1st trial means nothing.

As pzhon wrote
[ QUOTE ]
If a red comes up on a roulette wheel, this neither means reds are hot, nor that blacks are due.

[/ QUOTE ]

[DoomSlice]

Sober now, and I'm thinking about it a bit more. Let's say you play a game with your friend where you toss a coin and win $1 every time Heads comes up.

Then let's suppose you get extremely lucky and win 20 times in a row.

My (revised) hypothesis is that if you play this game every day until the day you die, there will be a non-zero probability that the mean will be centered around 20, and NOT zero. What I guess I'm saying is that things do not "regress" back to the mean, since you are as likely to go on another 20 game hot streak than you are to go on a 20 game loss (even over time).

I'm not sure about that, suppose you ignored the 20 that happened and stick it at the end of the sequence instead.