Assuming that is the case, my question is this: how the hell execute I convince someone that is the case?
They are smart and educated yet seem determined not to think about that I could be in the appropriate on this (argument).
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they are trying come assert the <...> if there have been 10 heads, climate the following in the succession will much more likely be a tail because statistics states it will counter in the end
There"s only a "balancing out" in a very certain sense.
If it"s a fair coin, then it"s quiet 50-50 in ~ every toss. The coin cannot recognize its past. The cannot understand there was an overabundance of heads. It cannot compensate for its past. Ever. It simply goes on randomly being heads or tails with continuous chance the a head.
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If $n_H$ is the number of heads in $n=n_H+n_T$ tosses ($n_T$ is the variety of tails), for a fair coin, $n_H/n_T$ will have tendency to 1, as $n_H+n_T$ goes to infinity .... However $|n_H-n_T|$ doesn"t walk to 0. In fact, it also goes come infinity!
That is, nothing plot to do them an ext even. The counts don"t have tendency toward "balancing out". On average, imbalance between the count of heads and also tails in reality grows!
Here"s the an outcome of 100 set of 1000 tosses, v the grey traces mirroring the distinction in number of head minus variety of tails at every step.
The grey traces (representing $n_H-n_T$) room a Bernoulli random walk. If friend think of a particle moving up or under the y-axis through a unit action (randomly with equal probability) at every time-step, then the circulation of the position of the fragment will "diffuse" away from 0 end time. That still has 0 supposed value, yet its meant distance native 0 grows together the square source of the variety of time steps.
The blue curve above is at $\pm \sqrtn$ and also the environment-friendly curve is at $\pm 2\sqrtn$. As you see, the typical distance between total heads and also total tails grows. If there was anything exhilaration to "restore to equality" - to "make increase for" deviations indigenous equality - castle wouldn"t have tendency to frequently grow further apart favor that. (It"s not hard to present this algebraically, yet I doubt that would certainly convince your friend. The an important part is that the variance the a sum of independent random variables is the amount of the variances $$ -- every time you include another coin flip, you include a constant amount top top the variance of the sum... For this reason variance must thrive proportionally v $n$. In turn the conventional deviation boosts with $\sqrtn$. The continuous that gets added to variance in ~ each action in this situation happens to be 1, yet that"s not critical to the argument.)
Equivalently, $\fracn_H-n_Tn_H+n_T$ does go to $0$ together the total tosses goes to infinity, however only because $n_H+n_T$ goes come infinity a lot quicker than $|n_H-n_T|$ does.
That method if we divide that cumulative count by $n$ at every step, it curves in -- the usual absolute distinction in counting is of the order of $\sqrtn$, however the typical absolute distinction in proportion need to then it is in of the bespeak of $1/\sqrtn$.
That"s every that"s going on. The increasingly-large* arbitrarily deviations indigenous equality are simply "washed out" by the even bigger denominator.
* raising in common absolute size
See the little animation in the margin, here
If your friend is unconvinced, toss some coins. Every time you gain say 3 heads in a row, obtain him or she to nominate a probability because that a head on the next toss (that"s less than 50%) the he thinks should be fair by his reasoning. Ask for them to give you the matching odds (that is, he or she need to be ready to salary a bit much more than 1:1 if you gambling on heads, because they insist that tails is an ext likely). It"s finest if it"s set up as a many bets each for a small amount of money. (Don"t it is in surprised if there"s some excuse as to why they can"t take it up their fifty percent of the gambling -- yet it does at least seem to dramatically reduce the vehemence with which the position is held.)