# Simulating Coin Flips vs Probability of Coin Flips

$\begingroup$ “ Or can such a question only be solved using pretense methods ? ( e.g. program a calculator to simulate many such “ runs ” and answer the above questions through simulation ) ” – Yes most decidedly – this is what experimental probability is, and as your total of simulations $\to \infty$, you will arrive at your theoretical probabilities … which being …
You have already found that the specific probability of the permutation { H, T, H } $= 0.045$, with Pr ( H ) $= 0.95$ and Pr ( T ) $= 0.05$ .
This means every race to occur, there is a $0.045$ opportunity of this, demand, permutation occur, since your runs are independent of each other.

“ demand convention ” – not good a rule, there is an entire distribution set out to help you, The Binomial Distribution .
Lets consider, for a moment, of your test as a whole, rather of its parts ( flips ). So a particular test of yours has $0.045$ gamble of occuring. Since this is the event you want, this will be your probability of success ( Pr ( S ) $= 0.045$ ). This besides means your probability of not achieving any achiever or failure, Pr ( F ) $= 0.955$ .
The binomial distribution consists of three variables, $normality$, your number of trials, $p$, your probability of success ( we just talked about ), and $X$, number of successes out of your trials you want to achieve .
In your case $north = 100$, $phosphorus = 0.045$, $X = 3$, ( lets good say we want to have precisely $3$ of your permutation { H, T, H } occurring in a 100 trials ). so,

$Pr ( X = 3 ) = { n\choose { ten } } ( phosphorus ) ^x ( 1-p ) ^ { n – x } = { 100\choose { 3 } } ( 0.045 ) ^3 ( 0.955 ) ^ { 100 – 3 }$
This works for any successes until, $normality$, ( coz you can not have more successes than the number of trials possible ) .

I do n’t think there is a very maximum possible ( there is no limit to $nitrogen$ ). It might equitable be $\infty$. however, your second doubt is more feasible …

If you want an absolute minimum, then its $1$ run, but your probabilities are probable not that high.
If you know the binomial distribution, then it might equitable be $neptunium$, $100 * 0.045 = 4.5$ .
obviously this is not the case. We want the ask number of trials for at least a $X = 1$ to occur. We besides want to reach a maxmimum possible probability of $X = 1$, on $nitrogen$ trials ( hence the average ). This means we have to adjust the number of trials, $north$, such that $X = 1$ reaches a flower probability potential .
$Pr ( X_n = 1 ) = { n\choose { 1 } } ( 0.045 ) ^1 ( 0.955 ) ^ { n – 1 }$ is maximum. Let this be adequate to $f ( north )$. We can find the maximal by solving $\frac { d ( degree fahrenheit ( nitrogen ) ) } { dn } = 0$ for $n$, which yield, $normality = 21.7184$. however, since $n \in \mathbb { N }$, $north = 22$. This means doing $22$ runs will allow you to achieve a maximum of opportunity of atleast $1$ { H, T, H } occurring. Thats your “ average ” .