The last banana: A thought experiment in probability - Leonardo Barichello

Prevodilac: Marija Kojić Lektor: Jelena Kovačević

You and a fellow castaway are stranded on a desert island

Vi i još jedan brodolomnik ste se nasukali na pusto ostrvo

playing dice for the last banana.

i bacate kockice za poslednju bananu.

You've agreed on these rules:

Dogovorili ste se oko ovih pravila:

You'll roll two dice,

Bacaćete dve kockice,

i ako je najveći broj jedan, dva, tri ili četiri,

and if the biggest number is one, two, three or four,

Prvi igrač pobeđuje.

player one wins.

If the biggest number is five or six, player two wins.

Ako je najveći broj pet ili šest, drugi igrač pobeđuje.

Let's try twice more.

Hajde još dvaput da pokušamo,

Ovde, prvi igrač pobeđuje,

Here, player one wins,

and here it's player two.

a ovde, drugi igrač.

So who do you want to be?

Dakle, ko vi želite da budete?

At first glance, it may seem like player one has the advantage

Na prvi pogled, možda se čini da prvi igrač ima prednost

budući da će ona pobediti ako je ijedan od četiri broja najveći,

since she'll win if any one of four numbers is the highest,

ali zapravo,

but actually,

player two has an approximately 56% chance of winning each match.

drugi igrač ima otprilike 56% šanse da pobedi u svakoj partiji.

One way to see that is to list all the possible combinations you could get

Jedan način da to uvidimo je da izlistamo sve moguće kombinacije

by rolling two dice,

bacanja dve kockice,

and then count up the ones that each player wins.

i onda da izbrojimo one koje donose pobedu svakom igraču.

These are the possibilities for the yellow die.

Ovo su mogućnosti za žutu kockicu.

These are the possibilities for the blue die.

Ovo su mogućnosti za plavu kockicu.

Each cell in the chart shows a possible combination when you roll both dice.

Svaka ćelija u tabeli pokazuje moguću kombinaciju

bacanja obe kockice.

Ako bacite četvorku, pa onda peticu,

If you roll a four and then a five,

we'll mark a player two victory in this cell.

označićemo drugom igraču pobedu u ćeliji.

A three and a one gives player one a victory here.

Trojka i jedinica donose prvom igraču ovde pobedu.

There are 36 possible combinations,

Postoji 36 mogućih kombinacija,

each with exactly the same chance of happening.

sve imaju podjednake šanse da se dese.

Matematičari ovo nazivaju jednako verovatnim događajima.

Mathematicians call these equiprobable events.

Now we can see why the first glance was wrong.

Sada možemo da vidimo zašto je prvi utisak bio pogrešan.

Even though player one has four winning numbers,

Iako prvi igrač ima četiri pobednička broja,

and player two only has two,

a drugi igrač ima samo dva,

the chance of each number being the greatest is not the same.

šanse da svaki od brojeva bude najveći nisu jednake.

There is only a one in 36 chance that one will be the highest number.

Postoji šansa samo 1 u 36 da će broj 1 biti najveći.

But there's an 11 in 36 chance that six will be the highest.

Međutim, šanse su 11 u 36 da će broj 6 biti najveći.

So if any of these combinations are rolled,

Dakle, ako se bilo koja od ovih kombinacija dogodi,

player one will win.

prvi igrač pobeđuje.

A ako se bilo koja od ovih kombinacija dogodi,

And if any of these combinations are rolled,

player two will win.

drugi igrač pobeđuje.

Out of the 36 possible combinations,

Od 36 mogućih kombinacija,

16 give the victory to player one, and 20 give player two the win.

16 obezbeđuje pobedu prvom igraču, 20 obezbeđuje pobedu drugom igraču.

You could think about it this way, too.

Možete to i ovako da posmatrate, takođe:

The only way player one can win

Jedini način da prvi igrač pobedi

is if both dice show a one, two, three or four.

je da obe kockice pokažu jedinicu, dvojku, trojku ili četvorku.

A five or six would mean a win for player two.

Petica ili šestica znače pobedu drugog igrača.

The chance of one die showing one, two, three or four is four out of six.

Šanse da će jedna kockica pokazati

jedinicu, dvojku ili četvorku je četiri u šest.

The result of each die roll is independent from the other.

Rezultat svakog bacanja kockice

je nezavisan od rezultata drugog bacanja.

And you can calculate the joint probability of independent events

Možete izračunati zajedničku verovatnoću

nezavisnih događaja

by multiplying their probabilities.

ako pomnožite njihove verovatnoće,

So the chance of getting a one, two, three or four on both dice

Dakle, šansa da dobijete jedinicu, dvojku, trojku ili četvorku na obe kockice

is 4/6 times 4/6, or 16/36.

je 4/6 puta 4/6, ili 16/36.

Because someone has to win,

Budući da neko mora da pobedi,

the chance of player two winning is 36/36 minus 16/36,

šanse da drugi igrač pobedi su 36/36 minus 16/36

or 20/36.

ili 20/36.

Those are the exact same probabilities we got by making our table.

To su potpuno iste verovatnoće

koje smo dobili kada smo pravili našu tabelu.

But this doesn't mean that player two will win,

Međutim, to ne znači da će drugi igrač pobediti,

niti da ćete, ako odigrate 36 partija kao drugi igrač, pobediti u njih 20.

or even that if you played 36 games as player two, you'd win 20 of them.

That's why events like dice rolling are called random.

Zato se događaji poput bacanja kockice nazivaju nasumičnim.

Even though you can calculate the theoretical probability

Iako možete da izračunate teorijsku verovatnoću

of each outcome,

svakog ishoda,

you might not get the expected results if you examine just a few events.

možda nećete dobiti očekivane rezultate ako ispitate samo par događaja.

But if you repeat those random events many, many, many times,

Međutim, ako ponovite te nasumične radnje mnogo, mnogo, mnogo puta,

the frequency of a specific outcome, like a player two win,

učestalost određenog ishoda, kao što je pobeda drugog igrača,

will approach its theoretical probability,

približiće se svojoj teoretskoj verovatnoći -

that value we got by writing down all the possibilities

vrednosti koju smo dobili, kada smo zapisivali sve mogućnosti

and counting up the ones for each outcome.

i izbrojali one koje određuju svaki ishod.

So, if you sat on that desert island playing dice forever,

Dakle, da sedite na pustom ostrvu bacajući kockice zauvek,

player two would eventually win 56% of the games,

drugi igrač bi na kraju pobedio u 56% svih partija,

and player one would win 44%.

a prvi igrač bi pobedio u 44%.

But by then, of course, the banana would be long gone.

Međutim, do tada, naravno, banane već odavno ne bi bilo.