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

You and a fellow castaway are stranded on a desert island

playing dice for the last banana.

You've agreed on these rules:

You'll roll two dice,

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

player one wins.

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

Let's try twice more.

Here, player one wins,

and here it's player two.

So who do you want to be?

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

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

but actually,

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

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

by rolling two dice,

and then count up the ones that each player wins.

These are the possibilities for the yellow die.

These are the possibilities for the blue die.

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

If you roll a four and then a five,

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

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

There are 36 possible combinations,

each with exactly the same chance of happening.

Mathematicians call these equiprobable events.

Now we can see why the first glance was wrong.

Even though player one has four winning numbers,

and player two only has two,

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

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

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

So if any of these combinations are rolled,

player one will win.

And if any of these combinations are rolled,

player two will win.

Out of the 36 possible combinations,

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

You could think about it this way, too.

The only way player one can win

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

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

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

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

And you can calculate the joint probability of independent events

by multiplying their probabilities.

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

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

Because someone has to win,

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

or 20/36.

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

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

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.

Even though you can calculate the theoretical probability

of each outcome,

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

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

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

will approach its theoretical probability,

that value we got by writing down all the possibilities

and counting up the ones for each outcome.

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

player two would eventually win 56% of the games,

and player one would win 44%.

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