Round The Island

[sea horse]

I learnt today of the Round The Island race.

Apparently before the participants start, a dice is rolled by the Oldest Inhabitant. If it's a One, then that means only one lap has to be run, i.e. it's the last lap. If any other number is thrown, then that many laps have to be run first, before the dice is thrown again.
And then, as before, if a One is thrown, it's means they then run the last lap, but any other number means the runners now have to do that number of laps, and then the dice is thrown yet again. And so on.

So I asked amongst the string of sea horses here, what is the likely number of laps to be run, based on probability. I received two different answers, 12 and 16, both with what sounded simple but plausible logic. But both can't be right.

Can anyone here more reliably answer this?

[Nurse Duckett]

Does anyone have any idea what this is about or how to do it?

Baffled Nursie

[dulcinea]

i guess probablities... get your mathematician's cap on..

[Nurse Duckett]

urchin said:

. I received two different answers, 12 and 16, both with what sounded simple but plausible logic. But both can't be right.

Can anyone here more reliably answer this?

Yes! 8 or 17.

Ta Daa.

[dulcinea]

... hmmmm is this turning into a sweepstake?

i pick 15

[Nurse Duckett]

8 or 17.

Or both.

[sea horse]

The answers so far have been 12, 16, 8, 17 and 15, ... (thanks dulcinea and Nurse D for being brave enough to have a go)

Then on Saturday, someone told me that the real mathematical answer was 12¼ laps!
Of course, this is nonsense because fractions goes against the whole idea of the dice rolling.

So I found a maths help website and asked them for help. This is their (very speedy) reply:


You require the expected total until the number 1 is thrown.

The expected number obtained on a die if we exclude 1 is

(1/5)[2 + 3 + 4 + 5 + 6] = 20/5 = 4

We can use a difference equation as shown below. The expected number, E, of laps is 1 with probability (1/6) or we return to (E+4) with probability 5/6

E = (1/6)(1) + (5/6).(E+4)

E = 1/6 + (5/6)E + 20/6

(1/6)E = 21/6

E = 21

So the expected number of laps to be run by the islanders is 21.


It seems we may all need a bit more stamina than we anticipated!

[Nurse Duckett]

Notwithstanding your elegant reasoning here, Sea Horse, I beg to disagree. You left one part out.

The correct answer is 8 if a woman and 17 if a man.

[Crusoe]

nurseduckett said:

The correct answer is 8 if a woman and 17 if a man.

I think I’m tempted to go along with your answer, Nurse D.

[Crusoe]

Sorry for re-opening this, Seahorse, but the more I think about it, the more convinced I become that the most likely number of laps is one (unless I’m misinterpreting the question, which is entirely possible).

A one will be thrown at the start of the race in a sixth of all instances. It seems to me that to get to any other number of laps requires an unlikely combinations of results: to run three laps requires a two to be rolled, followed by a one. This combination will only happen on one in thirty-six occasions and so is far less likely than one.

The probability of seven laps being run is relatively high (because there are a number of different combinations which will make seven laps) at nine in two hundred and sixteen but it’s still not a patch on one, which is four times more likely to occur.