As a rule, I don’t intend to use this blog to cover standard theory in probability or Statistics. But for a number of reasons, which will become clear in future posts, I’d like you to think about the following problem.

In the game of Monopoly, when you’re in jail, one way of getting out is to throw equal numbers – a so-called double – on the two dice. Actually, in standard Monopoly rules, you’re only allowed to try this method 3 times, and are then required to pay your way out, unless you have a ‘Get out of Jail Free’ card. But suppose there weren’t any such limit to the number of throws you could take, and just kept taking further turns until you rolled a double. In that case, how many rolls, on average, would it take for you to roll a double and get out of jail?

It’s very easy to show that the probability of throwing a double on any one throw is 1/6: let the score on the first dice be S; then the two dice will have the same score if the second dice also shows S. And the probability of that, by symmetry of the dice, is 1/6.

So actually, since 1/6 is also the probability of throwing a 6 on a single dice, this problem has an identical solution to the following one:

If I roll a standard dice until I get a 6, how many rolls of the dice will I need on average?

This is a standard problem in probability and statistics, so anyone who’s studied statistics to a reasonable level will automatically know the answer. But if you don’t know the answer, use your intuition to guess what it might be. Either way, please send me your answer via this survey.

I’ll discuss the solution – and your guesses – in a future post. Like I say, I’ll also be making use of the result in a couple of different ways, also in future posts.

## Published by Stuart Coles

Smartodds quant team member since 2004
View all posts by Stuart Coles

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