Two persons engage in a game of chance. The game is to nominate a sequence of three consecutive coin tosses [H or T].

Player one firstly nominates a sequence and then player two makes a nomination. The game finishes when the last three tosses match either one of the players' nominations.

How can player two be assured of winning most of the time?

Given the choices that can be made by player one, what are the odds of player two winning?

Oh, it doesn't matter who tosses the coin.

(In reply to solution by Charlie)

When I first saw Charlie's post I wondered at the table.
The second thing that I didn't understand was his closing two sentences.  2/3?  This was not in the scope of my source but I understand now that he puts into perspective the "global" odds, which may have been implied in the text, but that I had not been seeking.  

The first thing that baffled me was that I was expecting a simple rule for:
How can player two be assured of winning most of the time?

My second question was presented more as a bonus question to mean that: "For each event chosen by Player One, given 'winning strategy' of Player Two, what are the odds of a Player Two win?"

As I reviewed Charlie's reply I do indeed see a strategy for each event justified by an odds value.  I now realise that to get the exact outcome that I envisaged my wording was not good enough.  I wanted a simple rule for Player Two and would have like a simple ratio for each of the 'winning strategy' events.

While it wouldn't be hard to offer the ratios from Charlie's deriving the rule (based on the event pairings offered by that) might be just a mere bit challenging.

I cannot fault Charlie on his interpretations, I wouldn't have been able to derive odds akin to those that I sourced.

Posted by brianjn on 2013-06-14 05:15:52