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A party full of tossers

There are 1100 people at a party.
You make a bet with your friend that someone in the party is capable of calling correctly 10 consecutive coin tosses .
Your friend bets against this, confident of winning.
Unfortunately for him you know how to win the bet every time. How?

The answer will follow soon.

James Vandeleur
Thursday, February 14, 2002

Why does that headline read like a dirty magazine to me? :-)

This is a good puzzle too... I should probably link it up to all the other ones that are solved with similar thinking (not to ruin the answer).

Michael H. Pryor
Thursday, February 14, 2002


Hmm.  I suppose I could see how to win the bet most of the time, just not all of the time.

I suspect the bet might be run by gathering everyone up, and split up into two groups, those who call heads on the first toss, and those who call tails.  Toss the coin.  Everyone on the losing side is dismissed, and the same thing is repeated for the winning group, and so on.  After ten tosses, someone -might- be left over, who would have called every toss correctly.

This involves a bit of mob mentality, though.  Suppose in the first test, everyone decided to call heads.  Then everyone would be in the same group.  Chances are this won't happen, and furthermore, as everyone is milling around, everyone notices there are no tail-callers, and so a few people change their minds and call tails instead.  On every test, one might suppose this happens, so that every test has callers on both sides.  2 to the tenth power is 1024, which is less than 1100, so theoretically there should be enough people to make this pretty likely.

However, there might still be that one toss, say the eighth, where there are ten people left, and only one decides to call tails for whatever reason, and then the eighth toss is tails.  Now you have one person left and two tosses to go.  There's now a 75% chance you'll lose the bet.  So I'm not entirely sure this is the winning strategy.

I may muse on this some more today.

Paul Brinkley
Thursday, February 14, 2002

Paul, you've got to be a bit more authoritarian when you are fixing a bet. I mean, if you were fixing the World Series, you wouldn't simply let the teams play fairly would you? You have to tell them what to do otherwise the wrong team might win!

So at the party, if you have lots of time on your hands, you could go around to the first 1024 party-goers you can find and assign him or her a specific sequence to guess (eg: HHHHHHHHHH, HHHHHHHHHT, HHHHHHHHTH, HHHHHHHHTT, etc.) and you've covered all your bases (coin-wise).

But if you are short on time and your generous betting partner permits it, you can simply divide up the party goers before the first toss (you 512 people guess H, you 512 guess T, you 74 discuss amongst yourselves, the two of us will start exchanging bank account information), and after each result, redivide the correct guessers (you 256 guess H, you 256 guess T), etc. As Paul pointed out, even after 10 rounds, you will still have at least (exactly) one correct guesser since 2^10 = 1024.

Cheng Leong
Thursday, February 14, 2002

Well done Cheng.
That's a nice solution.
Have you got puzzle for me?

James Vandeleur
Thursday, February 14, 2002

this puzzle seemed simpler to me than the standard on this site.

shailesh kumar
Friday, February 15, 2002

I had thought of assigning an HT sequence to every party member before making the bet, and dismissed it as being the wrong solution.  It implies a LOT of preparation beforehand.  The problem is couched in a way that implies you just happen to notice there are over 1024 people at the party, not counting you and your bettor, and you suddenly decide to try this bet.

I realize I'm reading information into this problem that isn't stated explicitly.  However, that is virtually always a requirement in any word puzzle.  You have to sort of get used to the "expected context" of such puzzles.  In this case, I assume I don't get to talk to the partygoers before the bet.  If I could, the solution would be trivial - too trivial.

Paul Brinkley
Friday, February 15, 2002

You don't need to talk to any of the party goers before the bet. You could just bring along 512 coins , divide the party goers up into 512 groups of two and treat it like a tennis tournament. i.e the winners of the first round of coin tosses advances to the next round of 256 matches and so on . The winner of the final is the person who called correctly 10 times in a row. 

James Vandeleur
Thursday, February 21, 2002

I've heard this strategy used  by con men.  You
get yourself a list of 1024 people known to bet
heavily on sporting events.  Send everone a letter
explaining that you have a new "foolproof" method
of predicting the result of football games.  In half
the letters, you "predict" that team A will win in the
upcoming big game; and of course, the other half
get a letter predicting success for team B.  Continue
through the season, sending letters only to those
who have only witnessed your succesful predictions.
After the first eight or so games, the handful of folks
who are still getting your letters begin to think that
your method actually works.  Then you offer to sell
them the secret for some large amount of money.
(This is illegal, obviously.)

Jeff Hultquist
Saturday, March 16, 2002

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