Friday Puzzler -- Fair Game

Re: Friday Puzzler -- Fair Game

Bill Earl

Not exactly.

Re: Friday Puzzler -- Fair Game

Larry Barrett

I'd also say not exactly, but maybe not for the same reasons as Bill.

Thinking back to Proability 101, the probability of getting k heads in n flips of a coin is

n!/(k!(n-k)!) x 1/2^n. The key thing here is that the denominator is 2^n; no matter the numerator, the result will not be 1/3.

That said, this analysis is too much like Probability 101.

So on second thought, the answer to Alex's question is most likely yes, but I do not know how.

Good answr

Alex Y

Unfortunately, while I didn't mention it, you need to give an example, not just a "yes" or "no".

Re: Friday Puzzler -- Fair Game

Alex Y

Understand where you are coming from, but there is such a game.

An out-of-the-box solution

Alex Y

When I saw this posted there were several good outside the box solutions. My favorite was "Take the coin to a novelty store and buy a fair die..."

There were a couple of others that were only subtly against the rules. But what we are really looking for is a game using the coin to determine outcomes of a 50-50 event.

Re: Friday Puzzler -- Fair Game

Bill Earl

Could be done with three players.

Then you have a solution, don't you?

Alex Y

Just let one person play as player one, and the other play as players 2 and 3. Would that work in the solution you have in mind?

Re: Then you have a solution, don't you?

Bill Earl

Yes. Players 2 & 3 could be 'straw' players. Player 1 could actually do all the flipping.

Question

Larry Barrett

How is having 3 players flip a coin once different than one person flipping a coin 3 times?

Re: Question

Alex Y

no difference at all. But if I guess correctly the game Bill has in mind, three flips of the coin is not a complete description of the game.

Re: Question

Bill Earl

That is correct. The 'game' consists of one or more 'plays'. Each play would consist of 3 flips: A, B and C.

The game is over when a play results in 1 head and 2 tails. If the head is in position A, then player A wins.

Re: Question

Larry Barrett

Ah, good thinking. A can also win if one Tail and 2 Heads come up and the Tail belongs to A. This would shorten the game.

Could also do this with one player and three coins, as long as the coins were marked with a red, blue, or white dot.

Re: Question

Alex Y

Or just three successive tosses of the same coin

Correct!

Alex Y

What you have described is a slightly longer version of the classic way three friends decide who picks up the tab for their morning coffee: Odd man pays.

The key here is that the game is of indeterminate length. It is natural to think, as you and Larry were initially, of a fixed length game, in which case the probability will always be something over a power of 2

In addition to this game with "rounds" of three flips, there is a version in which each round is two flips (two heads I win, one head and one tail you win, two tails do-over) and a rather ingenious one where each flip is a potential determinant of who wins. I'll leave that as a followup puzzle if anyone want to work on it. I'll add it here next week if no takers.

And a couple of more

Alex Y

1) toss the coin into a circular container with lines drawn from the center to the edge, 120 degrees apart.

2) spin the coin, and when it stops note the orientation of the top of the coin relative to compass points (0 to 120 I win, or if no compass is available, relative to some reference line determined before the spin.

Obviously both of these requires something other than the coin, so they won't work. But I appreciated the creative thought that went into them.

Fair Game -- an elegant solution

Alex Y

Toss the fair coin until you get a heads, counting how many tosses it takes.

There is a 1/3 chance that the first heads will occur on an even-numbered toss.

If you are a computer-type, start your counting with 1 like the rest of us do, not with 0