Math Puzzle: YACTG (Yet Another Coin Tossing Game)
Level of Difficulty: Highschool
As my job interview nears, the deluge of probability puzzles continues! Another day, another coin tossing game.
Alice, Bob and Cindy play a game by tossing a coin. Alice goes first, followed by Bob, then Cindy, then it’s back to Alice and so forth. Each player in his or her turn tosses the coin. If it lands on the same side as it landed in the previous turn, that player wins. For example: Alice tosses the coin and gets tails (T). Bob tosses the coin and gets heads (H). Cindy tosses the coin and gets tails. Alice tosses the coin and gets tails - Alice wins.
What is the probability that Bob wins?
Hint: it’s not 1/3 (can you see why?).
EDITED, 5/June/2009: a solution has been posted here. Our faithful reader jbrydle has posted a solution using a different, more direct approach in the comments section - check out both to get the different views!
If I understand the rules correctly, it has to be at least 1/2. The first go around, Alice has no chance of winning. Then, whatever she flips, Bob has a 50% chance of repeating it, so there’s his 1/2 probability of winning. If he doesn’t win, then Cindy has 1/2 odds of winning on her turn (1/2*1/2, or 1/4 chance from the outset), and on it goes. Alice gets (1/2)^3 chance of winning on the next turn, then Bob gets (1/2)^4, etc.
The infinite series would be Sum[n=0-infinity](1/2)^(3n+1)
There was a time when I could calculate the series, but now I’ll have to look it up.
Yes, it is indeed greater than 1/2.
Unsummed infinite series are unacceptable :). However, if you would’ve summed it you would’ve gotten the correct answer, which is 4/7. But you don’t need to sum up an infinite series - as with all good puzzles, this one has a shortcut!
The series is, by the way, a geometric series, which are well understood and can be summed up and expressed in closed form.
Actually, I should add that your solution generalizes nicely to some cases whereas mine (which will be posted shortly) generalizes to others, so they’re both interesting and I definitely wouldn’t say your infinite series is inelegant (being geometric, and hence very manageable).
I concur with jbrydle equation. The series
\sum_{n=0}^{-infinity} {(1/2)^(3n+1)}
can be transformed into
1/2 * \sum_{n=0}^{-infinity} {(1/2^3)^n}
where the summation part is a geometric series, with sum = \frac{1}{1-r}, with r=1/2^3 = 1/8.
Therefore, the solution is
1/2 * 1/(1-1/8) = 1/2 * 1/(7/8) = 8/14.
oops. too late
jbrydle,
So, I’ve posted a solution, but I’m curious as to how you’d approach the following problem: you toss a coin repeatedly until either three heads come up in a row, in which case you win, or two tails come up in a row, in which case you lose. What is the probability of winning? That is, what’s the probability of getting 3 heads in a row before 2 tails show up?
I can solve this one using the same recursive approach used in the solution, but I was wondering whether you think you could extend your direct approach to this case as well - I’ve got no idea how to do that, as things get messy very quick.
I’ve put this new question in its individual post - you can access it here:
http://www.physicallyincorrect.com/2009/06/math-puzzle-a-heads-and-tails-race/