Courtesy LispClub.com:
Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6, the game is over. Sue rolls first. If she doesn’t roll a 6, Bob rolls the die; if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.
Bob rolls a 6 before Sue.
What is the probability Bob rolled the 6 on his second turn?
The answer is not 5/36.
I love puzzles which are simple to state but have a fiendishly tricky or counterintuitive answer. I just threw up a page on the xkcd IRC wiki to hold some of the better ones I’ve found over the years. I’ll be adding more over the next few weeks as I remember or find good ones. Feel free to add some of your own!
Edit: Buttons and then Daniel Barkalow got the correct answer first. Here it is, rot13‘d. Check your answer against this before posting smugly or people (I) will tease you: gjb friragl svir bire gjryir avargl fvk, be nobhg gjragl-bar cbvag bar creprag.
My first guess was 5/36
Now I guess I’ll have to simulate it.
A: Sue wins
B: Bob wins
C: Bob wins in his second turn
D: Sue does NOT win in her first turn
We want to find the probability that Bob wins in his second turn, given that he wins in first place:
P(C|B) = P(C^B) / P(B)
Since B follows logically from C:
P(C|B) = P(C) / P(B)
The probability that the winner is determined in the fourth die toss (Bob’s second) can be easily calculated:
P(C) = (5/6) * (5/6) * (5/6) * (1/6) = 125/1296
Calculating P(B) is a little bit trickier, but not so much. First, we note that, with probability 1, either Sue or Bob must win the game at some point. (The probability that the game hasn’t before the nth die toss converges to zero as n grows. This doesn’t mean that it’s impossible that the game never ends.) And, since they can’t both win at the same time:
P(A) + P(B) = 1
Given that Sue doesn’t win in her first turn, Bob’s chances are exactly the same that they were for Sue before her first turn. So we have:
P(A) = P(B|D) = P(B^D)/P(D)
Since D follows logically from B:
P(A) = P(B|D) = P(B)/P(D) = P(B) * (6/5)
Replacing for P(A) in P(A) + P(B) = 1 gives:
P(B) * (6/5) + P(B) = 1
P(B) = 5/11
(Kudos to Nathanael Nerode for calculating the probability that Bob wins using geometric series!)
So we can calculate the original conditional probability:
P(C|B) = P(C) / P(B) = (125/1295) / (5/11) = 275/1296
Could it possibly be 1/6? Since the previous rolls of the dice have no influence on the next roll, it would be just the vanilla chance, just like how flipping a coin 20 times, there’s still a 50% chance that it will be heads or tails, despite the other outcomes.
I’m sure that this takes more than my meager high-schools math, though.
Suppose you win when you toss two coins, and they both land heads (which you could interpret as throwing two coins simultaneously, or sequentially, or sequentially with the second toss skipped if you landed tails, which is all equivalent). Probability of winning: 1/2? Not quite; it is actually 1/4, no matter how independent your coins are (they are just combined by the game, which has nothing to do with the tosses themselves being independent). This puzzle is similar, just a bit more complex. There are actually more traps on your way if this comment helped, so don’t think you’re done just yet.
I’d have to go with 275/1296 (Bayes)
Although, seriously, life would be so much simpler if we could just go with 5/36.. too bad it’s wrong :(
Love the blog btw :)