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A client gave me this puzzle, which he claims I answered incorrectly. I'll give my answer and rationale, and his answer (for which he refuses to give a rationale).

The Puzzle: You have a marker on a number line, it starts at zero. You also have a fair coin, which you flip 100 times. For heads you move the marker one place to the right, for tails one to the left. Now suppose you are in Vegas. If you actually want to win, where should you place your bet (as to where the pin will be) after 100 flips?

My Answer: Zero. The probability of any result after 100 coin flips is a standard bell curve, centered on zero. Of any particular result after 100 coin flips, the highest absolute number of final events for any one number is at zero on the number line.

His Answer: You bet on the square root of the number of flips. That is, either 10 or -10.

Can anyone explain why he's right?

David M
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  • If you're betting on the location, then of course you're right, the most likely location is 0 (although it's still very unlikely). – quasi Mar 13 '17 at 21:29
  • The expected distance from zero is approximately 10, but that doesn't mean 10 or -10 are more likely than zero as final locations. – quasi Mar 13 '17 at 21:30
  • Well....let's do four flips. There are $16$ total cases, $4$ end at each of $\pm 2$, $1$ end at each of $\pm 4$ and $6$ end at $0$ so.... – lulu Mar 13 '17 at 21:31
  • http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/RandomWalk.htm might help. – Henno Brandsma Mar 13 '17 at 21:31
  • And lulu's comment is how you might convince your friend. Do an actual game with say 4 flips. He'll bet on 2 or -2 (but only one of them per game), you'll bet on 0. If the game is played for 1 dollar, he will almost certainly lose money (for any reasonably large number of games). That will either convince him that he was wrong, or else he'll rationalize it by saying you were just lucky. – quasi Mar 13 '17 at 21:36
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    "Can anyone explain why he's right?" — No, because he isn't. However your argumentation is incomplete: If it were $99$ flips, it also would be a bell curve centered at $0$. However betting on $0$ would be a very bad idea. – celtschk Mar 13 '17 at 21:40
  • @celtschk -- That's funny. And the friend would bet on 10, and every game (between the two of them) would be a tie. – quasi Mar 13 '17 at 21:42
  • Well, we know there are 100 choose 50 ways to end on zero (exactly 50 heads). There are 100 choose 55 ways to land on 10 (exactly 55 heads). Which is larger? 100 choose 50 - 100 choose 55 is 100! (1/50!50!-1/55!45!)=100!/50!45! (1/50...46 -1/55...51). Which is positive as 55...51 > 50...46. – fleablood Mar 13 '17 at 22:09

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