My travel schedule for the next several days will probably keep me from posting anything too substantial. So let me leave you with three lovely brain teasers to keep you occupied in the meantime.

1) (Hat tip to Ben Tilly): I have thought of two numbers, which I call A and B. You know nothing about how I came up with these numbers. I plan to flip a fair coin and then tell you the value of A if the coin comes up heads or B if the coin comes up tails. Your job is to guess whether the number I quote is the larger or the smaller of A and B. Devise a strategy that guarantees you a better-than-even chance of winning, no matter what A and B are.

(To make this more precise: Your probability P of winning is a function of A, B and your strategy. Devise a strategy S such that P(A,B,S) is greater than 1/2 for all A and B.)

2) (Hat tip to Stan Wagon): Alice and Bob ran a marathon (assumed to be exactly 26.2 miles long) with Alice running at a perfectly uniform eight-minute-per-mile pace, and Bob running in fits and starts, but taking exactly 8 minutes and 1 second to complete each mile interval (so, for example, it takes him exactly 8 minutes and 1 second to get from the 3.78 mile mark to the 4.78 mile mark, exactly 8 minutes and 1 second to get from the 3.92 mile mark to the 4.92 mile mark, etc.). Is it possible that Bob finished ahead of Alice?

3) (Hat tip to my old friend Steve Maguire): The border between Delaware and its neighbors includes a section with a circular arc: on the circle ten miles from a church in Dover Delaware. Can you name another state border that is partially defined by a circular arc?

I’ll continue to check comments while I’m on the road, but perhaps just a tad less diligently than usual.

#### 36 Responses to “Teasing Your Brain”

1. 1 1 RL

#2 seems easy enough. Alice has reached the end of mile 26 in 8×26 or 208 minutes, with .2 miles to go. Bob has reached the same point in 208 minutes 26 seconds. At an 8 minute mile, Alice finishes the last .2 miles in 1.6 minutes, or 96 seconds. If Bob can run the last .2 miles in less than 76 seconds, therefore, he wins. And by hypothesis he can do that. So Bob wins.

But here’s my question: If Bob takes 8min1sec to run ANY 1 mile interval, mustn’t it be true it takes him 8min1sec to run EVERY 1 mile interval, and isn’t that in contradiction to the fits and starts initial assumption?

Whenever I start a math problem, “…seems easy enough” this is actually a good indication I missed something.

2. 2 2 RL

Sorry. In line 4 above, that should be “70 seconds,” not “76 seconds.” This does not impact on the reasoning of the argument, however.

3. 3 3 Glen

Here’s my solution to #2.

Bob runs the first 0.2 miles in 1:09, the next 0.8 miles in 6:52, then the next 0.2 miles in 1:09, the next 0.8 miles in 6:52, etc. This gives him a total running time of 209:35, one second less than Alice’s 209:36.

For the regular mile intervals (0 to 1, 1 to 2, etc.), it’s obvious this will satisfy the requirement of 8:01/mile. But to make it work with *any* mile interval (e.g., 3.78 to 4.78), we need the additional assumption that Bob runs at a *uniform speed* within each 0.2-mile or 0.8-mile stretch as described above.

4. 4 4 Sprobert

RL, your solution would violate the assumption of 8 min. 1 sec. miles because if he’s running at a constant pace, then finishes the last .2 miles instantaneously, the interval from 25.2 to 26. 2 would presumably not take 8 min. 1 sec.
For my solution, I assume that Bob takes 0 sec. to complete the first and last fifth of each mile, and divides the remaining 3/5 of a mile evenly across his 8 min. 1 sec.
Thus his time by tenths of a mile are for the first mile:
0 + 0 + 80.167sec. + 80.167 + 80.167 + 80.167 + 80.167 + 80.167 + 0 + 0
His time from mile .1 to mile 1.1 then is: 0 + 80.167 + 80.167 + 80.167 + 80.167 + 80.167 + 80.167 + 0 + 0 + 0

Thus the times for the last mile, from 25.2 to 26.2 are:
80.167 + 80.167 + 80.167 + 80.167 + 80.167 + 80.167 + 0 + 0 + 0 + 0
which equals 8 min., 1 sec.
However, since this running will only produce 26 sequences of 6 straight 80.167sec/tenth of a mile, the total time is equal to 26 * 8min, 1 sec, or 208 min, 26 sec., which is less than Alice’s 209 min., 36 sec.

5. 5 5 Robert

I’m highly suspicious of question 1. Suppose I have a strategy S that does this. Suppose you have chosen A = 5, B = 7 and the coin lands heads. It must be the case that P(5, 7, S) > 1/2, i.e. the chance I say “greater” is greater than half.

Suppose instead you chose A = 5, B = 2 and the coin lands heads. Now P(5, 2, S) > 1/2 and the chance I say “lower” is greater than half. But in both cases the information I know is exactly the same: I know A = 5 and I know “heads”.

Is one of these situations impossible? Is there some way for the value of B to affect my strategy?

6. 6 6 Ben

Pick a number C, any number. If the number you announce is higher than C, I say that is your higher number. If it is lower than C I say that the other is your higher number. If equal, I choose randomly.

If C lies between A and B I win, otherwise I have evens chance.

The strategy assumes that the probability of C lying between A and B is nonzero, which is dubious.

Of course you might play the game by trying to predict my C and pick two numbers on the same side as C. If you succeed my strategy is evens.

Your best strategy (assuming you are playing against me) is to pick two numbers close together in a region you think I will not target.

2) Bob reaches the 26 mile mark 26 seconds after Alice. At this point Bob has 0.2 miles to run and Alice has 0.125 miles to run. Alice will complete the race in 60 seconds. Is it possible for Bob to cover 0.2 miles in less than 60 seconds? That’s a 5-minute mile speed which is not impossible but a stretch. We have been told he runs in fits and starts, so yes, it is possible.

7. 7 7 John Faben

Ben, your solution to 1 seems fine. Just pick a real number uniformly at random (or, in fact, from any distribution which gives positive probability to all of R) to be your ‘C’, and since there’s at least one real number between any two reals, you’re done (of course, if Steven is picking integers, it’s even easier.. just pick a half-integer at random)

Your solution to 2 has the same problem as the first one given. If Bob is running at exactly 8m1s per mile for the first 26 miles, he won’t do 25.2 to 26.2 in 8 minutes 1 sec if he suddenly speeds up. Glen’s answer seems to work fine though.

I like 3, except… are the great circles perfect circles? I thought the Earth was an oblate spheroid.

8. 8 8 Al V.

Looking at a map, I don’t see how the circular arc of Delaware’s border can be an arc of a circle centered on Dover. It looks to me like the circle would be centered on somewhere in Wilmington Manor.

9. 9 9 Phil

The correct answer to number 3, in my case: “no”.

10. 10 10 Al V.

Ben’s solution to #1 seems to be the best. However, if I know nothing about the domain of the numbers being selected, the difference between the probability of guessing correctly and 1/2 may be vanishingly small. Are the numbers natural, integers, or real? I could pick C to be 7, and Steve could select A = -13,178.26 and B = -13,178.29.

If, as Steve specifies, I know nothing about how he selected the numbers, I can only assume that the domain is all real numbers, and my best bet is to choose C=0. That way, I’m cutting the domain down the middle.

11. 11 11 Michael

Answer to question 3: The state bordering Delaware (specifically, Pennsylvania).

12. 12 12 MattF

For #3, there’s a number of states with stretches of border that coincide with the 49th parallel of latitude. A circle, ‘tho not a great one, truth to tell.

13. 13 13 Ben

2) OK I missed the point about ANY one mile interval.

Assume Bob runs 1 mile in 1 second then rests for 8 minutes. This satisfies all the criteria, since the interval from 0.5 miles to 1.5 miles he runs for 0.5 seconds rests for 8 minutes, then runs a further 0.5 seconds, also satisfying the criteria.

Clearly any pattern of running for varying speeds satisfy this criterion provided the pattern is repeated exactly every mile.

14. 14 14 David Sloan

1) As an earlier poster has mentioned, if the chosen pivot is between your numbers, I win every time. If it’s not, I break even. I can ensure “better-than-even” odds if my pivot has nonzero chance of landing between your numbers, no matter what they are. Someone suggested a uniformly random distribution over the reals, but none such exists. A normal distribution should do just fine though.

2) The requirement that Bob run any mile interval in the same length of time simply means that his “fits and starts” must be cyclical. For him to win the race, he must do the first fifth of a mile in a relatively short period of time, and use the rest of his 8min1sec for the remaining 4/5ths. Earlier posters have already provided detailed solutions that meet these criteria.

3) Any state with a border defined by a longitude or latitude line should work. Both types of lines are circular arcs.

15. 15 15 Gil

Regarding #3, the distance in question is 12 miles, not 10, and the center of the cirle is a cupola in the town of New Castle. Dover is too far south for a 10 mile circle to have anything to do with the Del/Pa border. Doesn’t really change the nature of the question though – Michael got that one.

16. 16 16 Jon Shea

2) If Bob runs the first 0.2 miles of the race in less than 70 seconds then he will beat Alice.

Bob’s first mile uniquely determines how he will run for the rest of the race. His pace-strategy for every mile section (after the first mile) must be an exact copy of his pace-strategy for the first mile. This is the only way that he can be always 1 mile ahead of where he was 8:01 minutes ago.

This isn’t too hard to see. Imagine that Bob runs the first mile however he likes. As he hits the 1-mile mark (at 8:01 seconds into the race) we put a clone of him (call it Bob2) at the start line. The clone will exactly copy how Bob ran the first mile. The clone represents where Bob was 8:01 minutes ago. Now Bob must always be exactly 1 mile in front of Bob2. If the course is straight then you could tie them to each end of a 1-mile long stick. The only way this would work is if Bob exactly copies how Bob2 runs. Therefore, Bob must copy how he ran the first mile. In this way Bob and Bob2 will always be 1 mile and 8:01 minutes apart.

If you’re not sure how things work when Bob gets to the 3-mile mark, then simply introduce a Bob3 at the start line…

Bob gets to the 26-mile mark at 208:26. Alice gets to the finish at 209:36. So that gives Bob 70 seconds to cover the last 0.2 miles. Equivalently, you could say it gives him 70 seconds to cover the first 0.2 miles since the one determines the other.

17. 17 17 ToddM

#3 seems like it’s missing essential information, but a lot of fun to think about! are you talking about the projection of the border onto a 2D surface (and, if so, if so, which projection)? that’s what I assume we’re talking about, if the arc at the top of Delaware counts. so it seems curvature in the Z-axis doesn’t really count as a part of an arc for this puzzle. on a globe, parallels are indeed parallel, and if you are look at them from directly above, they are not arcs. but lines of longitude are not parallel, and if you look at them from directly above, they are curved, so, say, the western border of PA, could be a candidate.

BTW Gil, I don’t think Michael got #3. The question is to find another “border”, not another state. His answer refers to the same border.

18. 18 18 Michael

I think all proposed solutions to #1 are wrong. As stated, the problem asks for a strategy s.t. for all A, B, P(A, B, S) > 1/2. This is not the same as asking for a strategy that offers a probability of winning > 1/2 when A and B are randomly drawn from some distribution. It is then insufficient to say that for some A, B; P(A, B, S) = 1/2 and for some other A, B; P(A, B, S) = 1.

Also, based on the wording of the question, it seems unlikely that any deterministic strategy can be correct (such as picking a point), because then P (winning) = 0 or 1 (for particular A, B), and 0 < 1/2, so our strategy must never have P (winning) = 0. Then such a deterministic strategy would lead to P (winning) = 1 for all A, B, and thus to a more irresistible riddle (though phrasing it this way would be extremely clever were it the case that we could do so much better, but that is not the standard practice).

How randomizing helps… I'm not sure yet.

19. 19 19 Michael

in the first paragraph I wrote above, it should say
It is then insufficient to say that for some A, B; P(A, B, S) = 0 …

20. 20 20 Al V.

@Michael @Gil – re. #3, I’m being picky, but the question was “Can you name another state border that is partially defined by a circular arc?” The object of the question is to name a border, not a state. The PA-DE border is the same border as the DE-PA border, so I don’t think that PA counts. If the question was “Can you name another state’s border …”, then I might give you the answer. However, I’m not an English teacher, so maybe my grammar is wrong.

21. 21 21 Ryan Mulligan

It might be the trivial answer to number 3 but, New York must also be defined by a circular arc since it shares that border with Delaware.

22. 22 22 Michael

wow, I forgot about the coin’s role in all this. Ignore most of what I wrote.

23. 23 23 Jeffrey

Solution to 1:

Let f(x) be any strictly increasing function over the real numbers with 0 < f(x) < 1 for all x. If you see x, guess that it's the highest value with probability f(x).

Suppose A 0, and these are better than even odds for all A and B.

24. 24 24 Jeffrey

Hmm … my greater thans, less thans, and B’s got eaten into accidental html.

Here’s a second try:

Let f(x) be any strictly increasing function over the real numbers with 0 < f(x) < 1 for all x. If you see x, guess that it's the highest value with probability f(x).

Suppose A is less than B. If you are shown B, you guess right with probability f(B). If A, the odds are 1 – f(A). So the chances that you guess right are 1/2 + 1/2 * (f(B) – f(A)). f is increasing, so f(B) – f(A) is positive, and the odd are better than even for all A and B.

25. 25 25 dave

so i finally got the book. im on chapter 4.
creation/math/physics/chemistry/biology seems to be the arguement for the ‘discovery’ of math as opposed to ‘invention’

what about zero and infinity? infinity is not a number? if division by zero is an error, something is wrong with division. zero certainly exists.

in my minds eye, there is no ‘multiverse’ there is only universe. it seems to be infinite, expanding, and everything outside it is zero.

contrary to the footnote on page 14, my observable universe is in alaska.

i looked at some pictures of black holes on wikipedia. ive never liked black holes. they scream ‘something is wrong with your reception of this picture’ at me.

i found the brain teasers way over my head (as usual) so i thought i would just ramble a reply. =]

for #3, the MT-WY border is on the 45th parallel. todd is right, they are parallel to the equator but they also ‘project’ circular arcs onto the surface of the sphere.

26. 26 26 dave

many state borders use parallels and longitudes so i guess neither is a candidate for the solution. like i said..way over my head. =/

27. 27 27 Jeffrey

For the first problem, let f(x) be a strictly increasing function. If shown x, guess its the larger one with probability f(x).

To prove that this works, suppose A is less than B. If shown B, the chance you win is f(B). If shown A, the chance is 1 – f(A). So the chance you win is .5 + .5 * (f(B) – f(A)). But f(x) is increasing, so this is better than even odds.

28. 28 28 Jeffrey

I meant to say, let f(x) be a strictly increasing function between 0 and 1.

29. 29 29 Jon Shea

Dave: I like the latitudinal lines, with some qualification.

If you claim the border lies on the surface of the Earth then you run into the problem that the Earth isn’t perfectly spherical. Since this would disqualify every border, including the given example, we should probably exclude local irregularities (hills and valleys). But there’s still the issue that the Earth isn’t really a sphere, but (to much higher approximation) an oblate spheroid. If we consider the Earth to be oblate then this disqualifies the longitudinal lines because they would be elliptical sections and not circle sections. But the latitudinal lines would still be circle sections, as they are drawn around the Earth’s oblate axis.

Of course, if you treat the Earth as non-sperical then the Delaware / Pennsylvania border is disqualified too, as it would be slightly taco or potato chip shaped. Maybe the latitudinal lines might be the best possible answer.

Finally, if we wanted to be really hard about it then we could include tidal distortions from the moon and the sun, which I’m pretty sure would make it impossible to draw a circle anywhere on the idealized surface of the Earth.

30. 30 30 ThomasH

It depends on Bob’s time x in the first .2 miles of each 1 mile segment.
Bob’s total time is 26/8.02+.2*x
Alice’s time is 26.2/8.0333 =3.2614 hours
Solving for x the time at which the times are the same:
.x =.2/(3.2614-26/8.0333) = 11.0086 mph

If Bob runs >11.0086 mph in the first .2 miles of each mile segment including 200-26.2 his time will be less than Alice’s.

31. 31 31 RL

Using the Delaware/Penn. border as an example of a curved border suggests the longitudinal line solution is not in the spirit of the puzzle. I think more likely the answer is the obvious one already provided, to wit the other side of the border, Pennsylvania.

32. 32 32 Neil

So if Jeffrey is right, I can use the standardized cumulative normal and guess larger if the number quoted is positive and guess smaller if the number quoted is negative.

33. 33 33 Jeffrey

I thought my first (two) posts were eaten by cyberspace, and hence my weird repetitiveness. I’m seriously not normally this computer illiterate.

Re: Neil

Not quite – you can use the cumulative distribution function for the normal curve, but what that would mean is guessing higher with probability

98% if quoted 2,
84% if quoted 1,
50% if quoted 0,
16% if quoted -1, etc.

That way if A = 1 and B = 2, you win 98% when shown 2, and 16% when shown 1. Overall, you win 1/2 * (98+16)% = 57% when A = 1 and B = 2.

Just guessing higher for positive numbers fails, for when A = 1 and B = 2, you don’t do better than even.

34. 34 34 Jonathan Kariv

For question 1. My stratergy is to choose the number I see if and only if it’s positive. You choose numbers according to some unknown (to me) distribution. Let p be the probability you choose a non-positive number.

The probability I win = p^2 (1/2) +2pq + (1-p)^2 (1/2)

i.e. I win if you choose 1 positive and 1 non-positive number and have a 50-50 shot otherwise.

I whacked myself in the head when someone pointed out the the state bordering deleware mnust also have the same border.

35. 35 35 donald

I got this far and got stuck.

It seems Bob cannot win, and his pace during each mile does not matter. According to the conditions, Alice must reach 25.2 miles before Bob (Alice covers this in 25.2 x 8 minutes, Bob in 25.2 x (8 minutes 1 second)).

Alice will run the last 1 mile in 8 minutes. Bob MUST run the total distance in 8 minutes 1 second, not matter how he varies his speed.

But Bob was behind Alice with 1 mile to go, exactly 25.2 seconds behind in fact. So, in order for Bob to beat Alice, he must cover the last mile in less than 480-25.2 seconds. Since the distance is a mile, Bob must take 481 seconds to cover the distance. Which places him 26.2 seconds behind Alice at the end. So he cannot win.

He could win if you allow his 481 seconds to cover a mile to include him going past the mile point, turning around and coming back. Thus, he runs each mile in 300 seconds, keeps going past the mile point for a distance, turns around and comes back, and the total time, including the extra distance run, is 481 seconds. Then he is still behind Alice at 25.2 miles, but he runs much faster, gets to the finish before Alice. Of course to satisfy the 481 second mile condition, he would have to keep going past the mile, then turn around and come back. For this to work, you have to define his time per mile based on when he gets back for all but the last mile. For the last mile you time only until he reaches the finish line the first time.

Of course, this is similar to saying that Alice runs at a constant speed, but wanders off course, and ends up running more than 26.2 miles, while Bob runs slower but on course, so he gets there first. Either of these solutions ignores the implications of the problem that both contestants are running in an otherwise conventional manner.