The Riddler - Series on FiveThirtyEight

[TheRealJoeFish]

The Riddler PUZZLE THREE - How Long Will Your Smartphone Distract You From Family Dinner?

---

This is the 3rd puzzle in FiveThirtyEight's The Riddler series, in which Oliver Roeder posts an interesting math/creative thinking-ish brain-teaser every Friday.  I'll post either my solution (if I can come up with one) or my initial thoughts (if I don't see how to solve it) in hide tags, and everyone is welcome to give it a go! I have promised a rep to people who solve a problem that I got wrong

I'm one for two; as noted above, I crashed and burned on the first one but was able to brute force the second.  I haven't looked at this one yet, so, let's see if this will join the "stumps Joe" category:

Now, here’s this week’s Riddler, which comes to us from Olivia Walch, a mathematics Ph.D. student and cartoonist:

You’ve just finished unwrapping your holiday presents. You and your sister got brand-new smartphones, opening them at the same moment. You immediately both start doing important tasks on the Internet, and each task you do takes one to five minutes. (All tasks take exactly one, two, three, four or five minutes, with an equal probability of each). After each task, you have a brief moment of clarity. During these, you remember that you and your sister are supposed to join the rest of the family for dinner and that you promised each other you’d arrive together. You ask if your sister is ready to eat, but if she is still in the middle of a task, she asks for time to finish it. In that case, you now have time to kill, so you start a new task (again, it will take one, two, three, four or five minutes, exactly, with an equal probability of each). If she asks you if it’s time for dinner while you’re still busy, you ask for time to finish up and she starts a new task and so on. From the moment you first open your gifts, how long on average does it take for both of you to be between tasks at the same time so you can finally eat? (You can assume the “moments of clarity” are so brief as to take no measurable time at all.)

Hmmm..............................  Initial thoughts below in hide tags, then I'll open the floor to discussion before I post an attempted solution or an admission of defeat!

[hide]
Interesting.  At first glance, this looks similar to Problem 2 in that it appears to be solvable by multiplying probabilities, to get P(1 Minute) + 2*P(2 Minutes) + 3*P(3 Minutes) + ... and so on.  Of course, we may need to use some sort of trick to calculate that infinite sum.  Hopefully, it will simplify to a very straightforward geometric sequence, so we can just add it up using (1 + n + n^2 + n^3 + ...) = 1/(1-n).

Another way of looking at it, which might be easier if the above math doesn't come easily or obviously, would be to look at the probability of changing between states.  There are essentially nine different states:

1) You're clear but your sister has 1 minute left;
2) You're clear but your sister has 2 minutes left;
3) You're clear but your sister has 3 minutes left;
4) You're clear but your sister has 4 minutes left;
5) You have 1 minute left but your sister is clear;
6) You have 2 minutes left but your sister is clear;
7) You have 3 minutes left but your sister is clear;
8) You have 4 minutes left but your sister is clear;
9) You're both clear.

If we view it this way, we'll be trying to figure out the probability of a given state 1-8 transitioning into state 9, and then adding up probabilities of that happening.

The ways of thinking about it above are, of course, similar, though one sort of moves minute-by-minute and the other moves task-by-task.  The minute-by-minute approach seems more natural, and is what I'll try first, but I'm not sure which will eventually yield the answer.  If we get an answer with one, it would behoove us to see if that makes sense when viewed through the lens of the other.

---

So, that's my thoughts on that! Time to put pen to paper.  THIS GOES FOR YOU ALL TOO.  REMEMBER, there is a SHINY NEW REP for someone who solves a problem I get wrong!!!

Good huntin'![/hide]

I know this is sort necropost but 1) it's my own thread I created for my own special purpose and 2) I've just thought about the problem for the first time since I posted it and 3) EvieAlasdair Ham necro'd it already a few weeks ago so it's his fault

So, here's my best shot?  Whether I'm right or wrong, I am absolutely certain there is an easier way to do this.  But this is the most *obvious* way to me, even if it's exceedingly tedious and involved:

---

The first person's task can be either 1, 2, 3, 4 or 5 minutes, and same for person #2.  There are 5 ways they can be the same ((1,1), (2,2)...), 8 ways they can be 1 minute off ((1,2), (2,3), (3,4), (4,5), (2,1)...), 6 ways they can be 2 minutes off, 4 ways for 3 minutes off, and 2 ways for 4 minutes off.  If they're 1 minute off, then there's a 20% chance the lower person matches up, a 20% chance they move to 1 minute off, a 20% chance they move to two minutes off, a 20% chance they move to 3 minutes off, and a 20% chance they move to 4 minutes off.  If they're 2 minutes off, there's a 40% chance they move to one minute off (if the next task is either 1 or 3 minutes), a 20% chance they match, a 20% they stay at 2 minutes off, and a 20% chance they go to 3.  If 3 minutes off, it's 20% match, 40% 1 minute, 40% 2 minutes.  If 4 minutes off, it's 20% match, 40% 1 minute, 20% 2 minutes, and 20% 3 minutes.

We're also going to keep a running count of the "total time", which we'll define as the amount of time committed by the lengthier person.  So, if Dude 1's first 3 tasks are 2, 5, 3 and dude 2's first 3 tasks are 1, 2, 5 we'll call the total time marker = 10.  This means that some changes (ie, if person 1's two minutes back and gets a 3 minute task) will change the "state" and the TT, some will change the TT but not the state, and some will change the state and not the TT.

So, the probabilities of changing from one phase to another phase are like this:

Start -> match, TT + 1 = 4%
Start -> match, TT + 2 = 4%
Start -> match, TT + 3 = 4%
Start -> match, TT + 4 = 4%
Start -> match, TT + 5 = 4%
Start -> 1 off, TT + 2 = 8%
Start -> 1 off, TT + 3 = 8%
Start -> 1 off, TT + 4 = 8%
Start -> 1 off, TT + 5 = 8%
Start -> 2 off, TT + 3 = 8%
Start -> 2 off, TT + 4 = 8%
Start -> 2 off, TT + 5 = 8%
Start -> 3 off, TT + 4 = 8%
Start -> 3 off, TT + 5 = 8%
Start -> 4 off, TT + 5 = 8%

1 off -> match, TT + 0 = 20%
1 off -> 1 off, TT + 1 = 20%
1 off -> 2 off, TT + 2 = 20%
1 off -> 3 off, TT + 3 = 20%
1 off -> 4 off, TT + 4 = 20%

2 off -> 1 off, TT + 0 = 20%
2 off -> match, TT + 0 = 20%
2 off -> 1 off, TT + 1 = 20%
2 off -> 2 off, TT + 2 = 20%
2 off -> 3 off, TT + 3 = 20%

3 off -> 2 off, TT + 0 = 20%
3 off -> 1 off, TT + 0 = 20%
3 off -> match, TT + 0 = 20%
3 off -> 1 off, TT + 1 = 20%
3 off -> 2 off, TT + 2 = 20%

4 off -> 3 off, TT + 0 = 20%
4 off -> 2 off, TT + 0 = 20%
4 off -> 1 off, TT + 0 = 20%
4 off -> match, TT + 0 = 20%
4 off -> 1 off, TT + 1 = 20%

Whew.

Now that we have that done... what next?  Well, I think we get to misbehave, and do something that feels wrong but isn't: add expected values together.  We can do this because, at every level, the probabilities are symmetrical.  So, from the beginning, the average state is: Match = 20%, 1 off = 32%, 2 off = 24%, 3 off = 16%, and 4 off = 8%, and the corresponding average TTs are Match = 3, 1 off = 3.5, 2 off = 4, 3 off = 4.5, and 4 off = 5.

At every new task, the chance of matching is 20%.  So, basically, we're going to have a series that looks like this: 20% * (TT after initial picks) + 80% * 20% * (average (delta)TT from stage 1 to stage 2) + 80%^2 * 20% * (average (delta)TT from stage 2 to stage 3) + ... and so on.

If we can simply figure out the average distribution of 1/2/3/4 at each stage, then we can get the average (delta)TT at that stage, and plug and chug, as it were.

The stage 1 distribution of 1:2:3:4 is 40%:30%:20%:10% (that's the 8 ways, 6 ways, 4 ways, 2 ways from above).  We also know the following (non-match) probabilities, along with average delta(TT):

1 -> 1 = 25%, TT + 1
1 -> 2 = 25%, TT + 2
1 -> 3 = 25%, TT + 3
1 -> 4 = 25%, TT + 4
2 -> 1 = 50%, TT + .5
2 -> 2 = 25%, TT + 2
2 -> 3 = 25%, TT + 3
3 -> 1 = 50%, TT + .5
3 -> 2 = 50%, TT + 1
4 -> 1 = 50%, TT + .5
4 -> 2 = 25%, TT + 0
4 -> 3 = 25%, TT + 0

So, every time we start with 1, we have an average TT increase of 2.5 (average of 1, 2, 3, 4).  With 2, we have 1.5.  With 3, we have .75, and with 4, we have .25.

The initial distribution goes from 40/30/20/10 to 10 (chance of going from 1 to 1)+15 (chance of going from 2 to 1)+10 (etc)+5/10+7.5+10+2.5/10+7.5+0+2.5/10+0+0+0 = ... wouldn't you know, 40/30/20/10.  This is PERFECT, because that means that the average TT increase is the same at each level because the average state distribution is the same at each level!!!

So, the average TT increase at each level, if there's no match, is = .4*2.5 + .3*1.5 + .2*.75 + .1*.25 = 1 + .45 + .15 + .025 = 1.625.  Because this happens 80% of the time, and the other 20% there's a match and the TT doesn't increase, the average TT increase at each level is 1.625*.8 = 1.3

The last thing to do is to calculate the first TT, the average TT after 1 pick: .2*3 + .32*3.5 + .24*4 + .16*4.5 + .08*5 = .6 + 1.12 + .96 + .72 + .4 = 3.8.

So, the average time is:

3.8 + .8*1.3 + .8^2*1.3 +... which is a perfectly straightforward geometric sequence, which sums to (1.3/(1-.8)) + 3.8 = (5.2 + 3.8) = 9 minutes.

---

Big Reveal, time to check if I'm correct:

YES, I'M CORRECT!
But that's small comfort, because the solution given in the Riddler (which I've put in the hide tags below) is basically  four lines long (mine in the hide tags above is 73 lines long):

---

The average time between moments of clarity is (1+2+3+4+5)/5 = (5+1)/2 = 3 minutes.
So, the probability that a particular minute will end with a moment of clarity is 1/3.
This is independent for both siblings.
So, on average, you sync up at the end of 1/9 of minutes.

---

Well... I got it right.  But my solution certainly won't be in "the book".