Poll

8 votes (47.05%)
6 votes (35.29%)
3 votes (17.64%)
2 votes (11.76%)
6 votes (35.29%)
2 votes (11.76%)
3 votes (17.64%)
2 votes (11.76%)
8 votes (47.05%)
6 votes (35.29%)

17 members have voted

Gialmere
Gialmere
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July 22nd, 2020 at 8:02:25 AM permalink
Quote: charliepatrick

I've have to resort to a spreadsheet but can see that the correct strategy is for Player #1 to look through 100/e (=37) and after that pick the box that's the higher than any seen so far. Probability of success = .371043 (slightly higher than 1/e).

So the second player has two options...
Scenario A - Player #2 looks at the first N/e = 37 then picks the second box higher than that seen so far (he knows the first one encountered wasn't highest).
Scenario B - Player #2 assumes the best box is in the first 37 and uses the same strategy for just those - Pr success ~= 1/e * 37/100 = .137286

(i) Player #1 - finds the best in the first 37 boxes, so cannot win (and therefore goes right through the boxes). Pr = 37/100
(This is the only case where the audience can help, let's assume they don't - so #2 will also lose.)

(ii) Player #1 - finds the 2nd best in the first N, only 1st is left, so will always win.
(iii) Player #1 - finds the 3rd best in the first N, so 1st and 2nd left, so has a 50/50 chance to win; if lose the 2nd competitor will always win.
(iv) Player #1 - finds the 4th best in the first N, 1st 2nd 3rd left, so has 1/3 chance to win, if lose the 2nd competitor has a 1/2 chance to win.
(lxx) 1-70 left, so has 1/70 chance to win, if lose the 2nd competitor has a 1/69 chance to win.

Assuming Player No2 adopts this strategy, the probabiity without knowing than Player No1 didn't win, is ....
Pr #1 or #2 wins if the 1st is in the first half = 0.
Pr #1 wins = Pr(finds the 2nd in the first 37)*1 + Pr(finds the 3rd in the first 37)*1/2 + Pr(finds the 4th in the first 37)*1/3 etc.
Pr #2 wins = Pr(finds the 2nd in the first 37)*0 + Pr(finds the 3rd in the first 37)*1 + Pr(finds the 4th in the first 37)*1/2 etc.

This works out to be .233740 (which is close to 1/e/(1-1/e)).

However the above includes situations where Player #1 would win. e.g. if the 2nd highest was in the first 37, Player No2 wouldn't be playing. So we have to exclude cases where Player #1 won. This is 1/e, so we have to multiply the probability by (1-1/e).

Gosh this means the chances that Player #2 wins as (.371631) about 1/e!!!

If the audience helps then you can add best box was in the first 37, and that's 1/e as well, so gives 2/e.

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Correct!
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ThatDonGuy
ThatDonGuy
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July 22nd, 2020 at 8:05:03 AM permalink
Could somebody check my math here? I think I found how to determine the solution, but I'm not quite sure I analyzed it correctly.

Let N be the number of secretaries - in our case, 100
Let S be the number at which you skip the first (S-1) secretaries, then choose the first one with a salary/ranking higher than the earlier ones (obviously, you don't choose a secretary that would be lower than an earlier one, as your selection would be incorrect)

The probability that secretary S is correct and is chosen is:
SUM(k=1..N) {P(#k is correct and is chosen)}
= SUM(k=1..N) {P(#k is correct) P(#k is chosen given that #k is correct)}
= SUM(k=S..N) {1/N P(#k is chosen given that #k is correct)}
(note P(#k is chosen given that #k is correct) = 0 for k = 1..S-1)
= SUM(k=S..N) {1/N P("the best of the first (k-1) secretaries is in the first (S-1)" given that #k is correct)}
(because if the best of the first (k-1) secretaries is not in the first (S-1), then that secretary will be chosen before #k)
= SUM(k=S..N) {1/N x (S-1)/(k-1)}
= (S-1) / N x SUM(k=S..N) {1/(k-1)}

The answer is the value of S where this is a maximum.
From Thomas Ferguson's article in Statistical Science, the general solution S = N/e apparently comes from the fact that, if x = LIMIT(n -> +INF) {r/n}, then, setting t = k/n (and dt =1/n), the sum becomes a Riemann approximation to the integral
x INTEGRAL(x, 1) {1/t dt}, which is -x ln x; the first derivative of -x ln x = (-1 - ln x), which is 0 when ln x = -1 -> x = 1/e -> -x ln x = -1/e * -1 = 1/e.

gordonm888
gordonm888
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July 22nd, 2020 at 11:35:40 AM permalink
Quote: ThatDonGuy

Could somebody check my math here? I think I found how to determine the solution, but I'm not quite sure I analyzed it correctly.


Let N be the number of secretaries - in our case, 100
Let S be the number at which you skip the first (S-1) secretaries, then choose the first one with a salary/ranking higher than the earlier ones (obviously, you don't choose a secretary that would be lower than an earlier one, as your selection would be incorrect)

The probability that secretary S is correct and is chosen is:
SUM(k=1..N) {P(#k is correct and is chosen)}
= SUM(k=1..N) {P(#k is correct) P(#k is chosen given that #k is correct)}
= SUM(k=S..N) {1/N P(#k is chosen given that #k is correct)}
(note P(#k is chosen given that #k is correct) = 0 for k = 1..S-1)
= SUM(k=S..N) {1/N P("the best of the first (k-1) secretaries is in the first (S-1)" given that #k is correct)}
(because if the best of the first (k-1) secretaries is not in the first (S-1), then that secretary will be chosen before #k)
= SUM(k=S..N) {1/N x (S-1)/(k-1)}
= (S-1) / N x SUM(k=S..N) {1/(k-1)}

The answer is the value of S where this is a maximum.
From Thomas Ferguson's article in Statistical Science, the general solution S = N/e apparently comes from the fact that, if x = LIMIT(n -> +INF) {r/n}, then, setting t = k/n (and dt =1/n), the sum becomes a Riemann approximation to the integral
x INTEGRAL(x, 1) {1/t dt}, which is -x ln x; the first derivative of -x ln x = (-1 - ln x), which is 0 when ln x = -1 -> x = 1/e -> -x ln x = -1/e * -1 = 1/e.



As an aside, we can see from ThatDonGuy's response that digital social media will be changing mathematical notation, probably forever.
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Gialmere
Gialmere
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July 22nd, 2020 at 12:43:37 PM permalink


The towns of Alpha, Beta, and Gamma are equidistant from each other.

You're driving to visit a friend in Gamma but have gotten lost. You come across a signpost stating that you are three miles from Alpha and four miles from Beta.

What then is your maximum possible distance from Gamma? (Assume the land is flat.)
Have you tried 22 tonight? I said 22.
CrystalMath
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Gialmere
July 22nd, 2020 at 2:12:43 PM permalink
Quote: Gialmere



The towns of Alpha, Beta, and Gamma are equidistant from each other.

You're driving to visit a friend in Gamma but have gotten lost. You come across a signpost stating that you are three miles from Alpha and four miles from Beta.

What then is your maximum possible distance from Gamma? (Assume the land is flat.)



Surprisingly, 7 miles. If you are the 7 miles from Gamma, then the distance between cities is about 6.0825. I don't see the logic that makes this an "easy" puzzle, though.
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charliepatrick
charliepatrick
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Gialmere
July 22nd, 2020 at 3:43:17 PM permalink
Quote: CrystalMath

Surprisingly, 7 miles. If you are the 7 miles from Gamma, then the distance between cities is about 6.0825. I don't see the logic that makes this an "easy" puzzle, though.

Using a spreadsheet I agree, but also haven't yet seen the logic. btw I get a very slightly different figure...
6.082 763
X - axis
Y - axis
A (aka Alpha)
0.000 000
0.000 000
B (aka Beta)
6.082 763
0.000 000
C (aka Gamma)
3.041 381
-5.267 827
Me
2.465 985
1.708 484
0.904 194
Cos C
Dist - Deltas
-0.575 396
6.976 311
7.000 000
Gialmere
Gialmere
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July 22nd, 2020 at 3:50:19 PM permalink
Quote: CrystalMath

Surprisingly, 7 miles. If you are the 7 miles from Gamma, then the distance between cities is about 6.0825. I don't see the logic that makes this an "easy" puzzle, though.


Quote: charliepatrick

6.082 763
X - axis
Y - axis
A (aka Alpha)
0.000 000
0.000 000
B (aka Beta)
6.082 763
0.000 000
C (aka Gamma)
3.041 381
-5.267 827
Me
2.465 985
1.708 484
0.904 194
Cos C
Dist - Deltas
-0.575 396
6.976 311
7.000 000


Correct!

Quote: CrystalMath

I don't see the logic that makes this an "easy" puzzle, though.


You have a point. This one just looks easy.
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Wizard
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July 22nd, 2020 at 4:46:08 PM permalink
Good problem. I don't immediately see why the answer is what CM said, not that I'm disputing it.

Reminder that math puzzles don't have to be posted here -- just easy ones that are not thread worthy. This one is thread worthy.
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charliepatrick
charliepatrick
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Gialmere
July 22nd, 2020 at 6:52:16 PM permalink
Just had a try with the sides 4 and 4 (rather than 3 and 4) and guess what the answer is!!
Yes you've guessed it, the answer is 8.

What's more interesting is the distance AB = SQRT(48) and, if AB is the x-axis, I am 2 above and G is 6 below. (Note that if the cities are ABG and I'm at M, AG^2+AM^2=SQRT(48)+4*4=8*8=GM^2; so the angle GAM is 90 degrees.)

Using the spreadsheet I can also see a pattern emerging (the length is always the sum of the two)...
0 4 SQRT(16)
1 4 SQRT(21) +5
2 4 SQRT(28) +7
3 4 SQRT(37) +9
4 4 SQRT(48) +11
4 5 SQRT(61)
3 6 SQRT(63) +2
2 7 SQRT(67) +4
1 8 SQRT(73) +6
0 9 SQRT(81) +8

The pattern can be found if you consider Cos(angle)=(a^2+b^2-c^2)/(2ab) as all the angles above are 120 degrees.

So all one has to do is show the angle is 120 degrees creates the maximum answer and this means length will be SQRT(a^2+b^2+ab).

If you carry on this logic you see the angles AMG and BMG both have to be 60 degrees - e.g. Cos(Theta)=(adj^2+adj^2-opp^2)/(2 adj adj) = 49+9-37)/(2*3*7) = 1/2 or (16+49-37)/(2*4*7) = 1/2.


Last edited by: charliepatrick on Jul 22, 2020
charliepatrick
charliepatrick
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July 22nd, 2020 at 8:24:09 PM permalink
btw I wonder whether this is connected to a puzzle where there are three cities and you want to join them by roads. The plan is to find a point in the middle for a junction and then build a road from each city to the junction.
Apart from silly cases, where they're in a straight line, or it's quicker to have the junction at one of the cities, you can prove the angles the roads meet at the junction are all 120 degrees. Some of the logic is based on symmetry that the forces must be equal. Thus I was wondering whether one can use this idea or to construct a point on GM extended.

Yes you can!

Consider locating a city D just Northish of "Me" on the line GM at a fixed distance from G. Then find the minimum "roads" joining A B and D. Since AM and BM are fixed, in this case 3 miles and 4 miles, we are minimizing DM. This therefore maximises MC.

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