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Author Topic: Riddle Me This! - Series  (Read 894 times)

DuDeR MaN

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« Reply #75 on: February 11, 2004, 10:22:00 AM »

QUOTE (katsumoto @ Feb 11 2004, 12:14 PM)
what position in line gives him the greatest chance of being the first duplicate birthday?

in front of the line, he won't get the free ticket but he has a better chance to be the first duplicate birthday.


and for that last one.... I thought about a blood test but I figured the one guy wouldn't do it since it would show he isn't the king's son.  Whatever.. keep them coming.
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Ripasser

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« Reply #76 on: February 11, 2004, 03:19:00 PM »

Yea thats what i was thinking too.
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cleary

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« Reply #77 on: February 11, 2004, 03:21:00 PM »

15-25th person in line.
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DuDeR MaN

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« Reply #78 on: February 12, 2004, 01:57:00 AM »

actually 2nd in line.. it doesn't hurt his chances of being the first duplicate and if by some wierd chance the guy in front of him is the same.. he gets a free ticket.
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katsumoto

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« Reply #79 on: February 12, 2004, 07:02:00 AM »

beerchug.gif

Here's why:

Suppose he was the Kth person in line. Then he wins if and only if the K-1 people ahead all have distinct birtdays AND his birthday matches one of theirs. Let
A = Fred's birthday matches one of the K-1 people ahead
B = those K-1 people all have different birthdays
Then
Prob(Fred wins) = Prob(B ) * Prob(A | B )

(Prob(A | B ) is the conditional probability of A given that B occurred.)

Now let P(K) be the probability that the K-th person in line wins, Q(K) the probability that the first K people all have distinct birthdays (which occurs exactly when none of them wins). Then

P(1) + P(2) + ... + P(K-1) + P(K) = 1 - Q(K)
P(1) + P(2) + ... + P(K-1) = 1 - Q(K-1)

P(K) = Q(K-1) - Q(K) <--- this is what we want to maximize.

Now if the first K-1 all have distinct birthdays, then assuming uniform distribution of birthdays among D days of the year, the K-th person has K-1 chances out of D to match, and D-K+1 chances not to match (which would produce K distinct birthdays). So
Q(K) = Q(K-1)*(D-K+1)/D = Q(K-1) - Q(K-1)*(K-1)/D
Q(K-1) - Q(K) = Q(K-1)*(K-1)/D = Q(K)*(K-1)/(D-K+1)

Now we want to maximize P(K), which means we need the greatest K such that P(K) - P(K-1) > 0. (Actually, as just given, this only guarantees a local maximum, but in fact if we investigate a bit farther we'll find that P(K) has only one maximum.) For convenience in calculation let's set K = I + 1. Then
Q(I-1) - Q(I) = Q(I)*(I-1)/(D-I+1)
Q(I) - Q(I+1) = Q(I)*I/D

P(K) - P(K-1) = P(I+1) - P(I)
= (Q(I) - Q(I+1)) - (Q(K-2) - Q(K-1))
= Q(I)*(I/D - (I-1)/(D-I+1))

To find out where this is last positive (and next goes negative), solve
x/D - (x-1)/(D-x+1) = 0

Multiply by D*(D+1-x) both sides:
(D+1-x)*x - D*(x-1) = 0
Dx + x - x^2 - Dx + D = 0
x^2 - x - D = 0

x = (1 +/- sqrt(1 - 4*(-D)))/2 ... take the positive square root
= 0.5 + sqrt(D + 0.25)

Setting D=365 (finally deciding how many days in a year!),
desired I = x = 0.5 + sqrt(365.25) = 19.612 (approx).

The last integer I for which the new probability is greater then the old is therefore I=19, and so K = I+1 = 20.

...viola!

EPISODE XVI

Land of the giants, it is often called,
Where warriors and wizards play with wolves;
Its suns and stars get hot and cold,
Whether one is at home or on the road.
Fourteen tribes live in the west, we are told,
One less than those in the eastern world;
The weakest thirteen will give up and fold.
While the strongest continue their quest for gold.
The kings are powerful and the magic is bold,
Others can fly and are a site to behold;
But when the season is past and the story is told,
New stars are born and old tribes evolve.

What is the Land of the Giants?

Have fun witht his riddle.  Good Luck!!! biggrin.gif
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DuDeR MaN

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« Reply #80 on: February 12, 2004, 11:28:00 AM »

QUOTE (katsumoto @ Feb 12 2004, 12:02 PM)
Well...clblanch1622 solved the problem.  Fred should try to be the 20th person in line.  Congrats. beerchug.gif

Here's why:

Suppose he was the Kth person in line. Then he wins if and only if the K-1 people ahead all have distinct birtdays AND his birthday matches one of theirs. Let
A = Fred's birthday matches one of the K-1 people ahead
B = those K-1 people all have different birthdays
Then
Prob(Fred wins) = Prob(B ) * Prob(A | B )

(Prob(A | B ) is the conditional probability of A given that B occurred.)

Now let P(K) be the probability that the K-th person in line wins, Q(K) the probability that the first K people all have distinct birthdays (which occurs exactly when none of them wins). Then

P(1) + P(2) + ... + P(K-1) + P(K) = 1 - Q(K)
P(1) + P(2) + ... + P(K-1) = 1 - Q(K-1)

P(K) = Q(K-1) - Q(K) <--- this is what we want to maximize.

Now if the first K-1 all have distinct birthdays, then assuming uniform distribution of birthdays among D days of the year, the K-th person has K-1 chances out of D to match, and D-K+1 chances not to match (which would produce K distinct birthdays). So
Q(K) = Q(K-1)*(D-K+1)/D = Q(K-1) - Q(K-1)*(K-1)/D
Q(K-1) - Q(K) = Q(K-1)*(K-1)/D = Q(K)*(K-1)/(D-K+1)

Now we want to maximize P(K), which means we need the greatest K such that P(K) - P(K-1) > 0. (Actually, as just given, this only guarantees a local maximum, but in fact if we investigate a bit farther we'll find that P(K) has only one maximum.) For convenience in calculation let's set K = I + 1. Then
Q(I-1) - Q(I) = Q(I)*(I-1)/(D-I+1)
Q(I) - Q(I+1) = Q(I)*I/D

P(K) - P(K-1) = P(I+1) - P(I)
= (Q(I) - Q(I+1)) - (Q(K-2) - Q(K-1))
= Q(I)*(I/D - (I-1)/(D-I+1))

To find out where this is last positive (and next goes negative), solve
x/D - (x-1)/(D-x+1) = 0

Multiply by D*(D+1-x) both sides:
(D+1-x)*x - D*(x-1) = 0
Dx + x - x^2 - Dx + D = 0
x^2 - x - D = 0

x = (1 +/- sqrt(1 - 4*(-D)))/2 ... take the positive square root
= 0.5 + sqrt(D + 0.25)

Setting D=365 (finally deciding how many days in a year!),
desired I = x = 0.5 + sqrt(365.25) = 19.612 (approx).

The last integer I for which the new probability is greater then the old is therefore I=19, and so K = I+1 = 20.

...viola!

 huh.gif  huh.gif  huh.gif
my head hurts now.  your "riddles" have gotten way too hard for me to comprehend... i give up
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cleary

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« Reply #81 on: February 12, 2004, 01:47:00 PM »

biggrin.gif
the nba ( national basketball association )
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katsumoto

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« Reply #82 on: February 13, 2004, 06:40:00 AM »

beerchug.gif   Okay, here's another one.  This may be easy.  Let's have fun!   biggrin.gif   Duder Man, don't give up yet.  The fun just begun... tongue.gif  

EPISODE XVII

Some know me as two of twenty-six
Distanced by seven,
Plus the first between odds,
Its left neighbor twice for eleven.

I can travel 3 states in a fraction of an hour,
Some 90 of your air bags, 70 of your thinking power.

What am I?  Good Luck!
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cleary

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« Reply #83 on: February 13, 2004, 05:54:00 PM »

biggrin.gif
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rastareaper

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« Reply #84 on: February 14, 2004, 09:52:00 PM »

i think its water because of this line.

"I can travel 3 states in a fraction of an hour," water can change to gas, liquid or solid in a fraction of an hour.
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rastareaper

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« Reply #85 on: February 14, 2004, 10:06:00 PM »

What is the difference between Michael Jackson and a grocery bag?
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cleary

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« Reply #86 on: February 14, 2004, 10:09:00 PM »

???
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rastareaper

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« Reply #87 on: February 15, 2004, 06:42:00 AM »

ones plastic and hazardous to children and the other holds groceries
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MeNaCe911

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« Reply #88 on: February 15, 2004, 10:22:00 AM »

nitrogen?
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katsumoto

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« Reply #89 on: February 19, 2004, 11:56:00 AM »

rolleyes.gif
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