ESD: Recitation #4
Birthday problem
An approximate method
• Bernoulli trials
• Number of trials to compare birthdays of
all people in the class:
n!
n!
N=
=
(n ! 2)!2! 2(n ! 2)!
• Probability that nobody has the same
birthday than someone else:
n!
! N $! 1 $ 0 ! 364 $ N ! 364 $ N ! 364 $ 2(n'2)!
P0 = # &#
&#
& =#
& =#
&
0
" 365 %
" 365 %
" %" 365 % " 365 %
The exact solution
• Probability that nobody has the same
birthday than anybody else:
"
i %
365!
P0 = ($1!
'=
n
#
&
365
365
(365 ! n)!
i= 0
n!1
What was the average travel
distance between two random
points in Budapest in the
1850s?
Budapest = Buda + Pest
Photo removed due to copyright restrictions.
The Danube River through Budapest, showing the two shores.
Only one bridge: Széchenyi Lánchid (Chain bridge)
Source: Wikipedia
Modeling
lP
Pest: λP
wP
u
wB
v
202 m
Buda: λB
lB
Within each city
• In Buda:
PB !B
2
#
&
w B .lB ."B
=%
(
w
.l
.
"
+
w
.l
.
"
$ B B B
P P
P'
1
DB = (w B + lB )
3
• In Pest:
PP !P
2
#
&
w P .lP ."P
=%
(
$ w B .lB ."B + w P .lP ."P '
1
DP = (w P + lP )
3
Between the two cities
• 4 cases:
(1)
(2)
(3)
(4)
Between (1) and (3)
• Probability:
w P .lP ."P .w B .lB ."B
u.v
P(1)!(3) = 2
2 #
(w B .lB ."B + w P .lP ."P ) lP .lB
• Average Distance:
1
1
D(1)!(3) = (w B + v) + (w P + u) + 202
2
2
Between (1) and (4)
• Probability:
w P .lP ."P .w B .lB ."B
u.(lB ! v)
P(1)!(4 ) = 2
2 #
lP .lB
(w B .lB ."B + w P .lP ."P )
• Average Distance:
1
1
D(1)!(4 ) = (w P + u) + (w B + (lB ! v)) + 202
2
2
And continue…
• Between (2) and (3)
• Between (2) and (4)
• Get the final answer…
More complications
• There is currently ten bridges on the
Danube.
• How does average traveling distance
change if we build another one?
Bertrand’s Paradox
Joseph Louis François
Bertrand
(1822-1900)
Wrote Calcul des
probabilités in 1888.
The question
• Consider an equilateral triangle
inscribed in a circle. Suppose a cord of
the circle is chosen at random.
• What is the probability that the chord is
longer than a side of the triangle?
Random endpoints
Figure by MIT OCW.
Random radius
Figure by MIT OCW.
Random midpoints
Figure by MIT OCW.
Barbershop
• One barber, two chairs for waiting customers.
• Prospective customers arrive in a Poisson
manner at the rate of λ per hour.
• It takes the barber 1/µ on average to serve a
customer.
• Prospective customers finding the barbershop
full are lost forever.
• What is the average number of customers?
Model
λ
0
λ
1
µ
λ
2
µ
λ = 2.µ
3
µ
Solving (1)
• What is the probability that N customers
are in the barbershop?
!
!
P1 = P0 ;P2 = P1
µ
µ
" ! %N
PN = $ ' P0
#µ&
!
1)
3
µ
* Pi = 1 ( P0 = " %3+1
!
i= 0
1) $ '
#µ&
Solving (2)
• Average number of customers:
!
1(
" ! %3
23
µ
PN = $ '
3+1 = 4
2 (1
"!%
#µ&
1( $ '
#µ&
3
34
E[Nb _ customers] = ) i.Pi =
* 2.2667
15
i= 0