Suppose that in the process one bets on cn. Cn is called a turning point if one of the following situations appears. (T) One lost the previous bet but wins on the bet with amount equal to cn. (T) One won the previous bet but loses on the bet with amount equal to cn.
Suppose that in the process one bets on cn. Cn is called a turning point if one of the following situations appears. (T) One lost the previous bet but wins on the bet with amount equal to cn. (T) One won the previous bet but loses on the bet with amount equal to cn.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 58E
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please send solution for Q1 complete
![[ Question 1 continues...]
Suppose that in the process one bets on Cn-
the following situations appears.
Cn is called a turning point if one of
(T) One lost the previous bet but wins on the bet with amount equal to Cn.
(T) One won the previous bet but loses on the bet with amount equal to cn .
For example, one starts the game with five consecutive losses [losing $ (1+1+2+
3+5) = $12], and one wins on the next bet of $ c6 = $ 8, making a profit equal to
$ c6 = $8. In this case c6 is the first turning point. Next, one bets with amount equal
to $ c4 = $3, and loses. It follows that c4 is the second turning point. Accordingly,
one bets with $ c3 = $5. Suppose one wins, then c, is the third turning point. One
continues with $ c3 = $2. Again one wins, and then bets with $ c = $1. If one
wins again, one finishes one round and stops. As the payoff equals to 1 : 1, we count
the net profit in this round:
$( 8 + 5 + 2 + 1 – 12 – 3) = $ 16 – $ 15 = $ 1.
In this round, there are three turning points in this example, namely, c6, C4, and
C5 (in order of appearance).
(i) Suppose that the first turning point is at $ CN1+1, where N1 2 1 is an
integer. Show that before the betting with amount equal to the turning point CN1+1,
one has lost
$(Ci + c2 + C3 + .+ CN1).
Here CN, is the N1-th Fibonacci number (the one just behind CN, +1).
( ii)
In a certain round, there are K turning points at
CN1: , CN; , ''', CNK
( in order of appearance).
Here K > 1 is an integer. Using mathematical induction, show that before the bet
with amount equal to the turning point CNK the net lost [after taking into account
on the profit(s) from the win( s)] is given by
$ (c + c2 + c3 + • • • + cNk -1
<-1 ).
- 1)- th Fibonacci number [ the one just behind
Here CNK -1 is the (NK
to (1.1)].
CNK
refer
( iii) Using (ii ), argue that, for the Fibonacci betting strategy described above,
in one round, the net profit is exactly equal to $1.
Note that if you use another method [ not from (ii)], you may not be awarded full
credit to this part.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedb31c39-2ab0-43ac-a5ed-7ff74e15773e%2Fdd974061-f82d-453b-a8d1-26e874f4cb97%2F6yjm98b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[ Question 1 continues...]
Suppose that in the process one bets on Cn-
the following situations appears.
Cn is called a turning point if one of
(T) One lost the previous bet but wins on the bet with amount equal to Cn.
(T) One won the previous bet but loses on the bet with amount equal to cn .
For example, one starts the game with five consecutive losses [losing $ (1+1+2+
3+5) = $12], and one wins on the next bet of $ c6 = $ 8, making a profit equal to
$ c6 = $8. In this case c6 is the first turning point. Next, one bets with amount equal
to $ c4 = $3, and loses. It follows that c4 is the second turning point. Accordingly,
one bets with $ c3 = $5. Suppose one wins, then c, is the third turning point. One
continues with $ c3 = $2. Again one wins, and then bets with $ c = $1. If one
wins again, one finishes one round and stops. As the payoff equals to 1 : 1, we count
the net profit in this round:
$( 8 + 5 + 2 + 1 – 12 – 3) = $ 16 – $ 15 = $ 1.
In this round, there are three turning points in this example, namely, c6, C4, and
C5 (in order of appearance).
(i) Suppose that the first turning point is at $ CN1+1, where N1 2 1 is an
integer. Show that before the betting with amount equal to the turning point CN1+1,
one has lost
$(Ci + c2 + C3 + .+ CN1).
Here CN, is the N1-th Fibonacci number (the one just behind CN, +1).
( ii)
In a certain round, there are K turning points at
CN1: , CN; , ''', CNK
( in order of appearance).
Here K > 1 is an integer. Using mathematical induction, show that before the bet
with amount equal to the turning point CNK the net lost [after taking into account
on the profit(s) from the win( s)] is given by
$ (c + c2 + c3 + • • • + cNk -1
<-1 ).
- 1)- th Fibonacci number [ the one just behind
Here CNK -1 is the (NK
to (1.1)].
CNK
refer
( iii) Using (ii ), argue that, for the Fibonacci betting strategy described above,
in one round, the net profit is exactly equal to $1.
Note that if you use another method [ not from (ii)], you may not be awarded full
credit to this part.
![Question 1
Consider the Fibonacci numbers
C1 = 1, c2 = 1, c3 = 2, C4 = 3, ·…, Cn +1 = Cn+Cn -1, *',
(1.1)
where n > 2 is an integer. In the Fibonacci betting strategy, one fixes a event
E in a game and always bets on the same event E. Here
0 < P(E) < 1,
and the payoff is
1:1.
[ That is, if one puts down one unit and wins, then one gets one unit as profit, and
collects back the one unit which is put down as the wager. On other hand, if one
puts down one unit and loses, then one forfeits the one unit which is put down as the
wager.]
The opening bet is equal to S cq = $1. Thereafter, one follows the procedures
described below [(a) - (h)].
(a) If one wins the opening game, one stops and makes a profit of $ 1. One round
is considered to be finished.
(b) If one loses the bet with $ c = $1, one bets on with the amount equal to the
next Fibonacci number, that is, $ c2 = $1.
(c) In case one wins on the bet of $ c2 = $ 1, one goes back to c and bets with the
amount equal to $c = $1. The process is stopped if one also wins the bet $ c = $1.
One round is considered to be finished. Otherwise, one follows step (b).
(d) In case one loses on the bet of $ c2 = $1, one continues the process, each time
bets with the amounts equal to the subsequent Fibonacci numbers ( c3, C4 , ………,
etc.), until a future win appears - say at the bet with amount equal to $ cn+1 (recall
that here n 2 2 is an integer).
(e) Suppose one wins on the bet of $ cn+1, then one crosses out the numbers cn
and cn-1 in the Fibonacci sequence. Next, one bets with the amount equal to $ cn-1
(n 2 2 is an integer). [ Corrected version. ]
(f) If one loses the next game with bet $ cn -1, one bets with the next Fibonacci
number, that is, $ cn, and continues with step (d) until one wins again, then back
to the instruction in step (e).
(g) If one wins after betting $ Cn -1, one crosses out the numbers cn -2 and cn -3
in the Fibonacci sequence, and next bets with $ cn-3, where n > 4. [ Corrected
version. ]
(h) The process continues until either one comes back to the number c =1 again,
and bets with $ c = $ 1, and wins; or one comes back to the number c2, and wins
the next two rounds with bets $ c2 = $1 and $ c = $1, respectively. In both cases,
after the last winning, one stops the process and finishes one round.
Q. 1 continues on the nert page](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedb31c39-2ab0-43ac-a5ed-7ff74e15773e%2Fdd974061-f82d-453b-a8d1-26e874f4cb97%2Fd2tck7pd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 1
Consider the Fibonacci numbers
C1 = 1, c2 = 1, c3 = 2, C4 = 3, ·…, Cn +1 = Cn+Cn -1, *',
(1.1)
where n > 2 is an integer. In the Fibonacci betting strategy, one fixes a event
E in a game and always bets on the same event E. Here
0 < P(E) < 1,
and the payoff is
1:1.
[ That is, if one puts down one unit and wins, then one gets one unit as profit, and
collects back the one unit which is put down as the wager. On other hand, if one
puts down one unit and loses, then one forfeits the one unit which is put down as the
wager.]
The opening bet is equal to S cq = $1. Thereafter, one follows the procedures
described below [(a) - (h)].
(a) If one wins the opening game, one stops and makes a profit of $ 1. One round
is considered to be finished.
(b) If one loses the bet with $ c = $1, one bets on with the amount equal to the
next Fibonacci number, that is, $ c2 = $1.
(c) In case one wins on the bet of $ c2 = $ 1, one goes back to c and bets with the
amount equal to $c = $1. The process is stopped if one also wins the bet $ c = $1.
One round is considered to be finished. Otherwise, one follows step (b).
(d) In case one loses on the bet of $ c2 = $1, one continues the process, each time
bets with the amounts equal to the subsequent Fibonacci numbers ( c3, C4 , ………,
etc.), until a future win appears - say at the bet with amount equal to $ cn+1 (recall
that here n 2 2 is an integer).
(e) Suppose one wins on the bet of $ cn+1, then one crosses out the numbers cn
and cn-1 in the Fibonacci sequence. Next, one bets with the amount equal to $ cn-1
(n 2 2 is an integer). [ Corrected version. ]
(f) If one loses the next game with bet $ cn -1, one bets with the next Fibonacci
number, that is, $ cn, and continues with step (d) until one wins again, then back
to the instruction in step (e).
(g) If one wins after betting $ Cn -1, one crosses out the numbers cn -2 and cn -3
in the Fibonacci sequence, and next bets with $ cn-3, where n > 4. [ Corrected
version. ]
(h) The process continues until either one comes back to the number c =1 again,
and bets with $ c = $ 1, and wins; or one comes back to the number c2, and wins
the next two rounds with bets $ c2 = $1 and $ c = $1, respectively. In both cases,
after the last winning, one stops the process and finishes one round.
Q. 1 continues on the nert page
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