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Application of Stochastic Process to Lottery in Nigeria
1
Olayinka Korede Peter,
2
Odukoya Elijah Ayooluwa,
3
Ajewole Kehinde Peter,
4
Akinyele Thomas Wale
1
Department of Statistics, Federal Polytechnic Ado-Ekiti, Ekiti State, Nigeria
2,4
Department of Statistics, Faculty of Science, Ekiti State University, Ado-Ekiti, Ekiti State, Nigeria
3
Department of Physical Sciences, Mathematics Programme, Landmark University, Omu-Aran, Nigeria
DOI: https://doi.org/10.51583/IJLTEMAS.2025.140500087
Received: 01 June 2025; Accepted: 05 June 2025; Published: 20 June 2025
Abstract: Lottery is one of the most popular forms of gambling which offers gamblers opportunity to win large cash prize for a
relatively low cost. Research on lottery gambling could examine the effect of mediators in the relationships between independent
variables and gambling behaviour. We consider the case of premier lotto and selected six (6) different players at random over a
number of years. The methods used in these researches are Markov Chain and Stationarity of the state process. For every player
in the system, we obtain the long run probability or likelihood of winning if every player continues playing indefinitely. We
obtain the inter state transition probabilities indicating the transition of each player from one year to other.
Keywords: Lottery, Markov Chain, Stationarity, Transition Matrix, Gambling
I. Introduction
Playing the lottery is one of the most popular forms of gambling in Nigeria. Part of the reason for this is that it offers an
opportunity to win a large cash prize for a relatively low cost. Tickets are also much more readily available than other types of
gambling, with places to buy a ticket on most street corners and more recently online in some jurisdictions. In heated debate over
the costs and benefits of gambling, proponents emphasized the economic regeneration, while opponents emphasized the social
costs-particularly in term of increased problem gambling-that increased gambling is argued to bring. In most cases, the debate
tended to be based on polarized opinions based on ethical and/or religious convictions rather than factual evidence. At its heart,
gambling is a rather paradoxical behavior because it is widely known that ‘the house always wins’. Whether you are gambling on
fruit machines, horse racing, Baba Ijebu, blackjack or roulette, the odds will have been meticulously arranged to ensure a steady
profit for the casino or the bookmaker. The only way to achieve this is for the gambler to make a steady loss. So, why do
gamblers and particularly problem gamblers continue to play when the overwhelming likelihood is that they will lose money, by
further understanding the breakdown of self-control in gamblers, this program of research carries important implications for the
treatment of problem gambling, using both pharmacological and psychological therapies. Moreover, the development of objective
task of gambling will provide more valid outcome measures for assessing the effectiveness of new treatments, By understanding
how subtle features of gambling games, future changes in gambling legislation may be in a better position to protect vulnerable
individuals, „Conventional models are already available to assess different replacement strategies for a group of similar
equipment of different ages considering and without considering the time value of money. Here, NPV criterion based on nominal
interest rates does not reflect the real increase in the value of money‟. A replacement model is developed taking into account, the
combined influence of predicted inflation based on real time data for Air Conditioners and the time value of money. Furthermore,
in stable economies maximum weightage may be given to the most recent data and less weightage may be given to the old data.
This is because; the fluctuations in economy may be gradual and not all of a sudden in those economies. The trends are
determined using the developed model and hence they will be reliable. Therefore, the model is intensified further by using
“Weighted Moving Transition Probabilities” technique and the decision is arrived at. WMTP technique, a parsimonious model
that approximates the higher order Markov chain is introduced to consider the spread of sizeable data instead of single period‟s
past data…among researcher that has use markov chain are Wai-Ki Ching, Eric,Fung and Michael (2004) „presented the higher
order Markov Chain model for categorical data sequences,, Liana Cazacioc and Elena Corina Cipu (2004) have developed
transition probabilities for second order and third order Markov Chains. Dastidar , et al. (2010) simulated the trends in rain fall
over Gangetic West Bengal in India during monsoon season with the application of two-state higher order Markov chain models,
Stelios H Zanakis and Martin W Maret (1980), applied Markov process for modeling manpower, Ying-Zi Li, Jin-cang Niu (2009)
„applied Markov chain based forecasting for Power generation of Grid connected Photovoltaic system. Shamsad , et al. (2005),
used first and second order transition probability matrices of Markov chain to predict the time series of wind speed values.
II. Research Method
Stationary Process
A stationary process (or strictly stationary process or strongly stationary process) is a stochastic process whose joint probability
distribution does not change when shifted in time.
Discrete value and continuous value process.
is a discrete value process if the set of all possible values of
at all times t is
a countable set Sx; otherwise,
is a continuous value process. Discrete Time and continuous time process: the stochastic
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process
is a discrete time process if
is determined only for a set of time instants, tn=nT, where T is a constant and n is an
integer; otherwise
is a continuous time process.
Random variable from random processes: consider a sample function
, each
is a sample value of a random variable. We
use
for this random variable. The notation
can be refer to either the random process of the random variable that
correspondence to the value of the random process at time t. Example: in the experiment of the repeatedly rolling a die, let
=
. What is the pmf of
? The random variable X
3
is the value of the die roll at time 3 in this case.
Px3 (X) = Ʃ 1/6 x = 1, ……, 6
0 otherwise
Expected value and correlation.
The expected value of process: the expected values of a process X(t) is the deterministic function
(t) = Ʃ(X(t).
_ Auto –Covariance: The autocovariance function of the stochastic process X(t) is Cx(t,r) = Cov[X(t),X(t,r)]
- Autocorrelation: The autocorrelation function of the stochastic process X(t) is RX(t,r) = E[X(t)X(t +r)]
- Autocovariance and Autocorrelations of a Random Sequence:
Cx[m,k] = Cov[X
m
,Xm+k] = Rx[M,k] – E[Xm]E[Xm+k]
Wherem and k are integers, the authocorrelation function of the random sequence X
n
is R
x
[m,k] = E[
+ k]
the input to a digital filter is an i.i.d. random sequence … …,
,
,
,…
with E[
] = 0 and Var[
] = 1. The output is also a random sequence …, Y-
1
, Y
0
, Y
1
…
the relationship between the input sequence and output sequence is expressed in the formula.
=
+
for all integer n.
Find the expected value function E[
] and auto-covariance function C
y
(m,k) of the output
Because
=
+
, we have E[
) = E[
] + E[
]=0. Before calculating
[m,k], we observe that being an random
sequence with E[
] = 0 and Var[
] = 1 implies
C
X
[m,k] = E[
] = { 1 k = 0
{ 0, otherwise
distribution of the time to the kth point in a Poisson process on (0, ∞) with rate .can be obtained as follows
Denote the time to the kth point by Tk. It has a continuous distribution, which is specified by a density function. For t> 0 and
small >0, P{t≤T<t + } = P {exactly k- 1 points in (0,t, exactly one point in (t, t+) } + smaller order terms the ‘smaller order
terms’ contribute probability less than P{2 or more points in (t, t +) { = P { Poisson (π≥2} = e
-
()
2
+ …2!
By the independence property (ii) for Poisson processes, the main term factorizes asp (exactly k – 1 points in [0, t] } P ( exactly
one point in [t, t + )]
= e
-
( t)
k-1
e
- t
()
1
(k – 1)! 1!
= e
- t
k
t
k-1
+ smaller orders terms
(k-1)!
That is, the distribution of T
k
has density
e
- t
k
t
k-1
for t > 0
Markov Chain
A Markov chain is a stochastic process where the probability of the next state given the current state and the entire past depends
only on the current state:
Transition Matrix: The entries in the first row of the matrix P in example 1 represent the probabilities for the various kinds of
weather following a rainy day.
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Using the tradition matrix P, we can write this product as
. The other two events also have probabilities that can be written
as products of entries of P. thus, we have P
(2)
13
=
+
+
.
This equation should remind the reader of a dot product of two vectors; we dotting the first row of P with the third column of P.
this is just what is done in obtaining the 1;3-entry of the product of P with itself. In general, if a Markov chain has r states, then
P(2)ij = Ʃ
r
k=1
pik[kj:
The following general theorem is easy to prove by using the above observation and induction.
Let P be the transition matrix of a Markov chain. The ijth entruy p(n) nij of the matrix Pn gives the probability that the Markov
shain,, staring in state si, will be in state sj after n steps.
Consider again the weather in the Land of Oz. we know that the powers of the transition matrix give us interesting information
about the process as it evolves. We shall be particular interested in the state of the chain after a large number of steps. The
program Matrix power computes the powers of P.
- is the probability of moving from state i to j
if i = j the gambler is stationary.
P=P
=P
n=1,2,…,m≥0 i,j,€ s
Then for the n years we have the transition probability matrix
. . .
. . .
p = . . . .
. . . .
. . . .
. . .
Equilibrium or steady State
The equilibrium state is defined as that condition in the long run where n state has been reached. That is at equilibrium or steady
state, the entisep
n
as n becomes larger and larger, the entry becomes a constant value in which each (column) has the same entry.
Let W be (w
1
, w
2
, . . . w
n
) be the steady state, then it can be shown that wp=w. to prove this, we obtain the following system of
equations;
Where: (W is the steady state distribution Markov stochastic process will be used to obtain transition probabilities and the steady
state probability of the system. The equilibrium / limiting state of the system could be obtained by solving the equation
. . .
. . .
. . . . .
. . . . . = .
. . . . . .
. .
+
+
+. . .+
=
+
+
+. . .+
=
+
+
+. . .+
=
.
.
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.
+
+
+. . .+
=
With a condition that
Presentation and Analysis of Data
The Method Described in The Previous Chapter Will Be to Analyseddata Presented Below
=
0.100
+ 0.104
+0.100
+ 0.094
+ 0.092
+ 0.512
=
(i)
0.107
+ 0.115
+ 0.106
+ 0.096
+ 0.092
+ 0.484
=
(ii)
0.102
+ 0.104
+ 0.099
+ 0.125
+ 0.076
0.494
=
(iii)
0.134
+ 0.107
+ 0.098
+ 0.099
+ 0.087
+ 0.475
=
(iv)
0.112
+ 0.104
+ 0.100
+ 0.085
+ 0.074
+ 0.446
=
(v)
0.109
+ 0.087
+ 0.106
+ 0.101
+ 0.100
+ 0.498
=
(vi)
With the condition that
+
+
+
+
+
=1 (vii)
Replacing one of the systems of linear equations by equation (vii) and solving.
=
=
0.100
+ 0.104
=
0.107
+ 0.115
=
+
= 0.1111
= 0.1111-
Subtracting equation (iii) from (i)
0.100
– 0.107
+ 0.104
– 0.115
=
-
- 0.007
– 0.011
=
-
Substitute for
-0.007 (0.1111-
) – 0.011
= 0.1111-2
-0.00078 + 0.0007
-0.011
= 0.1111-2
-0.00
+ 2
= 0.1111 + 0.00078
=
= 0.05605
= 0.1111 – 0.05605
= 0.05505
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=
0.100
+ 0.094
=
(i)
0.106
+ 0.096
=
(ii)
+
= 0.1111 (iii)
= 0.1111 -
(iv)
Subtract equation (ii) from (i)
0.100
– 0.107
+ 0.104
– 0.115
=
-
0.100
– 0.106
+ 0.094
-0.096
=
-
0.006
– 0.002
=
-
Substitute for
0.006 (0.1111 -
) – 0.002
= 0.1111-
-
0.000667 + 0.006
– 0.002
= 0.1111 + 0.000667
2.004
= 0.111767
=
= 0.05577
= 0.1111 – 0.05577
=0.05533
=
0.092
+ 0.512
=
(i)
0.092
+ 0.487
=
(ii)
+
= 0.1111 (iii)
= 0.1111 -
(iv)
Subtract equation (ii) from (i)
0.092
– 0.092
+ 0.512
– 0.487
=
-
0.025
=
-
Substitute for
0.025
= 0.1111 -
-
0.025
= 0.1111 - 2
0.25
+ 2
= 0.1111
=
= 0.05486
= 0.1111 – 0.05486
0.1111 – 0.05486
= 0.05624
=
0.102
+ 0.104
=
(i)
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0.134
0.107
=
(ii)
+
= 0.1111 (iii)
= 0.1111 -
(iv)
Subtract equation (ii) from (i)
0.102
– 0.134
+ 0.104
– 0.107
=
0.03
– 0.003
=
-
Substitute for
0.03 (0.1111 -
) – 0.003
= 0.1111 -
-
0.00333 + 0.03
– 0.003
= 0.1111 - 2
0.003333 + 0.027
= 0.1111 + 0.003333
0.027
+ 2
= 0.1111 + 0.003333
2.027
= 0.114433
=
=0.05645
= 0.1111 -
= 0.1111 – 0.05645
= 0.05465
=
0.099
+ 0.125
=
(i)
0.098
+ 0.099
=
. (ii)
+
= 0.1111 (iii)
= 0.1111 -
(iv)
Subtract equation (ii) from (i)
0.099
– 0.098
+ 0.125
– 0.099
=
-
0.001
+ 0.026
=
Substitute for
0.001 (0.1111 -
) + 0.026
= 0.1111 -
-
0.000111 + 0.025
= 0.1111 - 2
0.025
+ 2
= 0.1111 - 2
0.025
+ 2
= 0.1111 – 0.000111
2.025
= 0.110989
=
= 0.054809
= 0.056291
=
0.076
+ 0.494
=
(i)
0.087
+ 0.475
=
(ii)
+
= 0.1111 (iii)
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= 0.111 -
(iv)
Subtract equation (ii) from (i)
0.076
– 0.087
+ 0.494
-0.475
=
-
0.011
+ 0.019
=
-
Substitute for
0.111 (0.1111 -
) + 0.019
= 0.1111 -
-
0.0012221 + 0.011
+ 0.019
= 0.1111 -2
0.0012221 +0.03
= 0.1111 - 2
0.03
+ 2
= 0.1111 + 0.0012221
2.03
= 0.1123221
=
= 0.05533.
= 0.1111 – 0.05533
= 0.05577
=
=
0.112
+ 0.104
=
(i)
+
= 0.1111 (ii)
= 0.1111-
(iii)
Substitute for
in equation (i)
0.112 (0.1111 -
0.0124432 – 0.112
+ 0.104
= 0.1111 -
0.0124432 – 0.008
= 0.1111 -
0.992
= 0.0986568
From equation (iii)
= 0.1111 -
= 0.01165
= 0.09945
=
0.100
+ 0.085
=
(i)
+
= 0.1111 (ii)
= 0.1111 -
(iii)
Substitute for
in equation (i)
0.100 (0.1111 -
) + 0.085
= 0.1111 -
0.01111 – 0.100
+ 0.085
= 0.1111 -
0.01111 – 0.015
= 0.1111 -
- 0.015 +
= 0.1111 – 0.01111
0.985
= 0.09999
=
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= 0.1015
= 0.1111 – 0.1015
= 0.0096
=
0.074
+ 0.446
=
(i)
+
= 0.1111 (ii)
= 0.1111 -
(iii)
Substitute for
in equation (i)
0.074 (0.1111 -
) + 0.446
= 0.1111 -
.
0.0082214 – 0.074
+ 0.446
= 0.1111 -
0.0082214 + 0.372
= 0.1111 – 0.0082214
1.372
= 0.1028786
=
= 0.07498
From equation (iii)
= 0.1111 – 0.07498
= 0.03612
= 0.05505 + 0.05533 + 0.05624 = 0.16662
= 0.05605 + 0.05577 + 0.05486 = 0.16668
= 0.05465 + 0.056291 + 0.05577 = 0.166711
= 0.05645 + 0.054869 + 0.05533 = 0.166589
= 0.056102 + 0.055378 + 0.053355 = 0.05737
= 0.054998 + 0.055378 + 0.057748 = 0.27593
0.9999
III. Discussion of Results
The state of the system was derived by taking observations at different time (t) for every individual player included in the system.
We were able to arrive at a 6×6 probability or stochastic matrix which describes the probability of winning for each player over
the years Transition probability is the chance of transiting from a state i to j and it is shown in the matrix below
When i = j, the state is stationary
= 0.100 indicate the probability that player 1 will win in the first year
= 0.115 indicate the probability that player 2 will win
The condition that the vector (
,
,
,…,
) which is the limiting probability matrix of the system is
= 1
Therefore, haven’t performed the analysis, it shows that
+
+
+
+
+
= 1 i.e
0.16662 + 0.16668 + 0.166711 + 0.166589 + 0.05737 + 0.27593 = 0.9999 which is approximately 1
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= 0.16662 is the probability that in the long run ,player 1 will win while the probability at infinite steps that player 1 will loose
will be 1- 0.16662 = 0.83338
= 0.16668 is the probability that in the long run ,player 2 will win while the probability at infinite steps that player 2 will loose
will be 1- 0.16668 = 0.83332
= 0.166711 is the probability that in the long run ,player 3 will win while the probability at infinite steps that player 3 will
loose will be 1- 0.166711 = 0.833289
= 0.166589 is the probability that in the long run ,player 4 will win while the probability at infinite steps that player 4 will
loose will be 1- 0.166589 = 0.833411
= 0.05737 is the probability that in the long run, player 5 will win while the probability at infinite steps that player 5 will loose
will be 1- 0.05737 = 0.94263
= 0.27593 is the probability that in the long run ,player 6 will win while the probability at infinite steps that player 6 will loose
will be 1- 0.27593 = 0.72407
V. Conclusion
We conclude that every player in the system has a chance of winning in the long-run if they remain in the system. The
probabilities associated with each player differ from one year to the other. It also increases the power of prediction as one who
gambles takes to prediction, this can help in events prediction apart from the prizes one. A gambler’s scope is also widenew
extensively as he reads past events and chances of winning from history to make better forecast
References
1. Abbott,m.w. (2001). What do we know about gambling in New Zealand? Report number seven of the zealand Gaming
Survey. Wellington Department of Internal Affairs
2. Abbott,M.W., Volberg, R.A.,Bellringer, M.E., & Reith, G. (2004).A review of research on aspects of problem gambling.
London: Resonsibility in Gambling Trust.
3. Amey, B. (2001). People’s participation in and attitudes to gaming, 1985-2000. Wellington: Department of internal
Affairs
4. Bernstein,PL. Against the odds: the remarkable of risk. John wiley & sons, New York;1996
5. Doob, J.L. (1953), Stochastic processes. John Wisley & Sons, Newyork.
6. Nowatzi and Willians (2002) identify this as a significant issue for self-exclusion programmes
7. Williams, West, & Simpson (2007) state that self-exclusion programme are only as effective as their ability to monitor
