The probability of success for player i in competition with player j is also calculated by the support of third-party players for player i’s policies versus player j’s policies. Similar to finding the median voter position,
it is not only about which player’s policies the parties prefer, but also the third-parties’ salience on the issue and their capability or power are considered.
Equation (4) shows the probability of success for player i in competition with player j according to Expected Utility Model [3].
xxxxx
The numerator calculates the expected level of support for i. The denominator calculates the sum of the
support for i and for j so that the expression shows the probability of success for i, and it obviously falls in the
range of 0 and 1.
2.5. Probability of Status Quo (Q)
The calculation of the probability of status quo (Q) has never been talked about in any of Bruce Beuno de
Mesquita’s publications. He considers Q constant in some publications such as [9] in which he considers Q to be
1. This means that when i decides not to challenge j, i and j remain in stalemate with each other and the change
in their positions in responding to other players’ offers does not affect their situation against each other. In
different publications [10] and [11], he considers Q to be 0.5 which means whether the change in the position
affects players situations versus one another or not, is completely random. In this work, we have been able to
calculate this probability for each pair of players. According to Figure 2, when A does not challenge B, B is
challenged by other players and may lose and be forced to move. If B moves, its position changes and its
distance to A either decreases (with probability T) or increases (with probability 1-T). Therefore, the probability
of status quo in this situation is the probability that B does not move at all. This is the probability that B wins the
challenge with every other player except A in that round. This probability is calculted as follows:
The probability that player(i) wins every challenge against another player(k), is the sum of two probabilities.
first, the probability that player(i) challenges player(k) and player k does not challenge it back and surrenders
which is (1 − Sk ) . Second, the probability that player(i) challenges player(k) and player(k) does respond to its
challenge and again player(i) wins this confrontation which is Pi . The probability that player(i) wins against
every other player except player(j), is the multiplication of this sum for all players except player(j).
2.6. Probability of Positive Change (T)
According to Figure 2, when A decides not to challenge B, but B moves due to other challenges, its move either
improves or worsens the situation for A. B move would be towards the median voter position, so the positions of
A, B and the median voter ( µ ) versus one another determines whether B moves closer to A or further away
from it. If B moves closer to A, it improves the situation for A, so T = 1. If B moves further away from A, it