In this section we investigate whether it is possible to rank individual players as 4BBB players. Example 3. Suppose y = (p, 1 - 2p, p) and x = (p - É›, 1 - p + É›, 0) are two types of players where 0 < É› < p < 0.5. Consider a 4BBB match between pair A with characteristics {x, y} and pair B with characteristics {x, x}. It can be shown that pair A is always the better pair (better since pA > pB), and this therefore suggests that y is a better 4BBB player than x. However, when we consider a 4BBB match between pair A with characteristics {y, x} and pair B with characteristics {y, y}, we find that pair A is the better 4BBB pair provided É› is sufficiently small so as to satisfy the equation p4 - (2 + É›)p3 + (É› + 1)p2 + É›p - É› > 0. For example, when p = 0.2, É› needs to be less than 1/30. Thus, provided É› is sufficiently small, this result suggests that x is a better 4BBB player than y. Therefore, we have, for small É›, one result suggesting that x is a better 4BBB player than y, and another result suggesting that y is a better 4BBB player than x. The truth is that, for small É›, x combines more successfully with y than does either x with x, or y with y. These two examples with p = 0.2 and É› = 0.01 appear in table 1 as examples 3a and 3b respectively. As another example, suppose w = (p + É›, 1 - 2p - 2É›, p + É›), x = (p + É›, 1 - p - É›, 0), y = (p, 1 - 2p, p) and z = (p - É›, 1 - p + É›, 0) are four types of players where 0 < É› < p < 0.5. Consider a 4BBB match between pair A with characteristics {w, x} and pair B with characteristics {z, x}. It can be shown that pair A is always the better pair, suggesting that w is a better 4BBB player than z. However, when we consider a 4BBB match between pair A with characteristics {z, y} and pair B with characteristics {w, y}, we find that pair A is the better 4BBB pair provided É› is sufficiently small. (For example, when p = 0.2, provided É› is less than about 0.017 (to 0.018)). Thus, provided É› is sufficiently small, this second result suggests that z is a better 4BBB player than w. So, again we have a contradiction when É› is small. For the case in which p = 0.2 and É› = 0.01, see examples 3c and 3d in table 1. Thus, it can be concluded that it is not possible to strictly/generally rank individual 4BBB players in an order from best to worst. We will see however that it is possible to rank 4BBB pairs. Thus, it is clear that the sum of the birdie probabilities of the players in a pair of 4BBB players tells us a great deal about the strength of that pair, but it is not the complete picture. The following example highlights this point. Example 4. We start by assuming players B1 and B2 are the same as in example 1, that pair A’s characteristics are {(0.3, 0.5, 0.2), (0.1, 0.7, 0.2)}, rather than {(0.2, 0.6, 0.2), (0.2, 0.6, 0.2)}. Note that the sum of pair A’s birdie probabilities are the same as in example 1, as are the sum of pair A’s par probabilities, and the sum of pair A’s bogie probabilities. However, the interaction between player A1’s and A2’s birdie outcomes is now smaller than before (0.03 instead of 0.04). This reduces the likelihood of players A1 and A2 having birdies at the same time, and this increases their probability of winning an 18-hole match (with playoff) from 0.5 to 0.5238, as can be seen in example 4a of table 1. If pair A’s characteristics change to {(0.3, 0.6, 0.1), (0.1, 0.6, 0.3)}, their probability of winning an 18-hole match against pair B (with playoff) increases further to 0.5398. Note that the probability that both A1 and A2 have a bogie has decreased by 0.01 (from 0.04 to 0.03), as can be seen in example 4b of table 1. If pair A’s characteristics are changed to {(0.3, 0.6, 0.1), (0.1, 0.7, 0.2)}, their probability of winning an 18-hole match against pair B (with playoff) increases further to 0.5560. The probability that both A1 and A2 have a bogie has decreased by 0.02 (from 0.04 to 0.02), as can be seen in example 4d in table 1. If pair A’s characteristics are changed to {(0.3, 0.6, 0.1), (0.1, 0.8, 0.1)}, their probability of winning an 18-hole match (with playoff) increases further to 0.5724, as their probability of both having a bogie has decreased by a further amount 0.01 (from 0.02 to 0.01), as can be seen in example 4e of table 1. Finally, if pair A’s characteristics are changed to {(0.3, 0.6, 0.1), (0.1, 0.9, 0.0)}, their probability of winning an 18-hole match (with playoff) increases further to 0.5888, as can be seen in example 4f of table 1. Example 5. In this example we see that if the ‘gap in the performance characteristics’ between player A and player B is ‘sufficiently large’, player A can be a better 4BBB player than player B no matter what are the common characteristics of player A’s and player B’s partners. Supposing the two partners with common characteristics are denoted by X, pair {A, X1} can be represented by {(pa,-1, pa,0, pa,-1), (x-1, x0, x+1)}, and pair {B, X2} can be represented by {(pb,-1, pb,0, pb,-1), (x-1, x0, x+1)}. The probability pair {A, X1} wins a hole against pair {B, X2}, pA, can be shown to equal A corresponding expression can be written down for pB, the probability pair {B, X2} wins a hole against pair {A, X1}. It can be shown that pA > pB provided Noting that x+1 [0, 1], it can be seen, for example, that player (0.3, 0.6, 0.1) is always a better 4BBB player than player (0.1, 0.8. 0.1), and that player (0.3, 0.4, 0.3) is a better 4BBB player than player (0.1, 0.9, 0.0) provided x+1 for the players with the common characteristics is less than 20/27. Example 6a. As special case of example 5, we consider the important family of ‘symmetric’ players (p, 1-2p, p). Suppose player A has p = p1 and player B has p = p2, where p1 > p2 > 0. Then it can be seen that player A is always a better 4BBB player than player B (unless x+1 = 1, which is the (irrelevant) case in which the partners with the common characteristics always score bogies). Example 6b. As another special case of example 5, we consider the family of ‘equally non-symmetric’ players (θ¸p, 1 - (1 + θ¸)p, p), where θ¸ â‰ 1 and 0 < θ¸ < p-1 - 1. Supposing θ¸ is fixed, player A has p = p1 and player B has p = p2, where p1 > p2 > 0, it can be shown that player A is always a better 4BBB player than B, provided x1 < θ¸ and x-1 ≠1. Further, for all (fixed) θ¸ such that 0 < θ¸ < p-1 - 1 (i.e. the symmetric (θ¸ = 1) as well as the non-symmetric (θ¸ â‰ 1)cases), if players C and D have p-values equal to p3 and p4 where p1 > p2 > p3 > p4 > 0, and p1 + p4 = p2 + p3, it can be shown that the pair (A, D) is always a better 4BBB pair than (B, C). |