# Admittance to a Tennis Club

Below I apply the notion of mathematical expectation to the example of a novice player seeking admittance to a tennis club.

To be admitted, the fellow had to beat in two successive games members G (good) and T (top) of the club. With probabilities g and t (t < g) of winning against G and T, the fellow had to choose between to possible orders of games: GTG or TGT. Paradoxically, the second choice appeared to be preferable gaining the fellow the membership with the probability gt(2 - t) against the smaller gt(2 - g) for the sequence GTG.

We shall be looking for the expected number of wins. Using L for a loss and W for a win for the aspiring novice, we shall consider two sample spaces. Following Havil, the space consists of 8 possible outcomes of a sequence of three games:

LLL, LLW, LWL, LWW, WLL, WLW, WWL, WWW

However note that in the sequences LLL, LLW, WLL, WLW the third game is superfluous as the result of the first two make it impossible for the fellow to win two successive games, whereas the third game is unnecessary in the last two sequences WWL, WWW because the two first wins already gain the fellow admittance to the club. This makes possible and reasonable to consider a smaller sample space:

LL, LWL, LWW, WL, WW

For the sequence TGT we have the following probabilities:

Win/Loss sequence Probability LLL (1 - t)(1 - g)(1 - t) LLW (1 - t)(1 - g)t LWL (1 - t)g(1 - t) LWW (1 - t)gt WLL t(1 - g)(1 - t) WLW t(1 - g)t WWL tg(1 - t) WWW tgt

for the first sample space and

Win/Loss sequence Probability LL (1 - t)(1 - g) LWL (1 - t)g(1 - t) LWW (1 - t)gt WL t(1 - g) WW tg

for the second. In both cases, the probabilities add up to 1, as required. Choosing the easier way out, we verify this only for the latter:

 (1 - t)(1 - g) + (1 - t)g(1 - t) + (1 - t)gt + t(1 - g) + tg = (1 - t)(1 - g) + [(1 - t)g(1 - t) + (1 - t)gt] + t(1 - g) + tg = (1 - t)(1 - g) + (1 - t)g + t(1 - g) + tg = [(1 - t)(1 - g) + t(1 - g)] + [(1 - t)g + tg] = (1 - g) + g = 1.

Now we introduce the random variable N that denotes the number of wins for the candidate. In the first case, N may be 0, 1, 2, or 3; in the second case, the are only three possible values: 0, 1, 2. The expectations E1 and E2 are

 E1(N, TGT) = 0·(1 - t)(1 - g)(1 - t) + 1·(1 - t)(1 - g)t + 1·(1 - t)g(1 - t) + 2·(1 - t)gt + 1·t(1 - g)(1 - t) + 2·t(1 - g)t + 2·tg(1 - t) + 3·tgt = 2t + g

and, correspondingly,

 E2(N, TGT) = 0·(1 - t)(1 - g) + 1·(1 - t)g(1 - t) + 2·(1 - t)gt + 1·t(1 - g) + 2·tg = t + g + tg - t2g.

Similarly,

E1(N, GTG) = t + 2g and
E2(N, GTG) = t + g + tg - tg2.

Since t < g, we see that

E1(N, TGT) < E1(N, GTG),

as expected (pun intended). We also have

E2(N, TGT) < E2(N, GTG),

which ameliorates the paradoxical situation that arose from the pure count of probabilities. Although, the probability of gaining the membership playing the top guy first is larger than when playing first just a good member, the expected number of the wins is greater when postponing the confrontation with the top player.

### References

1. J. Havil, Nonplussed!, Princeton University Press, 2007  