Late in the fourth inning of the sixth game, the Braves will be beaten.

This is no conjecture. It is statistical fact.

Purely scientifically, let us take p as the probability of the better team's winning an individual game. According to statisticians, Q=1-p, and A is the average number of games won per Series by losers.

It follows, quite naturally, that A=4pq (1 + 2pq + 5p2q2).

Now the average number of games won the Series losers has been 1.5455. To quote from Charles F. Mosteller, professor of Mathematical Statistics:

"If we set this equal to A, we can solve directly a cubic in pq and then a quadratic in p, or we might go directly to the 6th degree equation in p." In any case, the probability of the better team winning an individual game is .65.

It need hardly be mentioned that p remains constant, as an analysis of data through 1951 suggests that "at home" or "away" games made little difference in World Series play,

Now, how often does the better team win? If p is .65, it is obvious--by binomial probability--that .8 is the answer.

If we estimate that the American League has had the better team a fraction of the time x, we can recall that up to 1951 this league had won 31 of 48 Series. Then, the equation 0.80x + 0.20 (1-x)=31/48 must tell us that x=0.743!

The Yankees, however, have had a better record than the AL as a whole--they have won 14 of their last 16 Series.

Extrapolating from these figures that the Bombers are 25 per cent better than the American League as a whole, we are forced to conclude that the Yanks must win just about all of the time, although the Braves will stretch this series to 4 + A, or 5.5455 games, that is, late in the sixth game's fourth inning.