330 
Francis B. Sumner 
Now in any given throw, the chance for either a »head« or » 
tail« is of course 1 out of 2 Our binomial thus becomes 
11 l\n 
This, upon expansion and simplification, resolves it self into: 
,(-i,)vi"p?i(-^)v"i»4i^>(i)', ..c. 
As a concrete illustration, suppose that the number of throws^ 
[n) = 4, then our equation becomes: 
The successive coefficients (1, 4, 6, 4 and 1) represent respectively 
the chances that we shall throw 4 heads, 3 heads + 1 tail, 2 heads 
+ 2 tails, 1 head + 8 tails, and 4 tails 
If, instead of 4 trials, we should take 11 trials, the chances of 
our throwing heads every time would be 1 in 2048; the chances of 
throwing 10 heads and 1 tail would be 11 in 2048; those of throwing 
9 heads and 2 tails would be 55 in 2048, etc. What, now, are the 
chances that we shall obtain as large a proportion of heads as 9 out 
of 11? To find this we determine the collective chances for 11^ 
in A Q I. A ' 1 + 11 + 55 67 • . 1 1 
10 and 9 beads, i. e. 20i8 ' ^048 ' approximately 
The same figure represents the chances that in as large a majority 
of our size-groups as 9 out of 11 the mean figure for a given character 
shall be greater in the warm-room descendants. 
Figure larger 
for Warm-room 
Figure larger 
for Cold-room 
Figures 
equal. 
Chances 
descendants 
descendants 
Tail 
8 
3 
i 
Sexes com- 
bined 
Foot 
Ear 
9 
9 
2 
1 
1 
All three characters 
26 
6 
1 
•2 43 5 
Tail 
14 
5 
Sexes separ- 
Foot 
16 
3 
452 
ately 
Ear 
13 
5 
1 
tV 
All three characters 
43 
13 
1 
I 9^0 2 
1) This may readily be verified by anyone who cares to figure out tlie 
number of possible combinations of H and T in four consecutive throws. We 
have H + H + H-J-H, H-fH+H-fT, etc. etc. It will be found that 16 such 
combinations are possible. 
