In [20]:
%matplotlib inline

In [21]:
# vector addition pt a = (a0,a1), pt b = (b0,b1)
a0,b0 = .5,.1
a1,b1=rand(2,1000)

In [22]:
idx = (a0+a1 > b0+b1)
fig,ax=subplots()
ax.plot(a1[idx],b1[idx],'o',alpha=.1)
ax.plot(a0,b0,'rs',ms=15,alpha=.3)
ax.plot(0,a0-b0,'r^',ms=15,alpha=.3)
ax.plot(1-(a0-b0),1,'r^',ms=15,alpha=.3)
ax.axis((0,1.2,0,1.2))
ax.set_aspect(1)
ax.set_xlabel('a',fontsize=18)
ax.set_ylabel('b',fontsize=18)
ax.plot(linspace(0,2,3),linspace(0,2,3),'k--',lw=3.)
ax.set_title('prob = %3.3f,phat=%3.3f'%(1-(1-a0+b0)**2/2.,idx.mean()))

Out[22]:
<matplotlib.patches.Rectangle at 0x90ffbd0>
In [34]:
b0,b1 = rand(2,1000)
a0,a1 = rand(2,1000)
mean((a0+b0) > (a1+b1))

Out[34]:
0.50800000000000001
In [45]:
idx = (a0>b0)
mean(a0[idx]+a1[idx] > b0[idx]+b1[idx])

Out[45]:
0.76771653543307083
In [33]:
np.logical_and( a0>b0, (a0+a1) > (b0+b1)).mean() # 3/8 is exact solution

Out[33]:
0.38
In [25]:
np.logical_and( a0>b0, (a0+a1) > (b0+b1)).mean()

Out[25]:
0.39400000000000002
In [26]:
(a0>b0).mean()

Out[26]:
0.52000000000000002
In [47]:
import sympy as S

In [52]:
a0,b0 = S.symbols('a0 b0')
expr = S.integrate((1-(1-a0+b0)**2/2),(a0,b0,1))
S.integrate(expr,(b0,0,1)) # prob of (a0>b0) AND (a0+a1>b0+b1)
print 'conditional probability P(a0+a1>b0+b1|a0>b0) = %3.3f' % (float(_)/(0.5))

conditional probability P(a0+a1>b0+b1|a0>b0) = 0.750

In [ ]: