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calculate the probability that the current server wins a p-point tiebreak, given server's probability of winning serve and return points, and the current score.
## calculate the probability that the current server wins a best-of-p tiebreak.
## some results shown here:
## http://summerofjeff.wordpress.com/2010/12/04/7-point-tiebreak-win-expectancy-tables/
def fact(x):
if x in [0, 1]: return 1
r = 1
for a in range(1, (x+1)): r = r*a
return r
def ch(a, b):
return fact(a)/(fact(b)*fact(a-b))
def tiebreakProb(s, t, v=0, w=0, p=7):
## calculate the probability that the current server wins a best-of-p tiebreak.
## s = p(server wins service point)
## t = p(current server wins return point)
## v, w = current score
## check if tiebreak is already over:
if v >= p and (v-w) >= 2:
return 1
elif w >= p and (w-v) >= 2:
return 0
else: pass
## re-adjust so that point score is not higher than p;
## e.g., if p=7 and score is 8-8, adjust to 6-6, which
## is logically equivalent
while True:
if (v+w) > 2*(p-1):
v -= 1
w -= 1
else: break
outcomes = {} ## track probability of each possible score
## this is messy and probably not optimal, figuring out
## how many points remain, and how many are on each
## player's serve:
for i in range((p-1)):
remain = p + i - v - w
if remain < 1: continue
else: pass
if remain % 2 == 1:
if (v+w) % 2 == 0: ## sr[rs[sr
if (remain-1) % 4 == 0: ## ...s
svc = (remain+1)/2
ret = (remain-1)/2
else:
svc = (remain-1)/2
ret = (remain+1)/2
else: ## ss[rr[ss[
if (remain-1) % 4 == 0: ## ...s
svc = (remain+1)/2
ret = (remain-1)/2
else:
svc = (remain+1)/2
ret = (remain-1)/2
else:
if (v+w) % 2 == 0: ## sr[rs[sr
svc, ret = remain/2, remain/2
else: ## ss[rr[ss[
svc, ret = (remain-2)/2, (remain-2)/2
if remain % 4 == 0:
svc += 1
ret += 1
else:
svc += 2
## who serves the last point?
if (v+w) % 2 == 0:
## if remain in [1, 4, 5, 8, 9, 12, 13, 16, 17, 20, 21]: ## pattern: remain % 4 in [0, 1]
if (remain % 4) in [0, 1]:
final = s
svc -= 1
else:
final = t
ret -= 1
else:
## if remain in [3, 4, 7, 8, 11, 12, 15, 16, 19, 20]:
if (remain%4) in [3, 0]:
final = t
ret -= 1
else:
final = s
svc -= 1
pOutcome = 0
for j in range(svc+1):
for k in range(ret+1):
if (j+k) == (p - 1 - v):
m = svc - j
n = ret - k
pr = (s**j)*(t**k)*((1-s)**m)*((1-t)**n)*ch(svc,j)*ch(ret,k)*final
pOutcome += pr
else: continue
key = str(p) + str(i)
outcomes[key] = pOutcome
if remain % 2 == 1:
if (v+w) % 2 == 0: ## sr[rs[sr
if (remain-1) % 4 == 0: ## ...s
svc = (remain+1)/2
ret = (remain-1)/2
else:
svc = (remain-1)/2
ret = (remain+1)/2
else: ## ss[rr[ss[
if (remain-1) % 4 == 0: ## ...s
svc = (remain+1)/2
ret = (remain-1)/2
else:
svc = (remain+1)/2
ret = (remain-1)/2
else:
if (v+w) % 2 == 0: ## sr[rs[sr
svc, ret = remain/2, remain/2
else: ## ss[rr[ss[
svc, ret = (remain-2)/2, (remain-2)/2
if remain % 4 == 0:
svc += 1
ret += 1
else:
svc += 2
## probability of getting to (p-1)-(p-1) (e.g. 6-6)
final = 1
x = 0
for j in range(svc+1):
for k in range(ret+1):
if (j+k) == (p - 1 - v):
m = svc - j
n = ret - k
pr = (s**j)*(t**k)*((1-s)**m)*((1-t)**n)*ch(svc,j)*ch(ret,k)*final
x += pr
else: continue
outcomes['+'] = (x*s*t)/((s*t) + (1-s)*(1-t))
## add up all positive outcomes
wtb = 0
for z in outcomes:
wtb += outcomes[z]
return wtb
@JeffSackmann

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JeffSackmann commented Apr 17, 2015

This (and other tennis code) now lives in this repo:
https://github.com/JeffSackmann/tennis_misc

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