Guesses, metrics, and basketballfolios.rmc.edu/davidclark/wp-content/uploads/sites/... · David...

Guesses, metrics, and basketball

David Clark

Randolph-Macon College

MAA MathFestAugust 4, 2012

David Clark (Randolph-Macon College) Guesses, metrics, and basketball August 4, 2012 1 / 29

Yahoo! Sports Pick ’Em

2011 NCAA Men’s Basketball Bracket:

Breaking a tie

Problem: To break a tie, we need to compare two (or more) final scoreguesses and say which is the best guess.

Scenario 1:Winning team Losing team

All the Singler Ladies 74 62Boom Shaka Laka 86 52Actual Score 85 72

Whose guess is better? Which is “closer” to the actual score?

Point: We want a metric on the set of final scores

S = {(w , `) ∈ N× N : w > `}

Breaking a tie

S = {(w , `) ∈ N× N : w > `}

Breaking a tie

All the Singler Ladies 74 62

Boom Shaka Laka 86 52Actual Score 85 72

S = {(w , `) ∈ N× N : w > `}

Breaking a tie

All the Singler Ladies 74 62Boom Shaka Laka 86 52

Actual Score 85 72

S = {(w , `) ∈ N× N : w > `}

Breaking a tie

S = {(w , `) ∈ N× N : w > `}

Breaking a tie

S = {(w , `) ∈ N× N : w > `}

Breaking a tie

S = {(w , `) ∈ N× N : w > `}

W LAll the Singler Ladies 74 62Boom Shaka Laka 86 52Actual Score 85 72

We can compute that

dm(ASL’s guess,Actual) = 21dm(BSL’s guess,Actual) = 21

We can compute that

Actual

0 20 40 60 80 100

Actualr=21

0 20 40 60 80 100

Actual

0 20 40 60 80 100

Actual

0 20 40 60 80 100

We can compute that

d(ASL’s guess,Actual) =√

221 ≈ 14.9d(BSL’s guess,Actual) =

√401 ≈ 20.0

Using the standard metric, All the Singler Ladies wins!

We can compute that

√401 ≈ 20.0

We can compute that

√401 ≈ 20.0

Scenario 2:W L

Again, we compute:

d(ASL’s guess,Actual) = 2√

2 ≈ 2.83d(BSL’s guess,Actual) = 2

√2 ≈ 2.83

Scenario 2:W L

Again, we compute:

√2 ≈ 2.83

Actual

0 20 40 60 80 100

Actual

0 20 40 60 80 100

Actual

60 65 70 75 80

Actual

60 65 70 75 80

Scenario 2:W L

Again, we compute:

√2 ≈ 2.83

We appear to have a tie! But should we?

The sports gambling world

There are two common ways to place a bet on a sporting event:

1 Betting the spread (point differential between winner and loser)2 Betting the over-under (combined score)

Let’s create a distance function with this information, instead of theindividual team scores:

Old: d =√

(∆w)2 + (∆`)2

New: d ′ =√

(∆spread)2 +

(∆combined)2

There are two common ways to place a bet on a sporting event:1 Betting the spread (point differential between winner and loser)

2 Betting the over-under (combined score)

Old: d =√

(∆w)2 + (∆`)2

New: d ′ =√

(∆spread)2 +

(∆combined)2

There are two common ways to place a bet on a sporting event:1 Betting the spread (point differential between winner and loser)2 Betting the over-under (combined score)

Old: d =√

(∆w)2 + (∆`)2

New: d ′ =√

(∆spread)2 +

(∆combined)2

Old: d =√

(∆w)2 + (∆`)2

New: d ′ =√

(∆spread)2 +

(∆combined)2

Old: d =√

(∆w)2 + (∆`)2

New: d ′ =√

(∆spread)2 +

(∆combined)2

Old: d =√

(∆w)2 + (∆`)2

New: d ′ =√

ks(∆spread)2 + kc(∆combined)2

A new metric on the set of final scores

Consider two scores (w1, `1) and (w2, `2).

Then define

d ′((w1, `1), (w2, `2)

)=√ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

s1 = w1 − `1 (the spread of the first final score)s2 = w2 − `2 (the spread of the second final score)c1 = w1 + `1 (the total of the first final score)c2 = w2 + `2 (the total of the second final score)

Consider two scores (w1, `1) and (w2, `2). Then define

d ′((w1, `1), (w2, `2)

)=√ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

d ′((w1, `1), (w2, `2)

)=√ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

d ′((w1, `1), (w2, `2)

)=√ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

d((w1, `1), (w2, `2)

∆w2 + ∆`2

d ′((w1, `1), (w2, `2)

(ks + kc)(∆w2 + ∆`2)− 2(ks − kc)∆w∆`

Thus, when ks = kc = 1/2, we have that d = d ′.

Picking ks > kc gives more weight to the spread.Example: If ks = 1 and kc = 1/4, then

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 4 =⇒ ASL wins

Picking kc > ks gives more weight to the combined score.Example: If ks = 1/4 and kc = 1, then

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 2 =⇒ BSL wins

d((w1, `1), (w2, `2)

∆w2 + ∆`2

d ′((w1, `1), (w2, `2)

(ks + kc)(∆w2 + ∆`2)− 2(ks − kc)∆w∆`

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 4 =⇒ ASL wins

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 2 =⇒ BSL wins

d((w1, `1), (w2, `2)

∆w2 + ∆`2

d ′((w1, `1), (w2, `2)

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 4 =⇒ ASL wins

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 2 =⇒ BSL wins

d((w1, `1), (w2, `2)

∆w2 + ∆`2

d ′((w1, `1), (w2, `2)

Picking ks > kc gives more weight to the spread.

Example: If ks = 1 and kc = 1/4, then

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 4 =⇒ ASL wins

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 2 =⇒ BSL wins

d((w1, `1), (w2, `2)

∆w2 + ∆`2

d ′((w1, `1), (w2, `2)

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 4 =⇒ ASL wins

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 2 =⇒ BSL wins

d((w1, `1), (w2, `2)

∆w2 + ∆`2

d ′((w1, `1), (w2, `2)

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 4 =⇒ ASL wins

Picking kc > ks gives more weight to the combined score.

Example: If ks = 1/4 and kc = 1, then

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 2 =⇒ BSL wins

d((w1, `1), (w2, `2)

∆w2 + ∆`2

d ′((w1, `1), (w2, `2)

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 4 =⇒ ASL wins

(ASL) d ′((72, 62), (70, 60)

(BSL) d ′((72, 58), (70, 60)

)= 2 =⇒ BSL wins

An objective choice?

What, besides instinct, can inform our choice of the weights?

In the context of NCAA men’s basketball championship games,which is harder to guess: spread or combined score?

What, besides instinct, can inform our choice of the weights?In the context of NCAA men’s basketball championship games,which is harder to guess: spread or combined score?

Year W L Combined Spread2011 53 41 94 122010 61 59 120 22009 89 72 161 172008 75 68 143 72007 84 75 159 92006 73 57 130 162005 75 70 145 52004 82 73 155 92003 81 78 159 32002 64 52 116 122001 82 72 154 102000 89 76 165 131999 77 74 151 31998 78 69 147 91997 84 79 163 51996 76 67 143 91995 89 78 167 111994 76 72 148 41993 77 71 148 61992 71 51 122 201991 72 65 137 71990 103 73 176 301989 80 79 159 11988 83 79 162 41987 74 73 147 11986 72 69 141 31985 66 64 130 21984 84 75 159 91983 54 52 106 21982 63 62 125 11981 63 50 113 131980 59 54 113 51979 75 64 139 111978 94 88 182 6

Year W L Combined Spread1977 67 59 126 81976 86 68 154 181975 92 85 177 71974 76 64 140 121973 87 66 153 211972 81 76 157 51971 68 62 130 61970 80 69 149 111969 92 72 164 201968 78 55 133 231967 79 64 143 151966 72 65 137 71965 91 80 171 111964 98 83 181 151963 60 58 118 21962 71 59 130 121961 70 65 135 51960 75 55 130 201959 71 70 141 11958 84 72 156 121957 54 53 107 11956 83 71 154 121955 77 63 140 141954 92 76 168 161953 69 68 137 11952 80 63 143 171951 68 58 126 101950 71 68 139 3

Mean 76.6 67.2 143.8 9.4StDev 10.8 9.5 19.3 6.5

What, besides instinct, can inform our choice of the weights?In the context of NCAA men’s basketball championship games,which is harder to guess: spread or combined score?

Looking at all NCAA championship games going back to 1950,we see:

The standard deviation of spreads is 6.5.The standard deviation of combined scores is 19.3.

Thus it appears that combined score is harder to guess!

What, besides instinct, can inform our choice of the weights?In the context of NCAA men’s basketball championship games,which is harder to guess: spread or combined score?Looking at all NCAA championship games going back to 1950,we see:

The standard deviation of spreads is 6.5.

The standard deviation of combined scores is 19.3.

So let’s choose kc > ks . How much bigger?

A natural choice might beto let

=std dev combined

std dev spread

6.5≈ 2.97.

This still doesn’t pin down our weights – we need one more equation.For a (secret1) reason from linear algebra, the other equation will be

kcks =1

Big bonus points if you can figure this out

So let’s choose kc > ks . How much bigger? A natural choice might beto let

=std dev combined

std dev spread

6.5≈ 2.97.

kcks =1

=std dev combined

std dev spread=

≈ 2.97.

kcks =1

=std dev combined

std dev spread=

6.5≈ 2.97.

This still doesn’t pin down our weights – we need one more equation.

For a (secret1) reason from linear algebra, the other equation will be

kcks =1

=std dev combined

std dev spread=

6.5≈ 2.97.

kcks =1

=std dev combined

std dev spread=

6.5≈ 2.97.

kcks =1

1Big bonus points if you can figure this outDavid Clark (Randolph-Macon College) Guesses, metrics, and basketball August 4, 2012 16 / 29

Now we just solve the system

= 2.97 kcks =1

to get ks ≈ 0.29 and kc ≈ 0.86.

After rounding off our weights, we finally have a well-defined metricon the set S of final scores:

d ′((w1, `1), (w2, `2)

0.29(s2 − s1)2 + 0.86(c2 − c1)2

= 2.97 kcks =1

to get ks ≈ 0.29 and kc ≈ 0.86.

d ′((w1, `1), (w2, `2)

0.29(s2 − s1)2 + 0.86(c2 − c1)2

= 2.97 kcks =1

to get ks ≈ 0.29 and kc ≈ 0.86.

d ′((w1, `1), (w2, `2)

0.29(s2 − s1)2 + 0.86(c2 − c1)2

= 2.97 kcks =1

to get ks ≈ 0.29 and kc ≈ 0.86.

d ′((w1, `1), (w2, `2)

0.29(s2 − s1)2 + 0.86(c2 − c1)2

Scenario 2:W L

Here, we compute that

d ′(ASL’s guess,Actual) ≈ 3.71

d ′(BSL’s guess,Actual) ≈ 2.15

So, using our new metric, Boom Shaka Laka wins!

Scenario 2:W L

Actual

60 65 70 75 80

Recall:

d ′((w1, `1), (w2, `2)

ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

So what do our rings look like?

Example: A radius 3 ring about (70, 60) satisfies:√K1((w − 70)2 + (`− 60)2) + K2(w − 70)(`− 60) = 3

Recall:

d ′((w1, `1), (w2, `2)

ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

Recall:

d ′((w1, `1), (w2, `2)

ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

Recall:

d ′((w1, `1), (w2, `2)

ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

Example: A radius 3 ring about (70, 60) satisfies:

√K1((w − 70)2 + (`− 60)2) + K2(w − 70)(`− 60) = 3

Recall:

d ′((w1, `1), (w2, `2)

ks(s2 − s1)2 + kc(c2 − c1)2

(kc + ks)(∆w2 + ∆`2) + 2(kc − ks)∆w∆`

Actual

60 65 70 75 80

Actual

60 65 70 75 80

Actual

60 65 70 75 80

Scenario 3:W L

All the Singler Ladies 85 75

Boom Shaka Laka 105 95Actual Score 95 85

Scenario 3:W L

All the Singler Ladies 85 75Boom Shaka Laka 105 95

Actual Score 95 85

Scenario 3:W L

BSLActual

60 70 80 90 100 110 120

BSLActual

r=10.73

60 70 80 90 100 110 120

BSLActual

r=10.73

60 70 80 90 100 110 120

BSLActual

r=10.73r=30

60 70 80 90 100 110 120

Out on a limb

BSL’s guess is farther from the mean, so it’s a bolder guess.

Point: A bold guess should be rewarded, since it is less likely to be agood guess by accident.

=⇒ Boom Shaka Laka wins!

Thus, we have devised system to deal with d ′ ties.

Out on a limb

Some statistics

P. C. Mahalanobis (1893-1972)

Question: how do you mea-sure the likelihood that a datapoint lies in a given distribu-tion?

Mahalanobis distance:

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ),

where S is the distribution’scovariance matrix.

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ),

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ),

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ)

0 20 40 60 80 100 120

Applications:

Pattern recognitionData miningPredicting proteinstructures

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ)

0 20 40 60 80 100 120

Applications:

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ)

0 20 40 60 80 100 120

Applications:

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ)

0 20 40 60 80 100 120

Applications:

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ)

0 20 40 60 80 100 120

Applications:

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ)

0 20 40 60 80 100 120

Applications:

Some statistics

DM(~x) =√

(~x − ~µ)TS−1(~x − ~µ)

0 20 40 60 80 100 120

Applications:

Some statistics

More generally,

dM(~x , ~y) =√

(~x − ~y)TS−1(~x − ~y)

gives the distance between two random points in the samedistribution.

Some statistics

Actual

0 20 40 60 80 100 1200

Questions

Is there a better metric?

How would a metric change if we knew which teams wereplaying?

How would it change if we were talking about a different event?A different sport?

How is all this related to probability theory?

Is this the way we should be gambling?

Thank You!davidclark@rmc.edu

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