3人ゲームの混合戦略ナッシュ均衡を求める

1. ήʔϜཧ࿦#4*$ԋशͷิ଍
ઓུ‫ܗ‬ήʔϜɿਓήʔϜͷࠞ߹ઓུφογϡ‫ߧۉ‬

2. ࠞ߹ઓུφογϡ‫ߧۉ‬ͷఆٛ
༗༻ͳఆཧ
ਓήʔϜͷφογϡ‫ߧۉ‬

3. ࠞ߹ઓུφογϡ‫ߧۉ‬
ϓϨΠϠʔ Ҏ֎ͷࠞ߹ઓུΛ ͱ͠‫ݻ‬ఆ͢Δ
Λຬͨ͢ͱ͖ ͸ ʹର͢Δ࠷ద൓Ԡઓུͱ͍͏
ઓུͷ૊ ͕ࠞ߹ઓུφογϡ‫͋Ͱߧۉ‬Δͱ͸
͢΂ͯͷϓϨΠϠʔʹରͯ͠
Λຬͨ͢ͱ͖Λ͍͏
ήʔϜཧ࿦#4*$ୈճࢀর
i p−i
∈ P−i
Fi
(p*i
, p−i
) ≥ Fi
(pi
, p−i
), ∀pi
∈ Pi
p*i
p−i
(p*1
, ⋯, p*n
) ∈ P
Fi
(p*i
, p*−i
) ≥ Fi
(pi
, p*−i
), ∀pi
∈ Pi

4. ࠞ߹ઓུʹؔ͢Δఆཧ
ఆཧ
ࠞ߹֦େͨ͠ήʔϜ ʹ͓͍ͯ
ͭҎ্ͷ७ઓུʹର͠
ਖ਼ͷ֬཰Λ༩͑Δࠞ߹ઓུ ͕
ଞͷϓϨΠϠʔͷઓུͷ૊ ʹରͯ͠࠷ద൓Ԡͱ͢Δ
͜ͷͱ͖ͦͷਖ਼ͷ֬཰Ͱ༻͍ΒΕΔ७ઓུ΋ଞͷϓϨΠϠʔͷઓུͷ૊ ʹର͢Δ࠷ద൓ԠͱͳΔ
ήʔϜཧ࿦#4*$ԋशୈճࢀর
(N, {Pi
}i∈N, {Fi
}i∈N) pi
p−i
p−i

5. ҎԼͷήʔϜʹ͓͚Δࠞ߹ઓུφογϡ‫ߧۉ‬Λ‫ٻ‬ΊΑ
ਓήʔϜͷφογϡ‫ߧۉ‬
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B

6. ҎԼͷήʔϜʹ͓͚Δࠞ߹ઓུφογϡ‫ߧۉ‬Λ‫ٻ‬ΊΑ
७ਮઓུφογϡ‫ߧۉ‬͸ B
C
D
͓Αͼ B
C
D
ਓήʔϜͷφογϡ‫ߧۉ‬
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B

7. ҎԼͷήʔϜʹ͓͚Δࠞ߹ઓུφογϡ‫ߧۉ‬Λ‫ٻ‬ΊΑ
֤ϓϨΠϠʔͷࠞ߹ઓུΛ
ͱ͢Δ
͜ͷͱ͖ͷϓϨΠϠʔͷ‫ظ‬଴རಘؔ਺͸
p1
= (p,1 − p) p2
= (q,1 − q) p3
= (r,1 − r) p, q, r ∈ [0,1]
F1
(p1
, p2
, p3
) = 1 ⋅ pqr + 2 ⋅ p(1 − q)r + 2 ⋅ (1 − p)(1 − q)r + 1 ⋅ pq(1 − r) + 3 ⋅ (1 − p)q(1 − r)
= q(3r − 2)p − 5rq + 2r + 3q
ਓήʔϜͷφογϡ‫ߧۉ‬
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
p
1 − p
q 1 − q
r 1 − r
Ͱ੔ཧ
p

8. ҎԼͷήʔϜʹ͓͚Δࠞ߹ઓུφογϡ‫ߧۉ‬Λ‫ٻ‬ΊΑ
֤ϓϨΠϠʔͷࠞ߹ઓུΛ
ͱ͢Δ
͜ͷͱ͖ͷϓϨΠϠʔͷ‫ظ‬଴རಘؔ਺͸
p1
= (p,1 − p) p2
= (q,1 − q) p3
= (r,1 − r) p, q, r ∈ [0,1]
F2
(p1
, p2
, p3
) = 1 ⋅ pqr + 2 ⋅ (1 − p)(1 − q)r + 2 ⋅ (1 − p)q(1 − r) + 1 ⋅ (1 − p)(1 − q)(1 − r)
= (4pr − p − 3r + 1)q + 1 − p + r − pr
ਓήʔϜͷφογϡ‫ߧۉ‬
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
p
1 − p
q 1 − q
r 1 − r
Ͱ੔ཧ
q

9. ҎԼͷήʔϜʹ͓͚Δࠞ߹ઓུφογϡ‫ߧۉ‬Λ‫ٻ‬ΊΑ
֤ϓϨΠϠʔͷࠞ߹ઓུΛ
ͱ͢Δ
͜ͷͱ͖ͷϓϨΠϠʔͷ‫ظ‬଴རಘؔ਺͸
p1
= (p,1 − p) p2
= (q,1 − q) p3
= (r,1 − r) p, q, r ∈ [0,1]
F3
(p1
, p2
, p3
) = 1 ⋅ pqr + 2 ⋅ (1 − p)qr + 1 ⋅ (1 − p)(1 − q)r + 2 ⋅ pq(1 − r) + 1 ⋅ (1 − p)q(1 − r) + 3 ⋅ p(1 − q)(1 − r)
= (2pq − 4p + 1)r + 3p + q − 2pq
ਓήʔϜͷφογϡ‫ߧۉ‬
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
p
1 − p
q 1 − q
r 1 − r
Ͱ੔ཧ
r

10.
ఆཧͷར༻Λߟ͑Δ
΋ࠞ͠߹ઓུ
͕φογϡ‫͋Ͱߧۉ‬Γ
͔ͭ
Λຬͨ͢ͷͰ͋Ε͹
֤ϓϨΠϠʔͷਖ਼ͷ֬཰Ͱ࢖ΘΕΔ७ઓུ΋ಉ͡‫ظ‬଴རಘΛ༩͑ΔͷͰ
ҎԼͷ͕ࣜಉ࣌ʹ੒Γཱͭɿ
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p1
= (p,1 − p) p2
= (q,1 − q) p3
= (r,1 − r)
p ≠ 0, 1 q ≠ 0, 1 r ≠ 0, 1
1 ⋅ qr + 2 ⋅ (1 − q)r + 1 ⋅ q(1 − r) = 2 ⋅ (1 − q)r + 3 ⋅ q(1 − r) ⇔ q(3r − 2) = 0
1 ⋅ pr + 2 ⋅ (1 − p)(1 − r) = 2 ⋅ (1 − p)r + 1 ⋅ (1 − p)(1 − r) ⇔ 4pr − p − 3r + 1 = 0
1 ⋅ pq + 2 ⋅ (1 − p)qr + 1 ⋅ (1 − p)(1 − q) = 2 ⋅ pq + 1 ⋅ (1 − p)q + 3 ⋅ p(1 − q) ⇔ 2pq − 4p + 1 = 0
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
Λऔͬͨͱ͖ͷརಘ
a1 Λऔͬͨͱ͖ͷརಘ
a2
Λऔͬͨͱ͖ͷརಘ
b1 Λऔͬͨͱ͖ͷརಘ
b2
Λऔͬͨͱ͖ͷརಘ
c1 Λऔͬͨͱ͖ͷརಘ
c2

11.
ఆཧͷར༻Λߟ͑Δ
͔ͭ ΑΓ
͔ͭ ΑΓ
͔ͭ ΑΓ
ͱͳΓ
ʹໃ६
Ώ͑ʹ
Λຬͨ͢Α͏ͳࠞ߹ઓུφογϡ‫ߧۉ‬͸ଘࡏ͠ͳ͍
ͭ·Γ
͍ͣΕ͔ͷϓϨΠϠʔ͕
७ઓུ Λදࠞ͢߹ઓུ
ΛऔΔΑ͏ͳφογϡ‫ߧۉ‬ͷΈ͕ଘࡏ͢Δ
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
q(3r − 2) = 0 q ≠ 0 3r − 2 = 0 ⇔ r =
2
3
4pr − p − 3r + 1 = 0 r =
2
3
4p
2
3
− p − 3 ⋅
2
3
+ 1 = 0 ⇔ p =
3
5
2pq − 4p + 1 = 0 p =
3
5
2
3
5
q − 4 ⋅
3
5
+ 1 = 0 ⇔ q =
7
6
1 q ∈ (0,1)
p ≠ 0, 1 q ≠ 0, 1 r ≠ 0, 1
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r

12.
ϓϨΠϠʔ͕ ΛऔΔͷ͕࠷దͱͳΔ৔߹ɿ ͷ৔߹
ͳΒ͹
͕࠷ద
ͳΒ͹
೚ҙͷ ͕࠷ద
ͳͷͰ ͕࠷ద
ͨͩ͠
ͷ৚݅΋ߟྀ͢Δͱӈਤͷ࠷ద൓Ԡ‫ۂ‬ઢ͕ॻ͚Δ
·ͨ͸
·ͨ͸
ӈਤΑΓ
φογϡ‫ߧۉ‬͸
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p1
= (1,0) q(3r − 2) ≥ 0
F2
(p1
, p2
, p3
) = (4r − 1 − 3r + 1)q + 1 − 1 + r − 1 ⋅ r = rq
r 0 q = 1
r = 0 q ∈ [0,1]
F3
(p1
, p2
, p3
) = (2 ⋅ 1 ⋅ q − 4 ⋅ 1 + 1)r + 3 ⋅ 1 + q − 2 ⋅ 1 ⋅ q = (2q − 3)r + 3 − q
2q − 3 0, ∀q ∈ [0,1] r = 0
q(3r − 2) ≥ 0
q(3r − 2) ≥ 0 ⇔ q = 0 r =
2
3
q 0, r
2
3
(p1
, p2
, p3
) =
(
(1,0), (0,1), (0,1)
)
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
r
q
1
1
ϓϨΠϠʔ
ϓϨΠϠʔ
2
3
φογϡ‫ߧۉ‬

13.
ϓϨΠϠʔ͕ ΛऔΔͷ͕࠷దͱͳΔ৔߹ɿ ͷ৔߹
ͳΒ͹
͕࠷ద
ͳΒ͹
೚ҙͷ ͕࠷ద
ͳͷͰ ͕࠷ద
ͨͩ͠
ͷ৚݅΋ߟྀ͢Δͱӈਤͷ࠷ద൓Ԡ‫ۂ‬ઢ͕ॻ͚Δ
·ͨ͸
·ͨ͸
ӈਤΑΓ
φογϡ‫ߧۉ‬͸
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p1
= (1,0) q(3r − 2) ≥ 0
F2
(p1
, p2
, p3
) = (4r − 1 − 3r + 1)q + 1 − 1 + r − 1 ⋅ r = rq
r 0 q = 1
r = 0 q ∈ [0,1]
F3
(p1
, p2
, p3
) = (2 ⋅ 1 ⋅ q − 4 ⋅ 1 + 1)r + 3 ⋅ 1 + q − 2 ⋅ 1 ⋅ q = (2q − 3)r + 3 − q
2q − 3 0, ∀q ∈ [0,1] r = 0
q(3r − 2) ≥ 0
q(3r − 2) ≥ 0 ⇔ q = 0 r =
2
3
q 0, r
2
3
(p1
, p2
, p3
) =
(
(1,0), (0,1), (0,1)
)
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
r
q
1
1
ϓϨΠϠʔ
ϓϨΠϠʔ
2
3
φογϡ‫ߧۉ‬
φογϡ‫ߧۉ‬

14.
ϓϨΠϠʔ͕ ΛऔΔͷ͕࠷దͱͳΔ৔߹ɿ ͷ৔߹
ͳΒ͹
͕࠷ద ͳΒ͹
೚ҙͷ ͕࠷ద
ͳΒ͹
͕࠷ద
ͷ܎਺͕ͭͶʹਖ਼ͳͷͰ
͕࠷ద
ͨͩ͠
ͷ৚݅΋ߟྀ͢Δͱӈਤͷ࠷ద൓Ԡ‫ۂ‬ઢ͕ॻ͚Δ
·ͨ͸
·ͨ͸
ӈਤΑΓ
φογϡ‫ߧۉ‬͸
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p1
= (0,1) q(3r − 2) ≤ 0
F2
(p1
, p2
, p3
) = (−3r + 1)q + 1 + r
r
1
3
q = 0 r =
1
3
q ∈ [0,1]
r
1
3
q = 1
F3
(p1
, p2
, p3
) = r + q
r r = 1
q(3r − 2) ≤ 0
q(3r − 2) ≤ 0 ⇔ q = 0 r =
2
3
q 0, r
2
3
(p1
, p2
, p3
) =
(
(0,1), (0,1), (1,0)
)
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
r
q
1
1
ϓϨΠϠʔ
ϓϨΠϠʔ
2
3
φογϡ‫ߧۉ‬
1
3
φογϡ‫ߧۉ‬

15.
ϓϨΠϠʔ͕ ΛऔΔͷ͕࠷దͱͳΔ৔߹ɿ ͷ৔߹
ͳΒ͹
͕࠷ద ͳΒ͹
೚ҙͷ ͕࠷ద ͳΒ͹
͕࠷ద
ͳΒ͹
͕࠷ద ͳΒ͹
೚ҙͷ ͕࠷ద ͳΒ͹
͕࠷ద
ͨͩ͠
ͷ৚݅΋ߟྀ͢Δͱӈਤͷ࠷ద൓Ԡ‫ۂ‬ઢ͕ॻ͚Δ
·ͨ͸ ·ͨ͸
ΑΓӈਤަ఺͸ຬͨͣ͞
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p2
= (1,0) 4pr − p − 3r + 1 ≥ 0
F1
(p1
, p2
, p3
) = (3r − 2)p − 3r + 3
r
2
3
p = 1 r =
2
3
p ∈ [0,1] r
2
3
p = 0
F3
(p1
, p2
, p3
) = (−2p + 1)r + p + 1
p
1
2
r = 0 p =
1
2
r ∈ [0,1] p
1
2
r = 1
4pr − p − 3r + 1 ≥ 0
4pr − p − 3r + 1 ≥ 0 ⇔ p =
3
4
p
3
4
, r ≥
p − 1
4p − 3
p
3
4
, r ≤
p − 1
4p − 3
4
1
2
2
3
−
1
2
− 3
2
3
+ 1 = −
1
6
0
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
r
p
1
1
ϓϨΠϠʔ
2
3
1
3
1
2
2
3
3
4
ϓϨΠϠʔ
φογϡ‫Ͱߧۉ‬͸ͳ͍

16.
ϓϨΠϠʔ͕ ΛऔΔͷ͕࠷దͱͳΔ৔߹ɿ ͷ৔߹
Ώ͑ʹ೚ҙͷ ͕࠷ద
ͳΒ͹
͕࠷ద ͳΒ͹
೚ҙͷ ͕࠷ద ͳΒ͹
͕࠷ద
ͨͩ͠
ͷ৚݅΋ߟྀ͢Δͱӈਤͷ࠷ద൓Ԡ‫ۂ‬ઢ͕ॻ͚Δ
·ͨ͸
ӈਤΑΓ
φογϡ‫ߧۉ‬͸ ͱ
ͱ
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p2
= (0,1) 4pr − p − 3r + 1 ≤ 0
F1
(p1
, p2
, p3
) = 2r p ∈ [0,1]
F3
(p1
, p2
, p3
) = (−4p + 1)r + 3p
p
1
4
r = 0 p =
1
4
r ∈ [0,1] p
1
4
r = 1
4pr − p − 3r + 1 ≤ 0
4pr − p − 3r + 1 ≤ 0 ⇔ p
3
4
, r ≤
p − 1
4p − 3
p
3
4
, r ≥
p − 1
4p − 3
(p1
, p2
, p3
) =
((
1
4
,
3
4)
, (0,1), (r,1 − r)
)
, r ∈
[
3
8
,1
]
(p1
, p2
, p3
) =
(
(p,1 − p), (0,1), (1,0)
)
, p ∈
[
0,
1
4]
(p1
, p2
, p3
) =
(
(1,0), (0,1), (0,1)
)
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
r
p
1
1
ϓϨΠϠʔ
ϓϨΠϠʔ
2
3
1
3
1
2
2
3
3
4
φογϡ‫ߧۉ‬
1
4
φογϡ‫ߧۉ‬
φογϡ‫ߧۉ‬
φογϡ‫ߧۉ‬
3
8

17.
ϓϨΠϠʔ͕ ΛऔΔͷ͕࠷దͱͳΔ৔߹ɿ ͷ৔߹
ͳΒ͹
͕࠷ద
ͳΒ͹
೚ҙͷ ͕࠷ద
ͳΒ͹
͕࠷ద ͳΒ͹
೚ҙͷ ͕࠷ద ͳΒ͹
͕࠷ద
ͨͩ͠
ͷ৚݅΋ߟྀ͢Δͱӈਤͷ࠷ద൓Ԡ‫ۂ‬ઢ͕ॻ͚Δ
·ͨ͸
ӈਤΑΓ
φογϡ‫ߧۉ‬͸
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p3
= (1,0) 2pq − 4p + 1 ≥ 0
F1
(p1
, p2
, p3
) = qp − 2q + 2
q 0 p = 1
q = 0 p ∈ [0,1]
F2
(p1
, p2
, p3
) = (3p − 2)q + 2 − 2p
p
2
3
q = 1 p =
2
3
q ∈ [0,1] p
2
3
q = 0
2pq − 4p + 1 ≥ 0
2pq − 4p + 1 ≥ 0 ⇔ p = 0 p 0, q ≥ 2 −
1
2p
(p1
, p2
, p3
) =
(
(p,1 − p), (0,1), (1,0)
)
, p ∈
[
0,
1
4]
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
q
p
1
1
ϓϨΠϠʔ
2
3
ϓϨΠϠʔ
2
3
1
4
1
2
φογϡ‫ߧۉ‬
φογϡ‫ߧۉ‬

18.
ϓϨΠϠʔ͕ ΛऔΔͷ͕࠷దͱͳΔ৔߹ɿ ͷ৔߹
ͳΒ͹
͕࠷ద
ͳΒ͹
೚ҙͷ ͕࠷ద
ͳΒ͹
͕࠷ద ͳΒ͹
೚ҙͷ ͕࠷ద
ͨͩ͠
ͷ৚݅΋ߟྀ͢Δͱӈਤͷ࠷ద൓Ԡ‫ۂ‬ઢ͕ॻ͚Δ
ӈਤΑΓ
φογϡ‫ߧۉ‬͸
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
p3
= (0,1) 2pq − 4p + 1 ≤ 0
F1
(p1
, p2
, p3
) = − 2qp + 3q
q 0 p = 0
q = 0 p ∈ [0,1]
F2
(p1
, p2
, p3
) = (−p + 1)q + 1 − p
p 1 q = 1 p = 1 q ∈ [0,1]
2pq − 4p + 1 ≥ 0
2pq − 4p + 1 ≤ 0 ⇔ p 0, q ≤ 2 −
1
2p
(p1
, p2
, p3
) =
(
(1,0), (0,1), (0,1)
)
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r
q
p
1
1
ϓϨΠϠʔ
2
3
ϓϨΠϠʔ
2
3
1
4
1
2
φογϡ‫ߧۉ‬
φογϡ‫ߧۉ‬

19.
Ҏ্Λ·ͱΊΔͱ
φογϡ‫ߧۉ‬͸
͸ଞͷφογϡ‫·ؚʹߧۉ‬ΕΔ͕͋͑ͯ‫ࡌه‬
F1
(p1
, p2
, p3
) = q(3r − 2)p − 5rq + 2r + 3q
F2
(p1
, p2
, p3
) = (4pr − p − 3r + 1)q + 1 − p + r − pr
F3
(p1
, p2
, p3
) = (2pq − 4p + 1)r + 3p + q − 2pq
(
(1,0), (0,1), (0,1)
) (
(0,1), (0,1), (1,0)
)
(
(p,1 − p), (0,1), (1,0)
)
, p ∈
[
0,
1
4]
((
1
4
,
3
4)
, (0,1), (r,1 − r)
)
, r ∈
[
3
8
,1
]
(
(0,1), (0,1), (1,0)
)
͕७ઓུD
1ʘ2 C C
B
B
͕७ઓུD
1ʘ2 C C
B
B
p
1 − p
q 1 − q
r 1 − r

20. ήʔϜཧ࿦#4*$ԋशͷิ଍
ઓུ‫ܗ‬ήʔϜɿਓήʔϜͷࠞ߹ઓུφογϡ‫ߧۉ‬