Question

What were the bids by Arun, Bankim, Charu and Dipak, respectively in the first round?

• A. Hi, Lo, Lo, Lo
• B. Lo, Lo, Lo, Hi
• C. Hi, Hi, Lo, Lo
• D. Hi, Lo, Lo, Hi
• A. First
• B. Fourth
• C. Third
• D. Second

Answer 1

The Correct Option is D

Lets follow the table : 

	R1	R2	R3	T1	RX	RY	RZ	T2
A	2 H	3 H	1 L	6	3 H			7
B	-2 L	-1 L	1 L	-2	-1 L			-1
C	-2 L	-1 L	1 L	-2	-1 L			-5
D	2 H	-1 L	1 L	2	-1 L			-1

Given the conditions for each scenario:

• A: R.x+R.y+R.z=1⇒R.y+R.z=−2
• B: R.x+R.y+R.z=1⇒R.y+R.z=2
• C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
• D: R.x+R.y+R.z=−3⇒R.y+R.z=−2

For scenario A, possible values for (R.y,R.z) are:

• (-3, 1)
• (-1, -1)

Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)

• Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.

Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)

• For B, C, D, valid combinations are:
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)

Based on these combinations, the bids should be:

• A: (L, H)
• B: (H, H)
• C: (L, H)
• D: (L, H)

This configuration satisfies all the given conditions and is therefore valid.

	R1	R2	R3	T1	RX	RY	RX	T2
A	2 H	3 H	1 L	6	3 H	-1 L	-1 H	7
B	-2 L	-1 7	1 L	-2	-1 L	3 H	3 H	-1
C	-2 L	-1 L	1 L	-2	-1 L	-1 L	-1 H	-5
D	2 L	-1 L	1 L	2	-1 L	-1 L	-1 H	-1

The bids by Arun, Bankim, Charu and Dipak, respectively in the first round are HLLH

So, the correct option is (D): Hi, Lo, Lo, Hi.

Answer 2

Lets follow the table : 

	R1	R2	R3	T1	RX	RY	RZ	T2
A	2 H	3 H	1 L	6	3 H			7
B	-2 L	-1 L	1 L	-2	-1 L			-1
C	-2 L	-1 L	1 L	-2	-1 L			-5
D	2 H	-1 L	1 L	2	-1 L			-1

Given the conditions for each scenario:

• A: R.x+R.y+R.z=1⇒R.y+R.z=−2
• B: R.x+R.y+R.z=1⇒R.y+R.z=2
• C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
• D: R.x+R.y+R.z=−3⇒R.y+R.z=−2

For scenario A, possible values for (R.y,R.z) are:

• (-3, 1)
• (-1, -1)

Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)

• Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.

Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)

• For B, C, D, valid combinations are:
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)

Based on these combinations, the bids should be:

• A: (L, H)
• B: (H, H)
• C: (L, H)
• D: (L, H)

This configuration satisfies all the given conditions and is therefore valid.

	R1	R2	R3	T1	RX	RY	RX	T2
A	2 H	3 H	1 L	6	3 H	-1 L	-1 H	7
B	-2 L	-1 7	1 L	-2	-1 L	3 H	3 H	-1
C	-2 L	-1 L	1 L	-2	-1 L	-1 L	-1 H	-5
D	2 L	-1 L	1 L	2	-1 L	-1 L	-1 H	-1

Arun bid Hi in R1, R2, R, X, R.z 

So, the correct answer is 4.

Answer 3

Lets follow the table : 

	R1	R2	R3	T1	RX	RY	RZ	T2
A	2 H	3 H	1 L	6	3 H			7
B	-2 L	-1 L	1 L	-2	-1 L			-1
C	-2 L	-1 L	1 L	-2	-1 L			-5
D	2 H	-1 L	1 L	2	-1 L			-1

Given the conditions for each scenario:

• A: R.x+R.y+R.z=1⇒R.y+R.z=−2
• B: R.x+R.y+R.z=1⇒R.y+R.z=2
• C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
• D: R.x+R.y+R.z=−3⇒R.y+R.z=−2

For scenario A, possible values for (R.y,R.z) are:

• (-3, 1)
• (-1, -1)

Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)

• Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.

Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)

• For B, C, D, valid combinations are:
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)

Based on these combinations, the bids should be:

• A: (L, H)
• B: (H, H)
• C: (L, H)
• D: (L, H)

This configuration satisfies all the given conditions and is therefore valid.

	R1	R2	R3	T1	RX	RY	RX	T2
A	2 H	3 H	1 L	6	3 H	-1 L	-1 H	7
B	-2 L	-1 7	1 L	-2	-1 L	3 H	3 H	-1
C	-2 L	-1 L	1 L	-2	-1 L	-1 L	-1 H	-5
D	2 L	-1 L	1 L	2	-1 L	-1 L	-1 H	-1

Bikram bid Lo in R1, R2, R3, R. X. 

So, the correct answer is 4.

Answer 4

Lets follow the table : 

	R1	R2	R3	T1	RX	RY	RZ	T2
A	2 H	3 H	1 L	6	3 H			7
B	-2 L	-1 L	1 L	-2	-1 L			-1
C	-2 L	-1 L	1 L	-2	-1 L			-5
D	2 H	-1 L	1 L	2	-1 L			-1

Given the conditions for each scenario:

• A: R.x+R.y+R.z=1⇒R.y+R.z=−2
• B: R.x+R.y+R.z=1⇒R.y+R.z=2
• C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
• D: R.x+R.y+R.z=−3⇒R.y+R.z=−2

For scenario A, possible values for (R.y,R.z) are:

• (-3, 1)
• (-1, -1)

Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)

• Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.

Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)

• For B, C, D, valid combinations are:
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)

Based on these combinations, the bids should be:

• A: (L, H)
• B: (H, H)
• C: (L, H)
• D: (L, H)

This configuration satisfies all the given conditions and is therefore valid.

	R1	R2	R3	T1	RX	RY	RX	T2
A	2 H	3 H	1 L	6	3 H	-1 L	-1 H	7
B	-2 L	-1 7	1 L	-2	-1 L	3 H	3 H	-1
C	-2 L	-1 L	1 L	-2	-1 L	-1 L	-1 H	-5
D	2 L	-1 L	1 L	2	-1 L	-1 L	-1 H	-1

All the players made identical bids in R3 and R.z

So, the correct answer is 2 round.

Answer 5

Lets follow the table : 

	R1	R2	R3	T1	RX	RY	RZ	T2
A	2 H	3 H	1 L	6	3 H			7
B	-2 L	-1 L	1 L	-2	-1 L			-1
C	-2 L	-1 L	1 L	-2	-1 L			-5
D	2 H	-1 L	1 L	2	-1 L			-1

Given the conditions for each scenario:

• A: R.x+R.y+R.z=1⇒R.y+R.z=−2
• B: R.x+R.y+R.z=1⇒R.y+R.z=2
• C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
• D: R.x+R.y+R.z=−3⇒R.y+R.z=−2

For scenario A, possible values for (R.y,R.z) are:

• (-3, 1)
• (-1, -1)

Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)

• Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.

Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)

• For B, C, D, valid combinations are:
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)

Based on these combinations, the bids should be:

• A: (L, H)
• B: (H, H)
• C: (L, H)
• D: (L, H)

This configuration satisfies all the given conditions and is therefore valid.

	R1	R2	R3	T1	RX	RY	RX	T2
A	2 H	3 H	1 L	6	3 H	-1 L	-1 H	7
B	-2 L	-1 7	1 L	-2	-1 L	3 H	3 H	-1
C	-2 L	-1 L	1 L	-2	-1 L	-1 L	-1 H	-5
D	2 L	-1 L	1 L	2	-1 L	-1 L	-1 H	-1

Deepak got exactly one point in only R3.

So, the correct answer is 1.

Answer 6

Lets follow the table : 

	R1	R2	R3	T1	RX	RY	RZ	T2
A	2 H	3 H	1 L	6	3 H			7
B	-2 L	-1 L	1 L	-2	-1 L			-1
C	-2 L	-1 L	1 L	-2	-1 L			-5
D	2 H	-1 L	1 L	2	-1 L			-1

Given the conditions for each scenario:

• A: R.x+R.y+R.z=1⇒R.y+R.z=−2
• B: R.x+R.y+R.z=1⇒R.y+R.z=2
• C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
• D: R.x+R.y+R.z=−3⇒R.y+R.z=−2

For scenario A, possible values for (R.y,R.z) are:

• (-3, 1)
• (-1, -1)

Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)

• Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.

Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)

• For B, C, D, valid combinations are:
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
  • (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)

Based on these combinations, the bids should be:

• A: (L, H)
• B: (H, H)
• C: (L, H)
• D: (L, H)

This configuration satisfies all the given conditions and is therefore valid.

	R1	R2	R3	T1	RX	RY	RX	T2
A	2 H	3 H	1 L	6	3 H	-1 L	-1 H	7
B	-2 L	-1 7	1 L	-2	-1 L	3 H	3 H	-1
C	-2 L	-1 L	1 L	-2	-1 L	-1 L	-1 H	-5
D	2 L	-1 L	1 L	2	-1 L	-1 L	-1 H	-1

R2 is correct answer

So, the correct answer is (D) : Second.