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The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo. The bids are made simultaneously. If all four bid Hi, then all four lose 1 point each. If three players bid Hi and one bids Lo, then the players bidding Hi gain 1 point each and the player bidding Lo loses 3 points. If two players bid Hi and two bid Lo, then the players bidding Hi gain 2 points each and the players bidding Lo lose 2 points each. If one player bids Hi and three bid Lo, then the player bidding Hi gains 3 points and the players bidding Lo lose 1 point each. If all four bid Lo, then all four gain 1 point each.

Four players Arun, Bankim, Charu, and Dipak played the Hi-Lo game. The following facts are known about their game:

  1. At the end of three rounds, Arun had scored 6 points, Dipak had scored 2 points, Bankim and Charu had scored -2 points each.
  2. At the end of six rounds, Arun had scored 7 points, Bankim and Dipak had scored -1 point each, and Charu had scored -5 points.
  3. Dipak's score in the third round was less than his score in the first round but was more than his score in the second round.
  4. In exactly two out of the six rounds, Arun was the only player who bid Hi.

In how many rounds did Bankim bid Lo?

Entered answer:

Solution

✅ Correct Answer: 4
Slide 1/13

Understanding the set :-

  1. This set tells us about a game called Hi - Lo, having 6 rounds.
  1. Rules for game :-

a) In every round, each player chooses to bid Hi or Lo

b) Let 'H' represents Hi and 'L' represents Lo.

Given if they bid

Case 1: HHHH then all players gets -1 points.

Case 2: HHHL => H gets +1 and L gets -3.

Case 3: HHLL => H gets +2 and L gets -2.

Case 4: HLLL => H gets +3 and L gets -1.

Case 5: LLLL => every player gets +1.

This is a game sand tournament + table set.

Most important thing in games and tournament sets is to read and understand rules of the game clearly.

Maybe you can jot down the rules as a short note in the corner of the paper.

R1R2R3T1R4R5R6T2
A67
B-2-1
C-2-5
D2-1

Here T1 = total of score till round 3

and, T2 = total of scores till round 6.

And, A, B, C, D are Arun, Bankim, Charu, and Dipak respectively.

R1R2R3T1R4R5R6T2
A67
B-2-1
C-2-5
Dxyz2-1

Clue 3 says,

Dipak’s score in the third round was less than his score in the first round but was more than his score in the second round.

Let Dipak's score in round 1 = x

in round 2 = y

and round 3 = z

=> y < z < x

R1R2R3T1R4R5R6T2
A23167
B-2-1
C-2-5
D2-112-1

Now, we know that D has scored a total of 2 from the first three rounds.

So, we can consider three possible cases for the values of x, y, and z:

Case - 1 : (x, y, z) = (3, -3, 2)

In this case the points of A in R1, R3, R2 will be (−1,−2/2,1)(-1, -2/2 , 1) in any possible combination the sum will not be 6.

So, this case is invalid.

Case - 2 : (x, y, z) = (2, -1, 1)

In this case the points of A in R1, R3, R2 will be (−2/2,−3/1,−1/3)(-2/2, -3/1, -1/3)

so, if the points in R1, R3, R2 are (2,1,3)(2,1,3)

Therefore, the case is valid and no other cases are possible.

Case - 3 : (x, y, z) = (3, -2, 1)

In this case the points of A in R1, R3, R2 will be (−1,1/−3/1,2/−2)(-1, 1/-3/1, 2/-2) in any possible combination the sum will not be 6.

So, this case is invalid.

.'. Points of A,D in (R1,R2,R3) are (2,3,1) and (2,-1,1) respectively.

R1R2R3T1R4R5R6T2
A23 - Hi167
B-1 - Lo-2-1
C-1 - Lo-2-5
D2-1- Lo12-1

Since A got +3 in R2,

=> he is only the one to bid Hi in R2

and points of B and C in round 2 are (−1,−1)(-1,-1)

=> they bid Lo, Lo.

R1R2R3T1R4R5R6T2
A2- Hi3 - Hi167
B-2 - Lo-1 - Lo-2-1
C-2 - Lo-1 - Lo-2-5
D2 - Hi-1- Lo12-1

Since A and D got 2 points each in R1,

=> C and B must have got -2, -2

i.e they bid Lo, Lo.

R1R2R3T1R4R5R6T2
A2- Hi3 - Hi1 - Lo67
B-2 - Lo-1 - Lo1 - Lo-2-1
C-2 - Lo-1 - Lo1 - Lo-2-5
D2 - Hi-1- Lo1 - Lo2-1

Also, Since A and D got 1 point in R3,

=> C and B must also have got 1 in R3

i.e they bid Lo, Lo.

R1R2R3T1R4R5R6T2
A2- Hi3 - Hi1 - Lo67
B-2 - Lo-1 - Lo1 - Lo-2-1
C-2 - Lo-1 - Lo1 - Lo-2-5
D2 - Hi-1- Lo1 - Lo2-1

Now, we know,

  1. for A - R1 + R2 + R3 = 1

  2. For B - R1 + R2 + R3 = 1

  3. For C - R1 + R2 + R3 = -3

  4. For D - R1 + R2 + R3 = -3

R1R2R3T1RxRyRzT2
A2- Hi3 - Hi1 - Lo63 - Hi7
B-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-1
C-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-5
D2 - Hi-1- Lo1 - Lo2-1 - Lo-1

Clue 4 says,

In exactly two out of the six rounds, Arun was the only player who bid Hi.

Let A bid Hi in R.x

(Here Rx is any of round 4, 5, or 6 - as we do not know the exact information of these three rounds)

=> B,C,D bid Lo.

R1R2R3T1RxRyRzT2
A2- Hi3 - Hi1 - Lo63 - Hi7
B-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-1
C-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-5
D2 - Hi-1- Lo1 - Lo2-1 - Lo-1

For A,

R.x + R.y + R.z = 1

=> R.y + R.z = -2

For B,

R.x + R.y + R.z = 1

=> R.y + R.z = 2

For C,

R.x + R.y + R.z = -3

=> R.y + R.z = -2

For D,

R.x + R.y + R.z = -3.

=> R.y + R.z = -2

Therefore, (R.y, R.z) for A can be (-3,1) or (-1,-1).

R1R2R3T1RxRyRzT2
A2- Hi3 - Hi1 - Lo63 - Hi7
B-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-1
C-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-5
D2 - Hi-1- Lo1 - Lo2-1 - Lo-1

Case - 1 :

If for A, (R.y, R.z)=(-3,1)

Since for both C,D: R.y+R.z=-2

We can't get any combination such that the total points of B,C,D are obtained.

R1R2R3T1RxRyRzT2
A2- Hi3 - Hi1 - Lo63 - Hi-1 - Lo-1 - Hi7
B-2 - Lo-1 - Lo1 - Lo-2-1 - Lo3 - Hi-1 - Hi-1
C-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-1- Lo-1- Hi-5
D2 - Hi-1- Lo1 - Lo2-1 - Lo-1 - Lo-1 - Hi-1

Case - 2:

If for A, (R.y, R.z)=(-1,-1).

the (R.y, R.z) of B,C,D can be (3,-1), (-1,-1), (-1,-1)

and they must have bid (H,H), (L,H), (L,H) respectively

while A must have bid (L, H)

Hence this case is valid.

R1R2R3T1RxRyRzT2
A2- Hi3 - Hi1 - Lo63 - Hi-1 - Lo-1 - Hi7
B-2 - Lo-1 - Lo1 - Lo-2-1 - Lo3 - Hi-1 - Hi-1
C-2 - Lo-1 - Lo1 - Lo-2-1 - Lo-1- Lo-1- Hi-5
D2 - Hi-1- Lo1 - Lo2-1 - Lo-1 - Lo-1 - Hi-1

Bankim bid Lo in R1, R2, R3, and one of R4/5 or 6

=> in 44 rounds

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