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 Dipak gain exactly 1 point?

The correct answer is 1.

To find the number of rounds in which Dipak gained exactly 1 point, we analyze the bidding configurations and points scored round-by-round based on the rules and constraints.

1. Analyzing the Round Scoring Totals
Let us look at the total points distributed among all four players in any single round based on the bid combinations:
- If 4 players bid Hi:
$4\times (-1)=-4$
points.
- If 3 players bid Hi and 1 bids Lo:
$3\times \left(1\right)+1\times (-3)=0$
points.
- If 2 players bid Hi and 2 bid Lo:
$2\times \left(2\right)+2\times (-2)=0$
points.
- If 1 player bids Hi and 3 bid Lo:
$1\times \left(3\right)+3\times (-1)=0$
points.
- If 4 players bid Lo:
$4\times \left(1\right)=+4$
points.

2. Deducing Rounds 1, 2, and 3
At the end of Round 3, the cumulative scores are:
- Arun: 6 points
- Dipak: 2 points
- Bankim: -2 points
- Charu: -2 points
The sum of all players' scores at the end of Round 3 is:
$6+2+(-2)+(-2)=4$
Since the round sums can only be -4, 0, or +4, the only way to get a total sum of +4 over three rounds is to have one round of "All Lo" (sum +4) and two rounds of sum 0.

According to Fact 3, Dipak's score in the third round was less than in the first round but more than in the second round. Let Dipak's scores in Rounds 1, 2, and 3 be
$D_1$
,
$D_2$
, and
$D_3$
respectively:
$D_1>D_3>D_2$
Since the cumulative score of Dipak after 3 rounds is 2 points:
$D_1+D_2+D_3=2$
This condition is uniquely satisfied by the individual round scores:
$D_1=2,D_2=-1,D_3=1$
This gives:
$2>1>-1$
which is consistent with the constraint. Let us determine the bids in these rounds:
- Round 3 (Dipak scores +1): Since all players must get +1 in an "All Lo" round, this round is indeed the "All Lo" round (LLLL). All four players (Arun, Bankim, Charu, Dipak) bid Lo and earn +1 point each.
- Round 1 (Dipak scores +2): For Dipak to score +2, the bidding combination must be "2 Hi, 2 Lo" (HHLL), where Arun and Dipak bid Hi (gaining +2 points each), and Bankim and Charu bid Lo (losing 2 points each).
- Round 2 (Dipak scores -1): For Arun to reach a total of 6 points after 3 rounds, he must score:
$6-(2+1)=3$
points in Round 2. This corresponds to the "1 Hi, 3 Lo" scenario (HLLL), where Arun is the only player to bid Hi (gaining +3 points), and Bankim, Charu, and Dipak bid Lo (losing 1 point each).

3. Deducing Rounds 4, 5, and 6
The cumulative scores at the end of Round 6 are:
- Arun: 7 points (gains
$7-6=1$
point in Rounds 4–6)
- Bankim: -1 point (gains
$-1-(-2)=1$
point in Rounds 4–6)
- Dipak: -1 point (gains
$-1-2=-3$
points in Rounds 4–6)
- Charu: -5 points (gains
$-5-(-2)=-3$
points in Rounds 4–6)
The sum of scores gained in Rounds 4–6 is:
$1+1+(-3)+(-3)=-4$
To get a sum of -4 over these three rounds, we must have one round of "All Hi" (HHHH) with a sum of -4, and two rounds of sum 0.

According to Fact 4, Arun was the only player to bid Hi in exactly two rounds. We already found that Arun was the only player to bid Hi in Round 2. Therefore, there must be exactly one round among Rounds 4, 5, and 6 where Arun is the only player who bids Hi (HLLL, earning +3 points, while the others earn -1 point each). Let us calculate the scores of the other round of sum 0:
For Arun:
$-1(\mathrm{All}\mathrm{Hi})+3(\mathrm{Arun}\mathrm{only}\mathrm{Hi})+a=1\Rightarrow a=-1$
For Bankim:
$-1(\mathrm{All}\mathrm{Hi})-1(\mathrm{Arun}\mathrm{only}\mathrm{Hi})+b=1\Rightarrow b=+3$
For Charu:
$-1(\mathrm{All}\mathrm{Hi})-1(\mathrm{Arun}\mathrm{only}\mathrm{Hi})+c=-3\Rightarrow c=-1$
For Dipak:
$-1(\mathrm{All}\mathrm{Hi})-1(\mathrm{Arun}\mathrm{only}\mathrm{Hi})+d=-3\Rightarrow d=-1$
Thus, the third round in this period has the scores: Arun (-1), Bankim (+3), Charu (-1), and Dipak (-1). This corresponds to a round where Bankim is the only player bidding Hi.

4. Summary of Dipak's Points Gained
The points gained by Dipak in each of the six rounds are:
- Round 1: +2 points
- Round 2: -1 point
- Round 3: +1 point
- Round 4: -1 point
- Round 5: -1 point
- Round 6: -1 point
(Note: Rounds 4, 5, and 6 can be in any order, with scores of -1, -1, and -1).

Dipak gained exactly 1 point only in Round 3. Therefore, the number of rounds in which Dipak gained exactly 1 point is 1.