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MathHardAdv 2022 • Paper 1

Match Probabilities of Player Scores in Two Player Die Rolling Game

Two players, P1P_1 and P2P_2, play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let xx and yy denote the readings on the die rolled by P1P_1 and P2P_2, respectively. If x>yx > y, then P1P_1 scores 55 points and P2P_2 scores 00 point. If x=yx = y, then each player scores 22 points. If x<yx < y, then P1P_1 scores 00 point and P2P_2 scores 55 points. Let XiX_i and YiY_i be the total scores of P1P_1 and P2P_2, respectively, after playing the ithi^{\text{th}} round.

List-IList-II(I) Probability of (X2≥Y2) is(P) 38(II) Probability of (X2>Y2) is(Q) 1116(III) Probability of (X3=Y3) is(R) 516(IV) Probability of (X3>Y3) is(S) 355864(T) 77432\begin{array}{ll} \text{List-I} & \text{List-II} \\ \text{(I) Probability of } (X_2 \ge Y_2) \text{ is} & \text{(P) } \frac{3}{8} \\[4pt] \text{(II) Probability of } (X_2 > Y_2) \text{ is} & \text{(Q) } \frac{11}{16} \\[4pt] \text{(III) Probability of } (X_3 = Y_3) \text{ is} & \text{(R) } \frac{5}{16} \\[4pt] \text{(IV) Probability of } (X_3 > Y_3) \text{ is} & \text{(S) } \frac{355}{864} \\[4pt] & \text{(T) } \frac{77}{432} \end{array}

The correct option is:

Options

A

(I) →\rightarrow (Q); (II) →\rightarrow (R); (III) →\rightarrow (T); (IV) →\rightarrow (S)

Correct
B

(I) →\rightarrow (Q); (II) →\rightarrow (R); (III) →\rightarrow (T); (IV) →\rightarrow (T)

C

(I) →\rightarrow (P); (II) →\rightarrow (R); (III) →\rightarrow (Q); (IV) →\rightarrow (S)

D

(I) →\rightarrow (P); (II) →\rightarrow (R); (III) →\rightarrow (Q); (IV) →\rightarrow (T)

Topics & Concepts

Probability (Advanced)Probability

Step-by-Step Solution

To find the correct matching between List-I and List-II, let us analyze the probability of score differences between player P1P_1 and player P2P_2 in each round.

Step 1: Single Round Probabilities

In any single round, players P1P_1 and P2P_2 roll a fair 6-faced die. The total number of equally likely outcomes is 6×6=366 \times 6 = 36.

  • Case 1: x>yx > y (P1P_1 wins the round) Number of outcomes where x>yx > y is (62)=15\binom{6}{2} = 15. P(x>y)=1536=512P(x > y) = \frac{15}{36} = \frac{5}{12} In this case, P1P_1 gets 5 points and P2P_2 gets 0 points. Score difference ΔD=Xi−Yi=+5\Delta D = X_i - Y_i = +5.

  • Case 2: x=yx = y (Tie) Number of outcomes where x=yx = y is 66. P(x=y)=636=16P(x = y) = \frac{6}{36} = \frac{1}{6} In this case, both get 2 points. Score difference ΔD=Xi−Yi=0\Delta D = X_i - Y_i = 0.

  • Case 3: x<yx < y (P2P_2 wins the round) Number of outcomes where x<yx < y is (62)=15\binom{6}{2} = 15. P(x<y)=1536=512P(x < y) = \frac{15}{36} = \frac{5}{12} In this case, P1P_1 gets 0 points and P2P_2 gets 5 points. Score difference ΔD=Xi−Yi=−5\Delta D = X_i - Y_i = -5.

Let dk=ΔDk5d_k = \frac{\Delta D_k}{5} be the normalized score difference in round kk. Then dk∈{+1,0,−1}d_k \in \{+1, 0, -1\} with probabilities: P(dk=+1)=512,P(dk=0)=16,P(dk=−1)=512P(d_k = +1) = \frac{5}{12}, \quad P(d_k = 0) = \frac{1}{6}, \quad P(d_k = -1) = \frac{5}{12}

Let Si=∑k=1idkS_i = \sum_{k=1}^i d_k be the total normalized score difference after ii rounds. Note that due to symmetry: P(Si>0)=P(Si<0)P(S_i > 0) = P(S_i < 0) Since P(Si>0)+P(Si=0)+P(Si<0)=1P(S_i > 0) + P(S_i = 0) + P(S_i < 0) = 1, we have: P(Si>0)=1−P(Si=0)2P(S_i > 0) = \frac{1 - P(S_i = 0)}{2} P(Si≥0)=P(Si>0)+P(Si=0)=1+P(Si=0)2P(S_i \ge 0) = P(S_i > 0) + P(S_i = 0) = \frac{1 + P(S_i = 0)}{2}

Step 2: Probability Calculations after i=2i = 2 Rounds

S2=d1+d2=0S_2 = d_1 + d_2 = 0 can occur in the following cases:

  1. (d1,d2)=(0,0)(d_1, d_2) = (0, 0): Probability =(16)2=136= \left(\frac{1}{6}\right)^2 = \frac{1}{36}
  2. (d1,d2)=(+1,−1)(d_1, d_2) = (+1, -1) or (−1,+1)(-1, +1): Probability =2×(512)×(512)=50144= 2 \times \left(\frac{5}{12}\right) \times \left(\frac{5}{12}\right) = \frac{50}{144}

Therefore: P(X2=Y2)=P(S2=0)=136+50144=4+50144=54144=38P(X_2 = Y_2) = P(S_2 = 0) = \frac{1}{36} + \frac{50}{144} = \frac{4 + 50}{144} = \frac{54}{144} = \frac{3}{8}

  • For (I) Probability of (X2≥Y2)(X_2 \ge Y_2): P(X2≥Y2)=1+P(S2=0)2=1+382=1116P(X_2 \ge Y_2) = \frac{1 + P(S_2 = 0)}{2} = \frac{1 + \frac{3}{8}}{2} = \frac{11}{16} This matches (Q).

  • For (II) Probability of (X2>Y2)(X_2 > Y_2): P(X2>Y2)=1−P(S2=0)2=1−382=516P(X_2 > Y_2) = \frac{1 - P(S_2 = 0)}{2} = \frac{1 - \frac{3}{8}}{2} = \frac{5}{16} This matches (R).

Step 3: Probability Calculations after i=3i = 3 Rounds

S3=d1+d2+d3=0S_3 = d_1 + d_2 + d_3 = 0 can occur in the following cases:

  1. All three zeros: (0,0,0)(0, 0, 0) Probability=(16)3=1216\text{Probability} = \left(\frac{1}{6}\right)^3 = \frac{1}{216}
  2. One +1+1, one −1-1, and one 00: Permutations of (+1,−1,0)(+1, -1, 0) in 3!=63! = 6 ways Probability=6×(512)×(512)×(16)=150864\text{Probability} = 6 \times \left(\frac{5}{12}\right) \times \left(\frac{5}{12}\right) \times \left(\frac{1}{6}\right) = \frac{150}{864}

Therefore: P(X3=Y3)=P(S3=0)=1216+150864=4+150864=154864=77432P(X_3 = Y_3) = P(S_3 = 0) = \frac{1}{216} + \frac{150}{864} = \frac{4 + 150}{864} = \frac{154}{864} = \frac{77}{432}

  • For (III) Probability of (X3=Y3)(X_3 = Y_3): P(X3=Y3)=77432P(X_3 = Y_3) = \frac{77}{432} This matches (T).

  • For (IV) Probability of (X3>Y3)(X_3 > Y_3): P(X3>Y3)=1−P(S3=0)2=1−774322=355864P(X_3 > Y_3) = \frac{1 - P(S_3 = 0)}{2} = \frac{1 - \frac{77}{432}}{2} = \frac{355}{864} This matches (S).

Conclusion

The matching list is:

  • (I) →\rightarrow (Q)
  • (II) →\rightarrow (R)
  • (III) →\rightarrow (T)
  • (IV) →\rightarrow (S)

This corresponds to Option A.

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