20 Sided vs 30 Sided Dice Game

Question

Alice and Bob have fair 30 sided and 20 sided dice, respectively. Alice rolls her die one time. However, Bob can roll his die twice and keep the maximum of the two values. In the event of a tie, Bob is the winner. Find the probability Alice is the winner.

My approach was as follows. 1/3 of the time, Alice gets a value between 21 and 30, so she wins automatically. 2/3 of the time, Alice's value is between 1 and 20. In this case, she wins in the event when her roll is strictly greater than both of Bob's rolls. The probability of Alice's roll being strictly greater is $\frac{(20^2-20)/2}{20^2}$ =$\frac{190}{400}$. The reasoning here is that there are 400 combinations, and 20 of these are times when Alice's and Bob's rolls are equal, in which case Alice loses. Hence, my total probability works out to be:

P = $\frac{1}{3}(1) + \frac{2}{3}(\frac{190}{400})^2$

This answer is wrong, and I suspect it is due to my probability of Alice's roll being strictly greater than both Bob's rolls.

Answer 1

Forming cases,

1.if Alice get $1$ then Bob wons in any case.

2.if Alice get $2$ then Bob wons in any case other than droping $(1,1)$.

3.if Alice get $3$ then Bob wons in any case other than droping $(1,1),(1,2),(2,1),(2,2)$.

4.if Alice get $4$ then Bob wons in any case other than combinations of $1,2,3$ comes in two dice roll i.e. $3^2$ combinations.

$........$

20.if Alice get $20$ then Bob wons in any case other than combinations of $1,2,3,...,19$ comes in two dice roll i.e. $19^2$ combinations.

21.if Alice get $21,22,23,24,...,30$ then Bob lost.

As fiar dices, Every posible outcomes of Alice's dice is $\frac{1}{30}$.

Bob's winning probability $$ \begin{aligned} &= \frac{1}{30} \cdot \frac{400-0}{400} + \frac{1}{30} \cdot \frac{400-1^2}{400}+\frac{1}{30} \cdot \frac{400-2^2}{400}+....+\frac{1}{30} \cdot \frac{400-19^2}{400} \\ &= \frac{1}{30} \left[\frac{20 \cdot 400 - (1^2 + 2^2 + 3^2 + \ldots + 19^2)}{400}\right] \\ &= \frac{1}{30} \left[20 - \frac{19 \cdot 20 \cdot (19 \cdot 2 + 1)}{400 \cdot 6}\right] \\ &= \frac{553}{1200} \end{aligned} $$

So, Alice's wining probability$$=1- \frac{553}{1200} =\bbox[10px, border: 2px solid black]{\frac{647}{1200}}$$

Proof of validitiy, (Python code for $10000$ trials.)

import random as r

def calculate_probability():
    # Initialize variables
    alice_wins = 0
    n = 10000

    for _ in range(n):
        # Calculate Alice rolling a specific value
        alice_roll = r.randint(1, 31)
        bob_roll = max(r.randint(1, 21), r.randint(1, 21))

        if alice_roll > bob_roll:
            alice_wins += 1

    # Return the probability that Alice is the winner
    return alice_wins / n

# Calculate and print the probability
probability = calculate_probability()
print(f"The probability that Alice is the winner is approximately {probability:.4f}")

#Output:The probability that Alice is the winner is approximately 0.5323
#Which good approximate of my answer for 10000 trials.