Optimal Strategy in a Dice Game

In this post, I talk about my attempt to find the optimal strategy to maximize the score in a dice game.

The Game

You roll a fair 10-sided die.

• If you roll a 10, the game ends, and you win nothing.
• If you roll any other number, that number is added to your score.
• You are not allowed to stop until your total reaches at least 10.
• Once your score reaches 10, you are given a choice and you can choose to stop or continue.

How do you play this game optimally?

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Strategy 1: Random Play

I first tried a randomized strategy — once I reached 10, I continued with 50% probability:

def strategy(res):
   return random.uniform(0,1) < 0.5

def roll_die(choices=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]):
    return np.random.choice(choices)

def game():
    res = 0
    while res < 10:
        roll = roll_die()
        if roll == 10:
            return 0
        else:
            res += roll

    play_again = strategy(res)
    while play_again:
        roll = roll_die()
        if roll == 10:
            return 0
        else:
        play_again = strategy(res)
            res += roll
    return res

After simulating thousands of games, the average score hovered around 11.6 to 12.0.

Strategy 2: Optimized Play

At this point, I asked myself can I calculate at each turn (after reaching a score of 10) what my score could be after the next roll.

Say, my score is 10, what possible values can my score take if I roll the die again?

Dice Throw Outcome	New Score	Probability
1	11	\(1/10\)
2	12	\(1/10\)
3	13	\(1/10\)
4	14	\(1/10\)
5	15	\(1/10\)
6	16	\(1/10\)
7	17	\(1/10\)
8	18	\(1/10\)
9	19	\(1/10\)
10	0	\(1/10\)

This means, my expected value at the end of the throw is \[ \sum_{i=11}^{19} \frac{i}{10} = 13.5 \]

This is a value greater than my current score. So, it makes sense for me to play the game once more.

So I updated my strategy as such.

def strategy(res):
    e = 0
    for i in range(1, 10):
        e += (res + i) / 10

    return e > res

What this does is ask the following question. Given my current score, do I expect a higher score if I play again?

At first, this confused me — res + i increases as res increases, so why does the gain shrink?

Turns out, the math reveals:

\[E(res)=0.9res+4.5⇒E(res)−res=−0.1res+4.5\]

So as score increases, the value of rolling decreases linearly. Once score reaches 45, the expected gain is 0. Beyond that, rolling again is actually worse than stopping.

I simulated this strategy using the code below and got a higher range for the confidence of the score (17.25, 17.95)

def simulate_game(n=10000):
    res_list = []
    for i in range(n):
        res_list.append(game())
    return res_list


def find_conf(n=10000, n_conf=10000, conf_per=95):
    mean_list = []
    for i in tqdm(range(n_conf), desc="num_simulations"):
        mean_list.append(np.mean(simulate_game(n)))
    lower = np.quantile(mean_list, (100 - conf_per) / 200)
    upper = np.quantile(mean_list, (conf_per + (100 - conf_per) / 2) / 100)
    return (lower, upper)