Let's assume, that there are only the two options from the header: 3 × 10% or 1 × 30%. But the following calculations should be applicable to all other cases too.

Next, we have to make an assumption about how much you can lose or win in every round. Since you are playing in two different currencies (weapon level vs in-game money), this isn't that easy. For purposes of simplifying the game, I'm going to assume that a weapon upgrade is about 200 coins worth.

Now, we have to calculate the expected value of your game. The expected value tells you, how much you win on average if you play the game very often. If the game is fair, it should be 0. If it's positive: good for you. If it's negative, good for the other one. We want to find out, which game option gives you the higher EV.

You calculate the expected value by multiplying the net win of every possible outcome with the probability for that outcome and then summing all those products.

For the 1 × 30% case that's easy. You have a 70% chance of losing 300 coins (-300 coins). And you have a 30% chance of losing 300 and earning 200 coins (-100 coins).

This gives us as expected value for the 30% case:

$$ E = 0.7 * (-300) + 0.3 * (-100) = -240 $$

This means, that on average you will lose 240 coins.

Now, let's look at the 3 × 10% case. Your question contains two different possible interpretations of that case:

• A: Once you win, you will stop playing.
• B: You will always play two rounds.

Let's consider each interpretation. I'm gonna start with A. First we have to look, which possible outcomes there are.

----x-------# win in 1. round (0.1)
    | loss in 1. round (0.9)
    |
    ------x----------# win in 2. round (0.1)
          | loss in 2. round (0.9)
          |
          ------x-------# win in 3. round (0.1)
                | loss in 3. round (0.9)
                #

As you can see, we have four different outcomes ("#"): win in 1st, 2nd or 3rd round or loss. They have these probabilities and net wins:

Based on that, we can calculate the expected value for the A interpretation:

$$ E = 0.1 * (+100) + 0.09 * (0) + 0.081 * (-100) + 0.729 * (-300) = -216.8 $$

Last, but not least, we're going to look at the interpretation B. Since you are always playing three round, you're always going to lose 300 coins. The question is, how often you win. If you win three times, you'll end up with +300 (-300 + +600), if you win two times, you'll end up with +100 (-300 + +400) and if you win one time, you'll end up with -100 (-300 + +100). If you lose every time, you'll end up with -300.

The probability to win three times is $0.1 * 0.1 * 0.1 = 0.001$. The probability to win two times is $3 * 0.9 * 0.1 * 0.1 = 0.027$. Note the 3, which is, because you have three options for choosing, which round you lose in. The probability to win one time is $3 * 0.1 * 0.9 * 0.9 = 0.243$. The 3 is here, because you have three options for choosing the one round you'll win in. The probability to lose every time is $0.9 * 0.9 * 0.9 = 0.729$.

Given that we can now calculate the EV for the last case.

$$ E = 0.001 * (+300) + 0.027 * (+100) + 0.243 * (-100) + 0.729 * (-300) = -240 $$

Well. That's interesting. The expected value for "1 × 30%" and "3 × 10%" is the same. Therefore, your choice between those doesn't really matter. However, the best strategy (of these), is to play up to three times, but stop if you win once.

posted over 3 years ago

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luap42‭ staff

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