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# Are 3 10% chances better than one 30% chance (when penalized by a variable for failures)?

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I'm playing a computer game in which you can spend 100 coins to be given a 10% chance to upgrade a weapon, or spend 200 coins for a 20% chance, or 300 coins for a 30% chance and so on, up to 1000 coins for a 100% chance.

There is no penalty for failure, only that you lose the coins spent.

My theory is that if you only have 300 coins to spend, it would be better to do 3 attempts since 10% + 10% + 10% = 30% and you have the added bonus of having a chance to spend less coins if it succeeds before the 3rd try.

Not sure if my theory is correct because a 100% chance is not the same as 10 10% chances. Also after getting failures, the odds remain the same. But I don't have the logic/math knowledge to have an answer and explanation.

Note:
For the purpose of the question, let's assume the game is somehow computing true random numbers and therefore able to give true 10%-100% odds
Upgrading weapons is something done very very often. Let's say the average player wants to upgrade ~1,000 times.

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As you seem to know (the $10\times 10\% \neq 100\%$-stuff - but continue doing anyway) and celtschk's answer shows you can't add the probabilities like that. (The events aren't independent)

The expected number of upgrades are the same, but what you really need to weigh against each other is the chance of not getting an upgrade at all ($70\%$ if you chose the $30\%$ strategy, $72{,}9\%$ if you use the $10\%$ strategy three times, and the possibility of getting more upgrades ($0$ if you use the $30\%$ strategy, $2{,}8\%$ if you use the $10\%$ strategy three times - and I did my math correctly). All of that is assuming you don't have any other use (like saving it) for the gold and chose to apply the $10\%$ strategy three times.

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It somewhat depends on your game situation.

If you have just that one weapon, need to upgrade it urgently, and are likely to have enough coins the next time you need to upgrade a weapon, then clearly the 30% option is better: It gives you a 30% probability of upgrade, while three 10% attempts give you a probability of $1-(1-0.1)^3=0.271$, that is 27.1%. Of course with the 30% strategyyou will lose 300 coins for sure, but by assumption, that is your lesser problem.

However I guess with upgraded weapons, you are also able to obtain more coins in the future, so maybe the question is which strategy allows you to get your updates as early as possible. In that case, the 10% strategy is the better one, as that way you can upgrade as soon as you've got 100 coins, instead of waiting for getting 300 coins, while maintaining the same average number of upgrades as described in the previous paragraph.

Of course all this assumes that all upgrades are of equal value. If you have one weapon where an upgrade gives you a massive advantage, and another one where the upgrade only gives you a marginal advantage, it might be a better idea to put your 300 coins in a 30% chance upgrade of that valuable weapon, instead of sacrificing some of that for a possible additional upgrade of the weak weapon.

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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:

var 1st 2nd 3rd loss
prob 0.1 0.9 × 0.1 = 0.09 0.9² × 0.1 = 0.081 0.9³ = 0.729
net win -100 + +200 = +100 -200 + +200 = 0 -300 + +200 = -100 -300

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.

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