Question related to binomial distribution? This may seem like an odd question but I appreciate any input.
I am playing a game where you are trying to promote a player's skill level.  You get tokens to attempt the promotion and you can attempt singly or in batches of up to 10. You can be promoted at most 5 times this way.
The probability of success is 0.02 per token, if you use 10 tokens your probability of success is 10 * 0.02 = 0.2
It has been stated that someone mathematically "proved" that it is better to attempt 1 token at a time rather than in batches of 10.  I disagree, my math/reasoning goes like this:
I see this as a binomial problem where the expected number of successes $E(X) = np$.  I will stop once I have 5 successes so $X >= 5$.
If you spend the tokens 1 at time $n * 0.02 = 5$ when $n = 250$.
If you spend the tokens in batches of 10 then $n * 0.2 = 5$ when $n = 25$.
In either case you still need on average a total of 250 tokens.
So am I correct - it doesn't make a difference or am I missing something?  The itch at the back of my brain is saying I am missing something and it has to do with stopping once I have 5 successes...
 A: It is hard to say which strategy is "better." I can give three different answers based on three interpretations of "better."
Suppose you have $n$ coins, and you use them in batches of $k$ for some $1\le k \le 10$. Let $p=0.02$. The number of levels you gain is distributed like $\operatorname{Bin}(n/k,kp)$. The expected number of levels you gain is equal to $np$, no matter what $k$ is, so in this sense the batch size does not matter. However, the variance of the number of levels gained is
$$
(n/k)\cdot kp\cdot (1-kp) = np(1-kp)
$$
As $k$ increases, the variance decreases. Therefore, using larger batches provides the same reward with less risk, so in a sense is a better strategy.
There is a third angle to look at. You say that you max out at level $5$. Suppose that you want to maximize the probability of reaching level $5$, starting from level $0$. Letting $X\sim \operatorname{Bin}(n/k,kp)$, then you want to maximize $P(X\ge 5)$. The distribution of $X$ is close to a normal bell curve as long as a $n$ is large enough, say $n\ge 30$. The mean of this bell curve is at $np$, and its width decreases as $k$ increases. Then it turns out the the optimal strategy depends on how $np$ compares to $5$:


*

*If $np\ge 5$, then you want the variance to be small, so you should use large batch sizes. You have a lot of money, so if you just play things safe you will be fine.

*If $np<5$, then you want the variance to be large, so you should use small batch sizes. Playing it safe is not good enough, you can only win if you get lucky.
In summary,

  
*
  
*With respect to expected gain, batch size does not matter.
  
*With respect to minimizing risk, larger batch sizes are better.
  
*With respect to maximizing the change of getting a certain number of levels, whether small or larger batches are better depends on context.
  

