How does the summation of a partial binomial should be resolved?

Question

I stumbled on a question that I cant do. Here is: In a casino roulette game, we have 18 red cases, 18 green ones and one blue (made up roulette game). We are interested on 2 players that, for every game, bet on the red cases. If the marble goes on one of the red cases, the player gets 2 times the amount he bets. Say one of the players has 10 dollars and plays 10 games and hes betting 1 dollar for each game on the red. What is the expectation of his money after 10 games?

1. I tried doing it with summation and the answer that it should give me is $2.10.\frac{18}{37} -10$ , but im stuck on this :

$$\Bbb E(X=i)= \sum_{i=0}^{20} i \binom{20}{i} (\frac{18}{37})^{i} (1- \frac{18}{37})^{20-i} $$ $$\downarrow$$

$$\Bbb E(X=i)= (\frac{19}{37})^{20} \sum_{i=0}^{20} \frac{20!}{(i-1)!(20-i)!} (\frac{18}{19})^{i} $$ $$\downarrow$$ $$ ?$$

I don't know how to get this done.

2. I didn't get the other way of doing it which is $$\Bbb E(X)=n(2p-1) $$

thanks for the help !

Answer 1

For one individual play, you bet one dollar. $-1$. And then with $p $ chance you get $2$ dollars back. Then the expected value for that play is $2p -1$.

If you play 10 times, and knowing that the outcome of one play does not depend on the others, the right answer will be arrived at by summing the expected value of one play, 10 times, hence making

$10\cdot (2p -1) = 2*10p - 10$

And the last formula is

$n\cdot (2p - 1)$

And not

$np2 - 10$

Because of the expected value for one single play.

Now on to the summation part. Set $X $ as his final money. Start by noticing that if the player lost $i$ times (and therefore won $10-i $) then he ended up with $10 - i + (10 - i) = 20-2i $ dollars. That is, we have $X \in \{0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20\} $.

When $X = 20 - 2i $ for some $i $, $i $ denotes the number of losses. The probability of having $X = 20-2i $ is given by the chance that we won exactly $10-i $ games times the chance that we lost exactly $i $ games times the number of permutations win/loss we can have to achieve that result. Hence we get

$$\Bbb P(X = 20-2i) = {10\choose {i}} \left (1 - \frac {18}{37}\right)^i\left(\frac {18}{37}\right)^{10-i} $$

And thus the expected final money ought to be the weighted sum of those probabilities, the value times the probability we got that value. Note that the expected money is initial money $+$ expected value.

$$\Bbb E(final\ money) = \sum_{i=0}^{10} (20 - 2i)P(X = 20 - 2i) = \sum_{i=0}^{10} (20 - 2i){10\choose {i}} \left (1 - \frac {18}{37}\right)^i\left(\frac {18}{37}\right)^{10-i}$$

I asked WolframAlpha how much this was and got the right answer: he said the expected money was $\frac{360}{37} $ and thus the expectation value for the 10 plays was $\frac{360}{37} - 10 = -\frac{10}{37}$

which matches the formula $20p - 10 = 20\cdot \frac{18}{37} - \frac{370}{37} = -\frac{10}{37}$