Game with 4 coins

Question

Consider the following game. We have 4 fair coins, they are tossed and you have 2 options, either take the amount of money equal to the number of heads that fell, or toss 4 coins again, after you toss 4 coins again you must take the money and the game ends. Obviously, the optimal strategy is: If on the first throw there are 2, 3 or 4 heads, then we take the money, but if the number of heads is 0 or 1, then we throw the coins again. How to calculate the expected value of this strategy?

I don't understand how to calculate expectations in such games, if anybody could explain on this example, I would appreciate that!

Answer 1

The expected payoff on the first round is the expectation of a Binomial distribution with parameters $(4,\frac 12)$, i.e., it is $2$. The strategy is to toss again only if the initial number of heads is less than $2$.

Let $N_1$ (resp., $N_2$) be a random variable that models the number of heads on the first (resp., second) round. $N_1$ and $N_2$ are independent, and each has distribution $B(4,\frac 12)$.

The total payoff is $N_1 1_{N_1\geq 2} + N_21_{N_1\leq 1}$, and the expected payoff is $$E[N_1 1_{N_1\geq 2}] + E[N_2] P(N_1\leq 1) = \frac {19}8 = 2.375.$$

Answer 2

Find the probability of each outcome. That is to say, find the probability we get 0 heads and toss again and get 0 heads, we get 0 heads, toss again and get 1 heads, etc. for every possible outcome. You can do this in a table or figure out if there's a nice expression. You can make this faster by finding the probabilities for 0, 1, 2, 3, and 4 heads in the first round and then multiplying these probabilities by the probability of 0 and 1 for the games that advance to the second round, according to your strategy. All together you should have 13 possible outcomes (I noticed in a comment you said you have 25. This is not true).

Then multiply the amount of money you get for each outcome by the outcome itself e.g. the probability of getting 4 heads the first round and stopping is $1/16$ and you get $4$ "monies" for this outcome so our product is $\frac{4}{16}=0.25$. Once you've done this for each outcome, add them all up and this is your expected value.

This comes from the formula $E[X]=\sum_x xp(x)$ where $x$ is the value of the outcome and $p(x)$ is the probability of said outcome.

Answer 3

A shortcut here is to notice that if you always decided to take the second throw, your expected win would be $2$, since this is the expected result of the second throw. But if you throw three heads in the first round, you instead win $3$, gaining $1$. And if you throw four heads in the first round, you win $4$, gaining $2$. Therefore the expectation is $2+p_3+2p_4$, where $p_i$ is the probability of throwing $i$ heads in the first round.