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I am flipping a coin n times. if it lands on heads, i win nothing but if it lands on tails on the m-th flip, i get m dollars. how much money can i expect to you at the end of the game.

I looked at each coin flip on the m-th try. For example,

using the mean of a random variable formula where mu = n*p

1st coin flip -> .5 chance of winning .50 dollar

2nd coin flip -> .5 chance of winning 1 dollar

3rd coin flip -> .5 chance of winning 1.5 dollars

4th coin flip -> .5 chance of winning 2 dollars ect.

For the example above, I can expect to win 5 dollars which is just the sum of the m-th flip divided 2. so in our example, I would expect to earn 10/2 dollars

so by the end of the game I should expect to win (m1+m2+m3+....+mn)/2 dollars. Is that correct or am I supposed to get a solid number, and not in terms of variables.

Any help would be appreciated.

user3239138
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2 Answers2

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You seem to be pretty much spot on.

Keep in mind that $1+2+3+4+...+x-1+x=x(x+1)/2$

This means that you can solve for an equation for your winnings dependant only on n (the number of flips). Without a value of n or a new rule (like the one suggested below) there is no way to get a solid number.

This all assumes that every times you flip the coin and it lands on tails you win and can keep playing until your nth flip. It might be a more interesting problem if you stop after it lands on tails. This would give you an exact number.

kaine
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For $i=1$ to $n$, let $X_i$ be the amount of money we win on the $i$-th toss. Then the total winnings $Y$ are given by $Y=X_1+\cdots+X_n$.

By the linearity of expectation, we have $$E(Y)=E(X_1)+\cdots+E(X_n).$$

Note that $E(X_i)=\frac{i}{2}$. Thus $$E(T)=\frac{1}{2}(1+2+\cdots +n)=\frac{n(n+1)}{4}.$$

André Nicolas
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