What is the value of a bet that pays zero if Albert loses overall and pays his winnings if he wins?

Question

Question: Albert and Bob are playing a coin flipping game. If the coin comes up heads, Albert wins $\$1$; if it comes up tails, Bob wins $\$1$. The coin is biased so that the probability of heads is $0.7$. Albert and Bob will play $4$ games. What is the value of a bet that pays zero if Albert loses overall and pays his winnings if he wins? Give your answer to two decimal places.

I was asked this interview question during interview.

I am not able to understand what they are asking for.

Any help is appreciated.

Answer 1

The value of a bet is not a fixed amount but a return rate of interest on the investment.

The odds to win overall for heads are 65.17% Those come out of the odds of loosing once or winning all: $4\cdot (0.3)\cdot (0.7)^3 +(0.7)^4=0.6517$

Therefore a bet against overall win on heads has 100%-65.17%=34.83% odds to win.

I guess that to a normalized bet representing 100%, the bet against overall winning for heads will pay 100/65.17 x 34.83 = 53.44%

Answer 2

The outcome of the "bet" referred to isn't incompletely specified, because:

• You're not told explicitly how much Albert loses if the coin comes up tails. Presumably, the two players are betting against each other, and so Albert would lose $\$1$ whenever the coin comes up tails.
• Even if this presumption is correct, there would still be a positive probability $\ 6\cdot0.7^2\cdot0.3^2\ $ that Albert breaks even, and you're not told what the bet will pay if that happens. Again, you might reasonably presume that the bet would pay zero also in the cases when Albert breaks even.

If the above two presumptions are correct then the bet will pay $\$4$ with probability $0.7^4$ (when four heads come up), $\$2$ with probability $4\cdot0.7^3\cdot0.3 \hspace{1pt} $ (when three heads come up)$\hspace{-1pt}\,^1$, and zero otherwise, so the expected return in dollars would be \begin{align} 4\cdot0.7^4 + 8\cdot0.7^3\cdot0.3&= 1.7836\\ &\approx 1.78\ \text{ (to two decimal places),} \end{align} which I would presume is what your interviewers meant by the "value" of the "bet".

$\hspace{-1pt}\,^1$ Thanks to KKKoo0 in the comments below for a correction to this quantity.

Answer 3

assume

• Albert wins $4 if there are 4 heads, the probability is \begin{align} 0.7^4 \end{align}

• Albert wins $2 if there are 3 heads, the probability is \begin{align} C(4,3)*0.7^3 * 0.3 \end{align}

• Albert wins $0 if there are 2 heads, the probability is \begin{align} C(4,2)*0.7^2 * 0.3^2 \end{align}

• Albert wins $0 if there are 1 heads as Albert loses overall, the probability is \begin{align} C(4,1)*0.7 * 0.3^3 \end{align}

• Albert wins $0 if there are 0 heads as Albert loses overall, the probability is \begin{align} 0.3^4 \end{align}

So overall, the value of the game is

\begin{align} $4*0.7^4 + $2* C(4,3)*0.7^2 * 0.3^2 = 1.78 \end{align}