In a coin tossing game that is made of two rounds, there are two cases.

In first round, if heads comes up, case 1 will be played in second round, if tails comes up, case 2 will be played in second round.

In case 1, two coins are tossed. If at least one of them is heads, player wins.

In case 2, two coins are tossed:

HH: Player wins,

HT and TH: Player loses

TT: Coin gets tossed again with the rules of case 2.

Now the probabilities of the first round are easy to calculate:

$$P(case_1) = P(case_1) = \frac{1}{2}$$

Probabilities of each individual cases are also easy:

For case 1:

$$P(win) = \frac{3}{4}$$

For case 2:

$$P(win) = \frac{1}{3} $$

Winning makes a profit of 1 dollar, and losing makes a loss of 1 dollar. What is the expected return of the game?

  • $\begingroup$ Im guessing you have to use expected value calculations, have you learned expected value? $\endgroup$ – user224997 Jan 10 '16 at 20:30
  • $\begingroup$ @Quantitative I can calculate that for roulette, but this game involve conditional probability. I can't do that :( $\endgroup$ – SarpSTA Jan 10 '16 at 20:34
  • $\begingroup$ Hint : The player can win in two ways. Multiply the probabilities in each way and add the two probabilities you get. The result is $\frac{13}{24}$. $\endgroup$ – Peter Jan 10 '16 at 20:39
  • $\begingroup$ @Peter I tried that but that didn't add up to 1. $\endgroup$ – SarpSTA Jan 10 '16 at 20:41
  • $\begingroup$ The sum is not $1$, it is the probability that $P$ wins. $\endgroup$ – Peter Jan 10 '16 at 20:41

Using the Law of Total Probability: $$\mathsf P(\mathrm{Win}) = \mathsf P(\mathrm{case}_1)\,\mathsf P(\mathrm{Win}\mid \mathrm{case}_1)+\mathsf P(\mathrm{case}_2)\,\mathsf P(\mathrm{Win}\mid \mathrm{case}_2)$$

You have evaluated $\mathsf P(\textsf{case}_1)$ and the two conditional probabilities, of a win given the case played.   Put it together and evaluate the expected value of return.

  • $\begingroup$ Hey there, thanks for the answer. What if an option like this is added to the game: In case 1, player can make a side bet that pays true odds. That is if he wins in case 1, he'll make a profit of 3 units and lose his side bet wager if his real bet loses. $\endgroup$ – SarpSTA Jan 10 '16 at 22:08

Let $R$ denote the return. You can condition on each case. For example $\mathbb E[R|case_1]$ denotes the expected return given that you are in case 1.

By the law of total expectation:

$\mathbb E[R]=\mathbb P[case_1]*\mathbb E[R|case_1] + \mathbb P[case_2] * \mathbb E[R| case_2]$


If we land on case 1 in round two, our game is guaranteed to end with either a win or a loss. The probability of a win is simply $\frac{1}{2} * \frac{3}{4} = \frac{3}{8}.$

However, if we land on case 2 in round two, our game could last for an infinite amount of time. To solve for the probability of a win, we sum the infinite geometric series $\frac{1}{2} * ((\frac{1}{4})^{1} + (\frac{1}{4})^{2} + (\frac{1}{4})^{3} + ... ) = \frac{1}{2} * \frac{1}{3} = \frac{1}{6}.$

Our total win probability is $\frac{3}{8} + \frac{1}{6} = \frac{13}{24}.$

The probability of loss is just one minus the probability of a win. This is just $1 - \frac{13}{24} = \frac{11}{24}.$

Our expected return is $\frac{13}{24} * 1 - \frac{11}{24} * 1 = \boxed{+\frac{1}{12}}.$


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