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In a certain game, you perform three tasks sequentially. First, you flip quarter, and if you get heads you win the game. If you get tails, then you move to the second task. The second task is rolling a single die. If you roll a six, you win the game. If you roll anything other than a six on the second task, you move to the third task: drawing a card from a full playing-card deck. If you pick a spades card you win the game, and otherwise you lose the game. What is the probability of winning?

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You win if and only if you satisfy one of the three mutually exclusive situations:

  • (1) You win on the first task by flipping a head

  • (2) You lose on the first task by flipping a tail followed by winning on the second task by rolling a six

  • (3) You lose on the first task by flipping a tail followed by losing on the second task by rolling a number other than six followed by winning on the third task by picking a spade

Case (1) occurs with probability $\frac{1}{2}$

Case (2) occurs with probability $\frac{1}{2}\times\frac{1}{6}$ (the 1/2 here referring to having lost the coinflip in the first task)

Case (3) occurs with probability $\frac{1}{2}\times\frac{5}{6}\times\frac{13}{52}$ (the 1/2 referring to having lost a coinflip and the 5/6 referring to having lost the dice roll)

Since these events are mutually exclusive and are the only ways in which you can win, adding these probabilities together gives the total probability of having won.

$$\frac{1}{2}+\frac{1}{2}\times\frac{1}{6}+\frac{1}{2}\times\frac{5}{6}\times\frac{13}{52}$$


An easier way to approach this calculation is to instead look at the probability that you lose instead and subtract it away from $1$. You lose if you failed the coinflip, failed to roll a six, and failed to draw a spade. That occurs with probability $\frac{1}{2}\times\frac{5}{6}\times\frac{39}{52}$. Subtracting away from $1$ gives the final total probability of winning as:

$$1-\frac{1}{2}\times\frac{5}{6}\times\frac{39}{52}$$

which equals the same as the above and simplifies to:

$$\frac{11}{16}$$

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