So this is a problem I'm stuck on,
You roll a fair 4-sided die and with probability 1/3 you get to roll once more, and with probability 2/3 you have to stop. Assume that you get as many coins as the sum of the rolls. What is the probability you will win an even number of coins?
Can somebody help?
PS: You can get as many extra rolls as possible!
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I'm assuming that the numbers on the die are 1,2,3,4. So there are two odd results on the die, and two even results.
Answer: The chance is $1/2 = 50\%$.
Reason: The situation is completely even-odd symmetric.
No matter what the current score (= number of coins won so far) is and how many rolls you already had, the chance to stop with an even score afterwards is $2/4 \cdot 2/3 = 1/3$, and the chance to stop with an odd score afterwards is $1/3$, too.
With a coin, at a probability of 1/4 you flip it and at a probability of 2/4 you stop. With a four-sided die, at a probability of 1/3 you get one more roll and at a probability of 2/3 you stop. If you multiply those together, you get 2/4*2/3=4/12=1/3, which is the probability of where you roll a four-sided die one more time. Also, 2/4 is equal to 1/2, which is the probability of flipping a coin.