March 2010: Problem and Solution
Problem of the Month - March 2010: eCoins
The expected value of an event or value with a given probability distribution
is equal to the weighted average of the expected outcomes, weighted by probability.
For example, if there is a 1/3 chance of catching 4 red balls and there is a 2/3
chance of catching 5 red balls, the expected number of red balls that will be caught
is the sum of 1/3 of 4 and 2/3 of 5.
Given this defininition of expected value, consider each of the following:
1. A coin is flipped until a heads appears. Determine, with justification, the expected
number of times the coin will be flipped.
2. A die is rolled until a composite number appears as its top face. Determine the
expected number of times the die will be rolled.
Finally, if you're willing to give it a shot, here's a more challenging problem:
3. A coin is flipped until two consecutive heads appear. Determine the expected
number of times the coin will be flipped.
March 2010: Solution
1. Suppose our desired answer is x. Then, we flip the first coin, and have
one-half probability of each of the outcomes occuring:
- If the coin is heads, we are done.
- If the coin is tails, we flip another x times, on average.
Thus, we have 1/2 chance of expecting 1 coin, and 1/2 chance of expecting 1 + x
coins. From our definition of x, we can write this as:
x=1/2 + 1/2 (1+x)
Which we solve to get x=2.
2. We use a similar method to the first problem, defining y to be the expected
number of rolls:
- If the coin is heads, we are done.
- If the coin is tails, we flip another y times, on average.
From the known probabilities and expected values, we write:
y=1/6 + 5/6 (1+y)
Which we solve to get y=6.
3. In solving this problem, we use a technique similar to before, declaring z
as our desired value, but in this problem, we also flip two coins to get things
started:
- Our first two flips are both heads; we are done.
- Our first flip is heads, but the second is tails; it should be clear that we must
flip, on average, another z times.
- Our first flip is tails but the second is heads; this is a special case, which we
will extend later.
- Our first two flips are both heads: we flip another z times.
Now we consider the third case: how many coins must we flip after that? We can just
provide another theoretical flip, and see that:
- If the third flip is heads, we are done.
- If the third flip is tails, we flip another z times.
We can set this all together in one equation:
z=1/4 (2) + 1/4 (2+ z) + 1/4 (2 + (1/2 (1) + 1/2 (1+ z ) )
+ 1/4 (2 + z)
Which, when solved, gives z=6.
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