Three Heads in a Row
Date: 01/25/2001 at 11:45:30 From: Rita Curtis Subject: Probability How many flips of a coin on the average will it take to hit/get three heads (or tails) in a row?
Date: 01/25/2001 at 12:34:11 From: Doctor Mitteldorf Subject: Re: Probability Dear Rita, The number you're looking for is sometimes called an "expectation value" of the number of flips. Let's look at the simplest case first. How many flips on average before you come up with your first head? We can call this number x, and write an equation about it. If the first flip is heads, then the answer is 1; on the other hand, if the first flip is tails, we've wasted one flip and we've still got x to go. We can combine these two to make an equation for x: x = (1/2)(1+x) + (1/2)(1) The second term on the right accounts for the fact that there's a 1/2 chance of getting it the first try; the first term says that if we get tails at first, we're in the same position as when we started: we've wasted one flip, but we still have x to go. This equation defines x in terms of itself, and with a little algebra, we can solve it for x: x = 2. This is the right answer for one head. Next let's move to the corresponding question about two heads. Again, if the first flip is tails, we have wasted one flip, so we still have a term (1/2)(1+x) on the right side. However, this time if we get heads on the first flip, we still have to think about what happens next: If we get a tail, we're back we're we started, having wasted two flips, but if we get a head, we're done, and we've accomplished our goal in only two flips. Putting all this together, we have: x = (1/2)(1+x) + (1/4)(2+x) + (1/4)(2) Solving this equation for x, we have x = 6, which is the right answer for two heads. The corresponding problem for three heads follows the same principles, but is a little longer, with more possibilities to account for everything up to the third step. First, thoroughly understand what I've done above; then develop your own equation for x in this case, by analogy. When you're all done solving for x, you should get x = 14. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum