You might add to the statement of the problem “in the order that they occurred”.

HTT is directional

HTH is not

If you allowed the pattern in either direction, Monday and Tuesday would be equal.

@emtucifor: I believe he intended that you start from scratch after finding the pattern. Each pattern search is a discrete event. So, on Monday, after you hit a sequence of HTH, you stop, write down the number of flips, rinse and repeat. So, to be somewhat long-winded: Because both patterns have the first two flips the same (H, T), the “steps toward success” as you say are equal for all except the patterns in question: H,T,H and H,T,T. If on Monday, you flip a H,T,H, you count your flips and start over. If you flip a H,T,T, you keep going, but you are starting from scratch because the previous two flips are T,T. If on Tuesday, you flip a H,T,T, you count your flips and start over. However, if you flip a H,T,H, you keep going, but you are starting with a H, so you already have 1/3 of the sequence. All of this hinges on the fact that it is implied in the problem (perhaps it should be more explicit) that as soon as you see the desired pattern, you stop flipping, record the number of flips it took to reach that pattern, and then START OVER and perform the trial from the initial point. The whole idea of the exercise is to see how many flips it takes before you see the pattern. The “HTH – one success. 1 step toward a success.” idea is meaningless in this context. Since you have already found out how many flips it takes to see the pattern, this iteration is done and the “1 step toward a success” is now irrelevant. Removing that and similarly “HTT – one success. 0 steps toward a success.” from Tuesday leaves you with:
Monday: HTT – one failure. 0 steps toward a success.
Tuesday: HTH – one failure. 1 step toward a success.
And these are NOT equally likely to produce the pattern. On Monday, you are back to the 1 in 8 chance of flipping a HTH, whereas on Tuesday, you now have a 1 in 4 chance of flipping the requisite T,T.

Q.E.D. ?

Step 1: For the initial 3 coins flipped, Monday and Tuesday each have exactly one success in 8 possible. Step 2: Correctly not singling out a success or failure so far, each of the 8 possible “lookeahead outcome” has an exact 1-1 correspondence between days.

So there is no bias, they have equal chances, and the correct answer is in fact not even offered: “Tuesday could be higher or lower with about equal chances.” We’d need to know the number of flips on each day and the help of a professional statistician to tell us the chances of the 2 days being exactly equal (quite slim, in my opinion).

Specifically, the “explanation” says that for Tuesday’s failure with HTT, “you are one-third of the way to achieving it again.” It ignores that for Monday’s success with HTH, you are one-third of the way to achieving it again, too!

Monday:
HTT – one failure. 0 steps toward a success.
HTH – one success. 1 step toward a success.

Tuesday:
HTH – one failure. 1 step toward a success.
HTT – one success. 0 steps toward a success.

See how these balance out? The writer only considered the first outcome for each day of these four possibilities.

For a deeper analysis of all outcomes, see http://forum.lessthandot.com/viewtopic.php?f=102&t=7271&p=36926#p36926

ducksauce: A few things about your program (and I am by no means a programmer)… First off, it is counting the number of times it finds the pattern, not how many flips it took to find the pattern. Granted you could get the average number of flips by dividing the number of tries by the number of finds. Secondly, it clears the array as soon as it finds it cannot make a match without taking into consideration that flips that break the previous sequence could be part of the next. So if your array has a 1, and the next flip is a 1, it clears the array and starts over. But that second 1 could have been the first 1 in the next sequence. You really need a FIFO queue of length three. You could flip the coin, pop the oldest flip off the end of the queue, push the new one on the top, and then examine the queue to see if it meets one of the patterns. There are better queue classes, but ArrayList can do it too. It can be instantiated as ArrayList(int initialCapacity). Then for each try, you generate your random flip, use remove(int index) method to remove the entry at index 0 (which then shifts index 1 → 0 and 2 → 1) and then put your new flip at index 2. Then you can use the inherited equal(Object o) to see if it now matches the requested pattern. I guess ‘o’ would be another ArrayList object populated by the method’s parameters, i.e. the intended pattern. Like I said, I’m a CS101 student so I could be way off here. :)

The problem is misleading because it says “coins” and not “coin,” while the solution treats it as a single coin.

If you’re flipping three coins, you have the same odds, because the odds of each flip is the same, and there is no “pattern” like the solution suggests.

I wrote a program to model this and the results came out even.

import java.lang.Math;
import java.util.ArrayList;
import java.util.List;

public class CoinFlip {