Question 573514: Here's another puzzle featuring squares -- but this time there's really no geometry involved, just a little game.
Suppose we play the following simple game on the 5-by-5 grid of squares shown at right.
• Begin with your game piece on the upper left square (marked "S" for "Start").
• You move your game piece by flipping a coin:
o If the coin comes up "heads", you move one square to the right.
o If it comes up "tails", you move one square down.
o If you are on the right edge and get "heads", you do nothing and wait for the next coin flip. Similarly if you are on the bottom row and get "tails".
• You keep moving until you either
o lose by landing on one of the red squares along the diagonal, or
o win by arriving safely at the lower right corner (marked "F" for "Finish").
Your challenge this week is to calculate the probability that you will win this game. Assume that the coin you flip is a fair one.
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! You lose if and only if the number of heads flipped equals the numbers of tails flipped (and does not equal 4). This means, a sequence such as HT or HHTT or HHTHTT loses.
We can easily find the probability of losing on the second coin flip. Out of 4 possible sequences HH, HT, TH, TT, two of them (HT, TH) lose. Therefore the probability is 1/2.
Finding the probability of losing on the fourth coin flip is a bit tricky. We need to find the probability of obtaining two H's and two T's, provided that the first two flips do not yield one head and one tail. Here, the only possible cases are HHTT and TTHH. There are 2^4 = 16 four-flip sequences, but we must ignore ones beginning with TH and HT (since these already lose), so there are only eight four-flip sequences. The probability here is 2/8 = 1/4.
Now we need to find the probability of losing on the sixth coin flip. Here, we need to obtain three H's and three T's, provided that the first two flips are not T,H and the first four flips are not T,T,H,H in some order. This appears tricky, but we can make it easier by analyzing the only possible sequences for the first four flips.
Note that we cannot begin with TH, HT, or TTHH, HHTT. Here, we must begin with either TTHT, TTTH, or HHTH, HHHT. Note that since each of these sub-sequences contains either three T's or three H's, the other two flips will be uniquely determined. Therefore the only possible losing sequences are
TTHTHH
TTTHHH
HHTHTT
HHHTTT (4 sequences)
There are 2^6 = 64 possible sequences of length six. However, we must subtract those sequences which begin HT, TH, HHTT, TTHH. These subsequences are disjoint, and it can be checked that there are 16, 16, 4, and 4 sequences with these characteristics respectively (see why?). Therefore the number of possible sequences of length six is 64 - 16 - 16 - 4 - 4 = 24. The probability of losing on the sixth coin flip is 4/24 = 1/6.
It is impossible to lose on the eighth coin flip because we either would have won, or would have been hitting the sides of the checkerboard. Therefore the probability of losing is 1/2 + 1/4 + 1/6 = 11/12, so the probability of winning is 1 - 11/12 = 1/12.
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