Probability of Same face in Coin flips

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The discussion focuses on calculating the probability of stopping on an even number of coin tosses when flipping a fair coin until the same face appears twice in a row. It is established that the probability of stopping on the nth toss is given by 1/2^(n-1). The probabilities for stopping on specific tosses, such as 2, 3, and 4, are calculated, revealing a pattern where the probability of stopping on an odd toss is half that of stopping on the preceding even toss. The relationship between odd and even stopping probabilities leads to the conclusion that the total probability of stopping on odd tosses is 0.5, thus making the probability of stopping on even tosses also 0.5. This analysis simplifies the problem significantly by leveraging conditional probabilities and established relationships.
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Homework Statement


You flip a fair coin until you get the same face twice in a row.
What is the probability that you stop on an even number of tosses?

Example: Probability of taking less than 6 tosses to stop is 15/16
In general, the probability of stopping on the nth toss is 1/2^(n-1)


Homework Equations




Example: Probability of taking less than 6 tosses to stop is 15/16
In general, the probability of stopping on the nth toss is 1/2^(n-1)

The sum of probabilities as n (the number of tosses) grows is 1.





The Attempt at a Solution



Pr(2 tosses) = 1/2
Pr(3 tosses) = 1/4
Pr(4 tosses) = 1/8

In general there are only two sequences for n tosses that end on the nth toss.

Either it starts with tails and alternates until the last 2 tosses or it starts with
heads and alternates until the last 2 tosses.

Example
HTHH
THTT
Only 2 sequences that end on the 4th toss.
 
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You are making this too hard. Don't worry about the sequences that end on 2 tosses, 3 tosses, ..., as you will quickly end up in a combinatoric nightmare. Look at things conditionally. For example, given that you have tossed 1000 alternating heads and tails, what are the odds the sequence stops on the 1001 toss? It becomes a simple proposition to compute the joint probabilities and then to compute the probability the sequence ends with an even number of tosses.

Edited to add

The problem statement already gives the joint probabilities. Add them up. (Show this is a valid thing to do.)
 
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This is indeed a very easy problem.. you knew the probability of stopping on the 3rd toss is half of stoppoing on the 2nd toss, and stopping on the 5th toss is, again, half of the 4th toss, therefore, we have a relationship:
P_3=\frac{1}{2}P_2
P_5=\frac{1}{2}P_4
...
In general,
P_{2n+1}=\frac{1}{2}P_{2n}
From the above equation, it is not difficult to draw the conclusion that the total probability of stopping on odd is half of stopping on even... applying P_{odd}+P_{even}=1, you will have the answer...
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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