Probability space: Let D_n be the probability of getting two heads in a row in n coin tosses

xmr

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Let D_n be the probability of getting two heads in a row in n coin tosses. Prove that D_n → 1 if n → ∞. Please explain in detail how to prove this.
 
This is a math help site. Do you really think that we will just give you the proof. Will that really be helpful?

Can you tell us the probability of getting two heads in a row in n tosses? Knowing this should help you greatly.
 
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I'm having trouble with this one.

Let's look at three cases.
Case 1: Getting two heads in a row after n tosses.
Case 2: Getting two tails in a row after n tosses.
Case 3: Not getting two heads or two tails in a row after n tosses.

Case 1 and Case 2 are basically the same.
There are exactly 2 ways for Case 3 to happen after n tosses.

Case 1 and Case 2 have a non-empty intersection.

There are 2^n - 2 ways of not being case 3.

So there are [2^n-2]/2 ways of being in Case 1

I then claim that Dn=[2^n-2]/2^(n+1) = 1/2 - 1/2^n which must be wrong as its limit is 1/2 not 1.
 
Let D_n be the probability of getting two heads in a row in n coin tosses. Prove that D_n → 1 if n → ∞. Please explain in detail how to prove this.
The first thing to make sure we know what this means.

I would guess that "getting two heads in a row in n coin tosses" means that we toss a coin n times, and somewhere in the sequence of n results, at least two successive tosses are heads. So it would be a succes (with n=5) if we tossed THHTT, or THHTH, or HHTHH, or HHHTH , or even HHHHH (and others).

I would tend to try the complement: What is the probability that there will never be two H's in a row? Then you need to count events like TTTTT, and THTHT, and HTTTH. Do you have any ideas about that? (Or any approach, for that matter?)

I suppose it's also conceivable that you could prove the claim without ever making a formula for D_n. What have you learned that might be useful either way? (Knowing your context can help us a lot, in knowing both what sort of help is appropriate, and how hard a problem to expect this to be!)
 
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