biased coin question

[tim]

Doubtless an embarrassingly easy question, but I'm not fully sure how it should be answered. The question, to give you the context, is from the section of a probability textbook on "conditional probability and independence", subsection on independence. It states,

A (possibly biased) coin is tossed twice. The result "heads" is noted f the coin lands heads up on the first toss and tails up on the second toss (HT). In a similar way, the result tails is "tails" is noted if the outcome is TH. HH and TT are ignored and the entire experiment is repeated until either HT or TH occurs. Show that the result heads is noted with probability 0.5 (however biased the coin).

Intuitively, I think I understand this. But I'm not sure what I'm supposed to show formally.

Since the same coin is tossed, probability of the event "heads or tails" on any single toss is 1. Since HH and TT results are ignored, the probability of the event " the result of the experiment is either HT or TH" is also 1. Now assuming that the results of the successive coin tosses are independent, and regardless of any bias, P(HT) = P(TH), and since P(HT) + P(TH) = 1, P(HT) = P(TH) = 0.5.

Is that adequate? It feels a bit too obvious. I've just read all these formula for disjoint and independent events and suspect I should frame my answer in terms of those...

Something along the lines of:

\(\displaystyle P(HT\cup TH)=1=P(HT)+P(TH)-P(HT)\cdot P(TH)\)

 

[JeffM]

Doubtless an embarrassingly easy question, but I'm not fully sure how it should be answered. The question, to give you the context, is from the section of a probability textbook on "conditional probability and independence", subsection on independence. It states,

A (possibly biased) coin is tossed twice. The result "heads" is noted f the coin lands heads up on the first toss and tails up on the second toss (HT). In a similar way, the result tails is "tails" is noted if the outcome is TH. HH and TT are ignored and the entire experiment is repeated until either HT or TH occurs. Show that the result heads is noted with probability 0.5 (however biased the coin).

Intuitively, I think I understand this. But I'm not sure what I'm supposed to show formally.

Since the same coin is tossed, probability of the event "heads or tails" on any single toss is 1. Since HH and TT results are ignored, the probability of the event " the result of the experiment is either HT or TH" is also 1. Now assuming that the results of the successive coin tosses are independent, and regardless of any bias, P(HT) = P(TH), and since P(HT) + P(TH) = 1, P(HT) = P(TH) = 0.5. I like this logic under the assumption that it is known that the experiment actually did end.

Is that adequate? It feels a bit too obvious. I've just read all these formula for disjoint and independent events and suspect I should frame my answer in terms of those...

Something along the lines of:

\(\displaystyle P(HT\cup TH)=1=P(HT)+P(TH)-P(HT)\cdot P(TH)\) Well I think it is obvious that
P(HT or TH) = P(HT) + P(TH) - P(HT and TH) = P(HT) + P(TH) - 0, which just takes you back to your original argument. So I do not see that this adds much to the argument because P(HT and TH) is trivially zero. Maybe I am missing something.

By the way I added a (perhaps idiotic) reply to your last comment under inductive proof on the calculus page.

 

[tim]

Cheers, Jeff.

A question: How can P(TH)*P(HT) = 0? This implies that either P(HT) = 0, or that P(TH) = 0, or that both P(HT) = 0 and P(TH) = 0. But these contradict my assumptions that P(TH) = P(TH) and P(HT) + P(TH) = 1.

 

[JeffM]

Cheers, Jeff.

A question: How can P(TH)*P(HT) = 0? This implies that either P(HT) = 0, or that P(TH) = 0, or that both P(HT) = 0 and P(TH) = 0. But these contradict my assumptions that P(TH) = P(TH) and P(HT) + P(TH) = 1.

What is confusing about this problem (and it may have confused me) is that the experiment ends as soon as you get a head first, tail second or tail first, head second. So, you either get HT or you get TH. They are mutually exclusive. They are also the only possible results.
So P(HT | TH) = 0.
Consequently, P(HT + TH) = P(HT | TH) * P(TH) = 0 * P(TH) = 0.
So 1 = P(HT or TH) = P(HT) + P(TH) - P(HT and TH) = P(HT) + P(TH) - 0 = P(HT) + P(TH) = 1. So I agree with your conclusion about that.

In other words, P(HT) * P(TH) is an irrelevancy.

Where independence is relevant is in the individual coin tosses.
Let p = the probability that a head results from a single toss.
And (p - 1) = the probability that a tail results from a single toss.
P(HT | HT or TH) = [p * (1 - p)] / 1.
P(TH | HT or TH) = [(1 - p) * p] / 1.
So, P(HT | HT or TH) = P(TH | HT or TH),
So P(HT) / 1 = P(TH) / 1.
So P(HT) = P(TH). So I agree with your conclusion about that.
P(HT) + P(TH) = 1 = 2 * P(HT)
P(HT) = 1/2. QED

That's my reasoning, but I have not studied probability theory in forty years so I may be wrong. Furthermore, the notation used in this problem seems confusing. Send PMs to royhaas or tkhunny to ask them to look at this thread. They seem to be the ones who most frequently give authoritative answers on probability theory. I usually limit myself to algebra and differential calculus, where I do not make a fool of myself TOO often. Subhotosh Khan, mmm, and galactus visit this topic sometimes so you might try PMing them too.