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January 21st, 2012, 10:34 PM  #1 
Joined: Jan 2012 Posts: 2 Thanks: 0  2Coin Toss Probability Proof
Q: "You flip a fair coin twice. The probability of getting heads is 0.5, and tails 0.5. If E represented an event and each of the two tosses was represented by the numbers 1 and 2 respectively (i.e. H1 means getting heads on the first toss), then the possible outcomes are: E1 = H1, H2 E2 = H1, T2 E3 = T1, H2 E4 = T1, T2 Since the coin is completely fair, we can assume that these pairs of outcomes are equally likely. That is to say, the probability of E1: P(E1) = 1 / 4 This mistake is commonly seen in introductory Probability and Statistics Textbooks." I talked to my professor and he said it was a really "subtle" question and not to over think it. Basically, the reasoning for why the answer is 1/4 is incomplete. That is to say there is something critical, that is missing from this argument. My guess is you cannot conclude that the "pairs of outcomes are equally likely" just because the coin is fair and each elementary outcome has a 0.5 chance of occurring. That said, even if my guess is correct I don't know what missing piece is so critical that there exists a situation where this assumption is false without it. Sorry if I broke any rules to this forum, first post. 
January 22nd, 2012, 04:53 AM  #2  
Joined: Jan 2012 Posts: 14 Thanks: 0  Re: 2Coin Toss Probability Proof Quote:
CB  
January 23rd, 2012, 02:27 AM  #3 
Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: 2Coin Toss Probability Proof
Interesting thought. The .50 probability of H/T not changing and being independent are indeed two different things.

January 23rd, 2012, 09:56 AM  #4  
Joined: Jan 2012 Posts: 2 Thanks: 0  Re: 2Coin Toss Probability Proof Quote:
Would something to the effect of: P(HnT) = P(HT)P(T) = P(H)P(T) so, P(HT) = P(HnT) / P(T) = P(H) and, P(TH) = P(TnH) / P(H) = P(T) suffice to complete the proof? Thanks a ton.  
January 26th, 2012, 12:09 AM  #5 
Joined: Jan 2012 Posts: 14 Thanks: 0  Re: 2Coin Toss Probability Proof
The definition of independednce is equvalent to the the reqirement that the probability of AB in that order is equal to the product of A and the probability of B. CB 

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