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    iandthou2's Avatar
    iandthou2 Posts: 5, Reputation: 1
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    #1

    May 3, 2005, 08:43 PM
    Flipping Coins, Seeking Heads
    Hi,

    I know if I flip a coin the chances are 50/50 that I will get either heads or tails. Let's say I get tails. What are the chances that I will get heads the second toss? I understand at one level it's 50/50 each toss... but still I feel there should be at least a slightly greater possibility of getting heads on the second try. Is that correct?

    What if I get two tails in a row, what are my chances for a heads on a third toss?

    Obviously I want to get a "heads." Perhaps less obvious: this has to do with a real-life and important issue my wife and I are dealing with! We will probably only get two flips (maybe only one!) so I'm particularly interested in the second toss.

    THANKS.

    Iandthou2
    CroCivic91's Avatar
    CroCivic91 Posts: 729, Reputation: 23
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    #2

    May 4, 2005, 02:30 AM
    Quote Originally Posted by iandthou2
    Hi,

    I know if I flip a coin the chances are 50/50 that I will get either heads or tails. Let's say I get tails. What are the chances that I will get heads the second toss? I understand at one level it's 50/50 each toss ... but still I feel there should be at least a slightly greater possibility of getting heads on the second try. Is that correct?

    What if I get two tails in a row, what are my chances for a heads on a third toss?

    Obviously I want to get a "heads." Perhaps less obvious: this has to do with a real-life and important issue my wife and I are dealing with! We will probably only get two flips (maybe only one!) so I'm particularly interested in the second toss.

    THANKS.

    Iandthou2
    This is a standard Probability question. You are trying to calculate probability of a "repeated experiment". You are "repeatedly" throwing coins. The model for such an experiment is like this:
    Let's imagine we code outcome of each experiment (either heads or tail) with 0 (heads) and 1 (tail). So if we say that in 3 repeated coin throws we got 011, it means that the outcome of first throw was heads, and the next two were tails.

    You are particularly interested in 2 coin throws (one after another), so I will explain that, and you can easily calculate the same thing for 3 throws.

    You want the heads to fall on the second try. That means that the first try can be either heads or tail. So you can get 00 or 10. Of course, if heads were to fall the first time around, it would mean success to you, but just follow me on this. You are actually asking yourself: "What's the chance of me winning in this game, if I can have 2 throws of the coin?". All outcomes of the 2 coin throws can be: 00, 01, 10, 11. Which of those would mean you success? 00, 01 and 10. That's 3 out of 4, which is 75%.

    3 coins? Outcomes: 000, 001, 010, 100, 011, 101, 110, 111. How many of them mean you success? 000, 001, 010, 100, 011, 101, 110. 8 out of 9!
    iandthou2's Avatar
    iandthou2 Posts: 5, Reputation: 1
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    #3

    May 4, 2005, 07:13 AM
    Thanks!
    Any other opinions?
    HANK's Avatar
    HANK Posts: 98, Reputation: 5
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    #4

    May 4, 2005, 02:05 PM
    Out Of 100:
    Let's consider the term "50%" a bit more closely. The symbol "%," of course, stands for per cent, which means "out of 100". When we say that there is a 50% chance that the coin will land heads up, we mean that, on the average, 50 tosses of the coin out of every 100 tosses will result in the coin landing heads up. If you believe this, there is no way to predict the outcome of a coin toss!

    HANK :eek:
    CroCivic91's Avatar
    CroCivic91 Posts: 729, Reputation: 23
    Senior Member
     
    #5

    May 5, 2005, 02:16 AM
    Quote Originally Posted by HANK
    Let's consider the term "50%" a bit more closely. The symbol "%," of course, stands for per cent, which means "out of 100". When we say that there is a 50% chance that the coin will land heads up, we mean that, on the average, 50 tosses of the coin out of every 100 tosses will result in the coin landing heads up. If you believe this, there is no way to predict the outcome of a coin toss!

    HANK :eek:
    You are attacking the foundations of Probability!

    And yes, I have to agree... it's all just a mathematical model, and since world is not perfect - you can never be sure.
    dmatos's Avatar
    dmatos Posts: 204, Reputation: 26
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    #6

    Jan 22, 2006, 07:29 PM
    Are we talking real world or ideal mathematics here? If it's ideal mathematics, then each coin toss is what is called an "independent event". That is, the outcome of each toss does not depend at all on all of the previous tosses. If you throw fifty heads in a row, the probability that the next toss is a tail is still 1/2 =50%.

    If we're talking real world, each toss gives you more information about the coin. After throwing fifty heads in a row, I'd assume that the coin was biased, an that there wasn't actually a 1/2 chance that the coin would come up heads. In fact, I'd check if the tosser were using a weighted coin, or even one with two heads :)

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