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If I flip a perfect coin, the chance of heads is obviously 0.5 with each toss. Equally obviously, if I toss 3 times in a row, the probability of getting 3 consecutive heads is 1/2.1/2.1/2. If I put the coin in my pocket, then take it out 12 hours later for a another flip, does this count as my fourth flip (i.e my probability of another head is 1/16) or does probability re-start so my chance and probability of a head are back to the beginning and are both 1/2? Thanks.

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4 Answers

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If you stand on a corner and watch a guy flip a coin (and you know the coin and the man are honest) and he flips three heads, and then he shouts out "What is the probability my next flip will be a head?" the answer is $\frac 12$; not $\frac 1{16}$.

You are confusing probability with conditional probability.

Consider this. What is the probability that two people in the same house are both Leo's. It's $\frac 1{144}$. Suppose Cynthia Merkowitz who was born on August 13, 1992 has a room-mate. What is the chance that the roommate is a Leo. Well, that is $\frac 1{12}$. But Cynthia is a Leo and we just said the probability of two leos is $\frac 1{144}$ so which is it? $\frac 1{12}$ or $\frac 1{144}$?

Well the answer is $\frac 1{12}$ because we have already been told that part of the condition has already occured. Cynthia is a Leo. There was a $\frac 1{12}$ probability that would happen... but it did. Now that is has there is a $\frac 1{12}$ chance her roommate is too.

So if you flip three coins .... and you wait $12$ hours. The probability of the 4th flip being a head is $\frac 12$ and it doesn't matter what the first three flips were.

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No the probability does not have memory. The independent events such as flipping the coin do not keep record of previous results so each time that you flip the coin you have probability of $\frac {1}{2}$ to get head. Probability of getting four heads in a row is $\frac {1}{16}$ because probability of getting a head each time is $\frac {1}{2}$ so $(\frac {1}{2})^4=\frac {1}{16}$

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I'll add my few cents to this

1. If you toss 4 fair coins the probability of any sequence of outcomes is 1/16
2. If you toss 4 coins but already know the outcomes of the first three, each outcome of the 4th has probability 1/2 because it is the only unresolved random event remaining

Also, a little bit of philosophy

• If you adhere to frequentist school of thought, any coins that have already been tossed have exactly the probability of 1 of being in the state they are and exactly 0 probability of being in any other state. If you happen not to know the results of some of the previous tosses, that does not make them probabilistic events
• If you adhere to bayesian school of thought, there is no such thing as objective knowledge of the outcome, only subjective. Your subjective knowledge about the coins that have already been tossed can be probabilistic if you don't know the outcomes, and can be updated once you do.

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The previous answers already pointed out where the misconception stands.

I would like to relate that to the old story of "delayed numbers in a lottery".

Let's put briefly as follows:

• one thing is to bet on a certain number in the ten (for example) extraction to follow (but you shall buy ten tickets).
• another thing is to bet now on a certain number (with one ticket), because it did not appear in the nine extractions done before.

In both cases, every single ticket has the same probability to win.

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