Calculating Probability of 4 Heads in a Fair Coin Flip

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Discussion Overview

The discussion revolves around calculating the probability of obtaining four heads in a series of four flips of a fair coin. It explores the independence of coin flips, the application of probability rules, and the comparison of different outcomes in sequences of flips.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant calculates the probability of getting four heads as P(H&H&H&H) = 1/(2^4) = 1/16, based on the independence of coin flips.
  • Another participant confirms the calculation and suggests verifying similar problems by counting the number of ways coins can land.
  • A participant expresses uncertainty about why their method of using tree notation works, while also inquiring about the probability of one specific pattern occurring before another.
  • One participant states that all 16 possible outcomes are equally likely, implying that predicting which pattern occurs first is not feasible.
  • A later reply emphasizes the need to differentiate between the probability of patterns occurring in discrete tuples versus continuous sequences of flips.

Areas of Agreement / Disagreement

Participants generally agree on the calculation of the probability for four heads, but there is uncertainty and differing views regarding the prediction of which pattern occurs first in sequences of flips.

Contextual Notes

Participants discuss the distinction between discrete and continuous interpretations of the probability of patterns occurring, indicating potential limitations in understanding the problem's scope.

Somefantastik
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Given 4 flips of a fair coin, what is the probability of {H,H,H,H}?

I thought since flips of a fair coin are independent, then P(H&H&H&H) = P(H)P(H)P(H)P(H) = 1/(2^4) = 1/16. Am I close?
 
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Since these are 4 independent trials, yes.

You can verify this & similar problems by figuring out how many ways 4 coins can "land."
 
thanks; you're right. I used the tree-notation and got N(ways HHHH)/N(all ways) = 1/16. Thanks again. I'm not entirely sure why that works.

Is it possible to compute the probability that one pattern comes before the other one? Like THHH will happen before HHHH?
 
Each of the 16 possible total outcomes (HHHH through TTTT) is equally possible, so there is no way to predict which will occur before another.

- Warren
 
Thanks. I should have known that.
 
Remember to distinguish the probability that "THHH" comes before "HHHH" in a series of discrete tuples of four flips from the probability that the subsequence "THHH" comes before "HHHH" in a continuous series of individual flips.
 

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