Tossing a Coin 100 Times: Probability of All Heads

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Discussion Overview

The discussion revolves around the probability of obtaining all heads versus obtaining fifty heads in a series of 100 coin tosses. Participants explore the implications of order in probability calculations and the intuition behind different outcomes, touching on concepts of run lengths and statistical simulations.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Experimental/applied

Main Points Raised

  • Some participants assert that the probability of obtaining all heads and fifty heads is the same, calculated as (1/2)^100, when considering specific orders.
  • Others challenge this by noting that there is only one way to achieve all heads, while there are many combinations to achieve fifty heads in 100 tosses.
  • A participant expresses confusion regarding the wording and clarifies that they are referring to a specific order of fifty heads followed by fifty tails, agreeing that the probability for that specific sequence is indeed (1/2)^100.
  • One participant suggests that intuitively, sequences with alternating heads and tails seem more plausible than sequences with long runs of the same outcome.
  • Another participant mentions mixing up the statistics of run lengths, proposing that longer runs are less likely to appear, which leads to their expectation that a sequence of all heads followed by all tails would be less likely.
  • A participant shares a computer simulation of 100 coin tosses repeated 100,000 times to analyze run lengths, indicating a desire to explore the statistical behavior of outcomes.
  • There is a discussion about the confusion between the statistics of run lengths and the maximum run length of a given outcome, with a participant clarifying that while many runs of length 1 are expected, only one possibility exists for the longest run starting with heads.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views on the interpretation of probabilities related to specific outcomes and the implications of run lengths in coin tosses.

Contextual Notes

Participants express uncertainty regarding the relationship between specific outcomes and their probabilities, particularly in the context of run lengths and statistical expectations. The discussion highlights the complexity of interpreting probability in sequences of random events.

jk22
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Suppose I toss a fair coin 100 times. If I consider the order of apparition, then obtaining all head, or fifty times head has the same probability, namely (1/2)^100 ?
 
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jk22 said:
Suppose I toss a fair coin 100 times. If I consider the order of apparition, then obtaining all head, or fifty times head has the same probability, namely (1/2)^100 ?
How did you get that answer for the second case, getting 50 heads?
 
jk22 said:
Suppose I toss a fair coin 100 times. If I consider the order of apparition, then obtaining all head, or fifty times head has the same probability, namely (1/2)^100 ?
No, because there is only one way in which you can toss all heads, but there are many more ways in which you can obtain 50 heads in 100 tosses.
 
I'm considering one particular order hhhh...ttttt... fifty time each.
 
jk22 said:
I'm considering one particular order hhhh...ttttt... fifty time each.

Yep, in that case the probability is the same.
 
But, intuitively, it seems to me that the serie htht... were more plausible than hhhh...ttt... ?
 
jk22 said:
I'm considering one particular order hhhh...ttttt... fifty time each.
I'm a bit confused with your wording there, did you mean getting 50 consecutive heads followed by 50 consecutive tails and you want to know the chance of getting that particular order? If that's the case, it should be indeed (1/2)^100.
jk22 said:
But, intuitively, it seems to me that the serie htht... were more plausible than hhhh...ttt... ?
Then you may want to refine your intuition, getting one particular order is as hard as getting another different particular order.
 
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Yes. I think I'm mixing with the statistics of the length of runs. The probability to obtain k times the same is 1/2^(k+1) hence if i write a program computing that statistics I should obtain an exponential decaying curve.

But that's where my problem arises above, since run length that are longer have less chance to appear, I would expect the case hhhh...tttt... were less likely to appear.

I did a computer simulation of throwing 100 times a coin, and repeating 100'000 times the experiment, storing run length values :
https://drive.google.com/file/d/0B9pGxyM9yy2fYlpIV08wMF9jVk0/view?usp=sharing
 
Last edited:
jk22 said:
But that's where my problem arises above, since run length that are longer have less chance to appear, I would expect the case hhhh...tttt... were less likely to appear.
I think you may be confusing the statistics of run length and the maximum run length of a given outcome. On average, you will get many runs of length 1, but you have only one possibility (if you start by h) that the longest run is of length 1.
 

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