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Coin is tossed $100$ times. For $1\leq i\leq 94$, let $A_i$ be the event that the $i^{th}$ toss is a tail and the following $6$ tosses are heads. Calculate the number of sample points in the event $A_i$ and hence determine its probability.

So the way I see it is $7$ coins have to be fixed for event $A_i$ to occur i.e. we get THHHHHH somewhere in the row of $100$, the T being one of the coins between and including coin $1$ and coin $94$. That means the other $93$ coins are unfixed for each $i$, so $93!$ possibilities where $A_i$ is satisfied. Overall we have $100!$ possible ways $100$ coin flips can turn out.

Probability = # (outcomes that satisfy Ai)/(Total outcomes) so is the answer $\frac{93!}{100!}$? The probability of six heads occuring in a row seems kinda small if this is the right answer...what am I doing wrong? I think I must be thinking about it the wrong way and would really appreciate it if someone could explain this to me. Thank you.

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  • $\begingroup$ Well first off there are only $2^{100}$ possibilities, not $100!$. $\endgroup$ – Gregory Grant Mar 11 '15 at 21:43
  • $\begingroup$ Basically you have to specify seven of the flips and the other 93 can be anything. Therefore for each position $i=1,\dots,94$ you can have $2^{93}$ other configurations. $\endgroup$ – Gregory Grant Mar 11 '15 at 22:02
  • $\begingroup$ 2^93 because each position can either be heads or tails, that makes sense. So the probability is 2^93/2^100 which cancels down to 1/2^7. Thanks. But then what was I doing wrong when I thought it was actually 93! over 100!? What kind of situation would I have for that to be the answer? I just want to be clear on the difference. $\endgroup$ – Wolverine Mar 12 '15 at 13:15
  • $\begingroup$ $100!$ would be something like if you had 100 different cards and you lay them out in a row. Then there are 100 ways to choose the first card, once the first is chosen there are 99 ways to choose the 2nd card, once the 1st and 2nd are chosen there are 98 ways to choose the 3rd, etc.. So there are $100!$ different ways to lay 100 different cards in a row. Meanwhile coin flipping is a totally different thing where each step involves exactly one of two choices. $\endgroup$ – Gregory Grant Mar 12 '15 at 15:02
  • $\begingroup$ @GregoryGrant I understand now, thanks for your help. $\endgroup$ – Wolverine Mar 12 '15 at 17:15
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Your hypothesis is that in the last seven tosses you obtain a specific sequence, and you do not ask any other about the first 93 tosses.

You have to remember that for Bernoulli theorem the single extraction have not memory, so the probability to obtain THHHHHH in the last seven tosses is:

$$ P = \frac{1}{2^7} = \frac {1}{128} $$

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