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Looking at it another way: Each time you set out to throw a coin four times you have a 1/16 chance of throwing four heads and therefore a 15/16 chance that you won't throw four heads. In a sequence of 100 throws you start again 97 times- so the chances that you won't throw a sequence of four within that must be (15/16)^97.
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so you don't get 97 attempts at all
Of course you do- each throw represents the beginning of a new attemt regardless of the prior esquence. Each attemt is therefore independent. |
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Try solving the problem for a total of 5 tosses... by hand and via your formula.
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Tigertiger,
That's pretty uncanny. I was going to suggest something very similar. By Aye Robot's logic, the probability of not getting 2 heads in a row in 3 tosses is: (1 - 0.25) ^ 2 = 0.5625 whereas, of course, the true answer is 5/8. |
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Try solving the problem for a total of 5 tosses... by hand and via your formula.
O.k- I see the problem. Give me a minute though because I'm sure there's a clean solution to this. |
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I think (and im on the edge of my maths here so dont shout too loud if i**k it up ) that it's easier to see if you take it to "chance of getting two heads in three attempts)
You would assume you have two 25% shots However as you prepare to flip the third coin (your second 25% shot) you know you're currently on one of the following sequences TT HT TH (HH would have declared a winner) so you have a 33%*50% chance of winning the second one, not a 50%*50% one. |
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I type too slow.
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Aye robot,
The problem with your approach is that it ignores the fact that a partial success in an uncompleted nth attempt (3 out of 3 heads) partially contributes to the probability of the success of the n + 1th attempt. If I've got 3 in a row in my attempt that started 3 flips ago, then obviously the probability of my attempt that started 2 flips ago is much higher than it would be if I were starting from scratch. |
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on the upside, I got the right answer
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