Probability Problem - Coin Tossing

[EnumaElish]

I don't see how that works. All the 1's add up to more than 5?

In this part I am adding 5 cells at a time. The downside with this is it will count a 6-run as two 5-runs.

 

[D H]

The algorithms in posts #4 and #13 are essentially the same algorithm, just looking at things in different ways. Neither counts a 6 run as two five runs.

 

[techmologist]

Sure. The procedure is the same as in post 4, except that you don't have to worry about double counting, making things a lot easier.

Going back to excel, in cell A1 =0.5^H
The value in each cell below this would be exactly half of this, as the sequences is always 'one longer'. For instance, when looking at at least 5 consecutive heads in 20 tosses, we're interested in 0.5^5 (HHHHH) in cell A1 and 0.5^6 (THHHHH) in cells A2 -A16. The sum of A1-A16 is the average number of occurences.

You start with [0.5^H] . As the other values will always be exactly half of this, you simply count how many there are of the 'half values' (N-H=15), you multiply this by half because it is half as likely as [0.5^H], you then add one before multiplying the first part by [0.5^H] to get a correct weighting [(N-H)*(0.5) applies to cells 2-16 +1 adds cell one which is not multiplied by .5 because the value is 2x times as high]

I'm sure there is a much clearer way of putting it, hope it's ok though :)

Oh, I see. Indicator random variables for the win! That counts a sequence like THHHHHHHHH as one run of at least 5 heads, if I understand correctly. Very nice. Thanks.

 

[scalpmaster]

http://www.bjmath.com/bjmath/probable/flips.htm
http://en.wikipedia.org/wiki/Feller's_coin-tossing_constants

In the above 2 links, it's mentioned that if we toss a fair coin ten times then the exact probability that NO pair of heads come up in succession (i.e. n = 10 and k = 2) is p(10,2)= 0.140625.

Therefore, conceptually, the probability of at least 2 consecutive heads coming up in 10 coin tosses = 1-0.14 = 0.86. Will a spreadsheet simulation show roughly the same probability 0.86 after many 10tosses intervals range run?

Let 1 cycle= at least 2 consecutive heads coming up at least once in 10 coin tosses.
Is the probability of 4 cycles appearing consecutively = 0.86*0.86*0.86*0.86 = 0.54 ? Or how should it be calculated? If it is 0.54, does it mean that the probability of winning after winning 4 times in a row is reduced below 1/2?