probability of a run

A randomiser of n bits (base 2 unsigned)

When n ≥ 2, we have 2ⁿ possible ways to store numbers.

Number of consecutive Heads:

10 heads + 90 random (1st ~ 10th try, 2nd ~ 11th try ... 91st ~ 100th try)

91 tries × 2⁹⁰ = 112 652 543 574 969 605 015 820 304 384

11 heads + 89 random (1st ~ 11th, 2nd ~ 12th ... 90th ~ 100th try)

90 tries × 2⁸⁹ = 55 707 301 767 842 112 370 460 590 080

12 heads + 88 random

89 tries × 2⁸⁸ = 27 544 165 874 099 711 116 505 513 984

...

The information above is an assumption; we do it by working backwards and finding a pattern.

97 heads + 3 random (97H, T, T, T; 97H, T, T, H; 97H, T, H, T; 97H, T, H, H  )

4 tries

98 heads + 2 random (98 heads, 1 tail, 1 head or 98 heads, 2 tails)

2 tries

99 heads + 1 tails

1 try

100 heads

1 try

                       100H   99H         98H            97H        96H            95H                  11H                  10H

Thus, we have 1 + (2¹ - 2⁰) + (2² - 2¹) + (2³ - 2²) + (2⁴ - 2³) + (2⁵ - 2⁴) + ... + (2⁸⁹ - 2⁸⁸) + (2⁹⁰ - 2⁸⁹) = 2⁹⁰ = 

                        ||

                        2⁰

1 237 940 039 285 380 274 899 124 224 ways.

Do some cancellation. Remove the brackets as there is only + outside of brackets, and a + (b - c) = a + b - c.

Thus, the probability of getting at least 10 consecutive heads is:

Percentage : \(8.0779356694631608 \times 10^{-28} %\)

Fraction : \(1 \over 1 237 940 039 285 380 274 899 124 224\)         

My Calculator

Brand: Casio

Model: fx-96SG PLUS

Type: Natural-V.P.A.M.

Digit Display: 10 digits

Calculation Input: 15 digits

Calculation Range: \(1 \times {10}^{-99} ≤ |x| < 1 \times {10}^{100} \, or \, 0\)

MWizzard has asked me to comment on this answer.  At least I think that is what I was expected to do ???

Anyway.

what is the probability of at least 10 consecutive heads if a fair coin is tossed 100 times?

I am not really sure...  maybe

The chance of getting all heads in any 10 consecutive rolls is     0.5^10

0.5^10 = 0.0009765625

There are a total of 90 groups of 10 consecutive rolls so maybe the prob is

90*0.5^10

90*0.5^10 = 0.087890625

I maybe be double counting here.    ://

BUT

The probability has to be greater than 0.0009765625  so I do not think yours is correct.

Maybe another mathematician could comment here.  

Alan, could you run this through a simulator maybe, I don't know how to do that but maybe your do????

If I could edit my answer I would.

I am quite sure that I have double counted however I still say the answer has to be greater then    0.5^10

I've just done a Monte-Carlo simulation using 10^6 trials (i.e. tossing 100 coins, one million times and counting in how many trials at least ten consecutive heads appeared).   The result is a probability of approximately 0.044  (i.e. 4.4% of the time you get at least ten consecutive heads).

Thanks very much Alan that is about what I would have expected since I knew I had double counted.

I have done questions like this recently and they were correct but I do it by logic not by formula.

I think this one might be too long to do that way.

Nauseated probably has a formula that will work for this but I don't know it. :/

Nauseated's formulas are excellent especially if you want to program them for different values.

I like Alan's method.  Simulations are great.  They can be used to check logic and formula outputs. Always a good thing when dealing with probablility.