Fooled by randomness

[tooting]

Hi all. I’m trying to write an article on something but as a failed gambler have long since forgotten the maths. Anyone help?

In 100 random ‘heads and tails’ tosses, what is the percentage likelihood of there being 8 successive heads, (or tails) ?

Cheers…

[Tuffers]

I think it is 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 (sorry no calculator handy)

[RedRum77]

Really, I thought it was 2 to the power of 100 divided by 2 to the power of 8 but I might be getting confused.

[aji]

The probability of n successive tosses with the same outcome is

1/(2 power n)

so of 8 successive heads is

1/(2 power = 1/256 = 0.00390625

[tooting]

Cheers guys, but not quite answering what I’m asking…

[Drone]

A straightforward reciprocating of 2^n may seem the logical solution but in fact the maths governing sequences of heads/tails is rather more involved, and as the first article below points out, counter-intuitive

see:

http://mathworld.wolfram.com/CoinTossing.html

http://mathworld.wolfram.com/Run.html

[Relkeel]

Is it too simple to say that, if you toss a coin 100 times, there are 93 sequences of 8 (1-2-3-4-5-6-7-8, 2-3-4-5-6-7-8-9, 3-4-5-6-7-8-9-10…upto…93-94-95-96-97-98-99-100)?

Each of those sequences has a 1/256 chance of being all heads, therefore the chance of any 8-head sequences is 93/256 (36%).

If you’re looking for a sequence of eight heads OR tails, I reckon the chance doubles to 72%.

[tooting]

Thanks you two – that’s what I was after.

Relkeel – that’s how I worked it out myself, when I was trying to remember. I rather fear Drone’s pages of equations is more accurate though!

Thanks – very useful.

[Venusian]

Relkeel, I think you’re right (but not entirely sure!).

For heads OR tails, basically a sequence of 7 rather than 8, it comes to 94/128 which is 0.734.

[Glenn]

0.726 (93/128) is the expected number of eight-in-a-row tosses.

It’s not the probability of getting 8-in-a-row.

If they were independent events you’d raise 1-p to the power of 93, not multiply by 93. That would make it 48%.

But they’re not independent events. If you’ve got eight-in-a-row once, you’re almost evens to get it twice (ie 9 in a row), almost 3-1 to get it three times etc

I make the true chance of success a shade under 32%.

[ricky lake]

Sir , this post is a touch geeky ….having just returned from Gotham(city) about 4 miles north of Nottingham , where they are just about to have a sale in logarithm calculators …..obv they are all tuned into this forum……its a concern for sure …

Now where was I …….is Paul Roy gone yet >>>>

Ricky

[Dolus]

The probability of n successive tosses with the same outcome is

1/(2 power n)

so of 8 successive heads is

1/(2 power = 1/256 = 0.00390625

I agree with aji

Then multiply 0.00390626 x 100 to get the percentage = 0.39%

Tuffers way of doing it gets the same answer.

Another way is to work out how much an 8 horse accumulator would return if each winner was 1/1

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256 return (odds 255/1)convert to a percentage 100/256 = 0.39%

[Glenn]

Sir , this post is a touch geeky ….having just returned from Gotham(city) about 4 miles north of Nottingham , where they are just about to have a sale in logarithm calculators …..obv they are all tuned into this forum……its a concern for sure …

Now where was I …….is Paul Roy gone yet >>>>

Ricky

I read somewhere that a tosser’s lounge was the answer to Gotham’s troubles – protecting the squares from themselves. Reading this thread it seems such a lounge may, conversely, provide the keys to the safe.

I have received a call from Ratman on the ole Batphone and he suggests that a certain liquid refreshment, if not imminenet, may be on the cards sometime soon. I will keep you posted.

[Marginal Value]

Hi all. I’m trying to write an article on something but as a failed gambler have long since forgotten the maths. Anyone help?

In 100 random ‘heads and tails’ tosses, what is the percentage likelihood of there being 8 successive heads, (or tails) ?

Cheers…

In general, if you are looking for X number of successes in Y tries, and the probability of each separate success is P, then the total probability is:

Y!/(X!(Y-X)!) * P^X * (1-P)^(Y-X)

where ! means Factorial, * means multiply, and ^ means raised to the power of

and Factorial is multiplying together all the cardinal numbers from your chosen number down to one, so Factoral 5 = 5 * 4 * 3 * 2 * 1 = 120

But your question is a little unclear about exactly how many times a sequence of 8 heads or tails is acceptable. Can you accept two or more separate sequences of 8, and a sequence of 9 is, in effect, two sequences of 8. It’s complicated, isn’t it?

However, if we assume that your “…what is the percentage likelihood of there being 8 successive heads, (or tails)…” means “… of there being AT LEAST one occurance of 8 successive…”, it becomes a little simpler, because we can then look at it in a very different way. We can find the probability of zero occurances of 8 successive heads/tails. And if we then subtract that from 1 we will have the probabilty of non-zero occurances (meaning “at least one occurance, but maybe more). Which is, I hope, what you mean.

As others have said the probabilty of getting 8 successive heads or tails with just 8 tosses of a fair coin is 0.00309625.

You are having 93 tries at getting 8 successive heads/tails by tossing a coin 100 times.

The probabilty of 1 or more occurances of an event with probability P happening, given N tries is

1-((1-P)^N)

If P = 0.00390625 (probability of the event happening)

If N = 93 (number of tries)

The overall probabilty of the event happening in 93 tries is

1 – ((1 – 0.00309625)^93)

= 1 – (0.99690375)^93

= 1 – 0.749463279

= 0.250536721

or about 25%

I did the calculations on a normal spreadsheet, so the rounding inaccuracies can be blamed on that. All other innaccuracies of knowledge and logic are purely down to me.

[Glenn]

Holy algorithims Batman.

The probability of failing to get 8 consecutive heads or tails on the first 8 tosses is 127/128.

The probability of failing to get an 8th consecutive head/tail on the ninth toss g

iven that you failed on the 8th

is higher than this.

Specifically there are 512 sequences of nine tosses. Four can be ruled out knowing that you failed on the 8th:

HHHHHHHHT
HHHHHHHHH
TTTTTTTTH
TTTTTTTTT

This leaves only two of the remaining sequences that give you success out of 508 possible sequences, so the chances of failure is 506/508.

I won’t go into the probabilities for tenth tosses on but they’re very similar to the ninth toss, with just a shade lower chance of failure (as you rule out some sequences that slightly favour you this time).

So the chances of failure can be approximated as

(127/128)*((506/508)^92)

=0.69

So the chances of success is 31% approximately. The real figure is a few ticks higher owing to the fact that we’ve mady a tiny overestimation of failure for the tenth toss on.

[tooting]

Since posting this on here, I’ve searched the net as well, and it’s amazing how many different completely credible sounding answers I now have.

Marginal Value – I particularly like your solution, mainly because I actually understand it. Then again, Glen is the resident board genius on these things…

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