# What are the odds that the pattern, win lose lose, will happen 23 times in a row (69 rounds)?

In a game of 50/50 (this example could be a coin flip). Before any flips, What are the odds that the pattern win lose lose (heads tails tails if your choice was heads each time.) will happen 23 times in a row (69 rounds)?

What about after the chain starts? After the initial wll, what do the odds become for the next "Set" of wll conditions? And the third, up to the 23 occurrence? Again, keeping with your initial choice (option) each time.

To put a twist on the question, how does the solution change if the odds are based off the roulette "50/50" options? (With pesky zeros added, or coin landing on its edge)

Would that change the odds since it's not about the number but rather pattern if the option won or lost?

Can this also be broken down to what are the odds for each round (or flip) for round 1 to be a winner round 2 will lose round 3 will lose, (rnd 4 will win 5 will lose 6 will also lose etc to rnd 69?) wllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwllwll

I have seen similar questions about odds and statistics asked, and one of those formulas may work for this riddle, but my knowledge of mathematics are limited. I greatly appreciate the helpfulness of the stack network communities and the cumulative knowledge we share.

The following Stack Exchange links are very informative on the subject as well.

What is the probability of a coin landing tails 7 times in a row in a series of 150 coin flips?

Measuring Roulette sequence probability

Roulette outcome probability

If I flip a coin $100$ times, why do the results trend towards $50-50$?

Roulette Conditional Probability

Roulette betting system probability

Simple coin flip probability

Probability of $5$ fair coin flips having strictly more heads than $4$ fair coin flips

## migrated from stats.stackexchange.comJun 25 '15 at 18:06

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• This problem actually is much easier than any of the related ones. Your questions are answered simply by applying the definition of independence (which is what you must assume to answer them without further assumptions): the probabilities of independent events multiply. So go ahead and do the multiplications. You will notice that although the patterns will determine the probabilities, they do not change the fact that in every case you just multiply them all, so there really is nothing here involving combinatorial issues. – whuber Jun 11 '15 at 14:49
• (Migrating to Mathematics by request of the OP.) – whuber Jun 25 '15 at 18:06
• Thank you: we appreciate your respect for not double-posting on SE sites. – whuber Jun 25 '15 at 18:09