Understanding expected odds?

Ok Im really confused. What's the point of expected odds?

For example if I toss a standard coin with "heads" and "tails" my odds are 1/2 or 50% chance of having one or the other come up.

The coin has NO MEMORY of previous results. My odds will ALWAYS remain 1/2 every toss independent of previous tosses.

My expected odds of me tossing 10 "heads" in a row is supposedly
(1/2)^10 = 0.0009765625 or 0.097%

(1/2)^n = x
as "n" approaches infinity, x approaches 0

Supposedly the more I toss the coin the more i'm EXPECTED to reach an average of 1/2. There is no physical law that prevents me of getting 10 heads in a row. Or even 1,000,000 "heads" or 1,000,000,000 "heads" or infinite "heads" in a row.

So if I toss the coin 9 "heads" in a row my 10th toss my odds will still be 1/2

So (1/2)^n should always equal 1/2. Shouldn't "n" be irrelevant because "n" is expected previous results?

Someone help me understand im a tad bit confused!

THANKS!

Quote: danepeterson

So if I toss the coin 9 "heads" in a row my 10th toss my odds will still be 1/2

So (1/2)^n should always equal 1/2. Shouldn't "n" be irrelevant because "n" is expected previous results?

Someone help me understand im a tad bit confused!

THANKS!

After the first 9 "heads" in a row, the formula for flip #10 is (1/2)^1=.5, since the 10th flip has a 50% chance of being heads.

If you want to know the odds that your next 10 flips will all be heads, the formula is (1/2)^10=.000977, or 1:1024. that formula has nothing to do with any previous flips - it's always forward looking.

Quote: rdw4potus

After the first 9 "heads" in a row, the formula for flip #10 is (1/2)^1=.5, since the 10th flip has a 50% chance of being heads.

If you want to know the odds that your next 10 flips will all be heads, the formula is (1/2)^10=.000977, or 1:1024. that formula has nothing to do with any previous flips - it's always forward looking.

The big difference is the pre-established definition of the number of flips (in the case above, 10) to be observed. As soon as you do that, the question is no longer, "What are the odds of any single trial?" It is now, "What is the expected outcome of a series of n trials?" These are two very different questions.

Edit: Actually, thanks the Professor Persi Diaconis at Stanford, we now know that flipping a coin is not a 50/50 proposition. It will end the same way it started about 51% of the time. Link to the 31 page .pdf of the report here.

>Ok Im really confused.
So am I.

> What's the point of expected odds?
Expected odds are a forward looking statement concerning averages of several trials over a long term.

>For example if I toss a standard coin with "heads" and "tails" my odds are 1/2 or 50% chance of having one or the other come up.
YOU GOT IT!!!!

>The coin has NO MEMORY of previous results. My odds will ALWAYS remain 1/2 every toss independent of previous tosses.
YOU REALLY GOT IT!!! You are not confused at all.

My expected odds of me tossing 10 "heads" in a row is supposedly
(1/2)^10 = 0.0009765625 or 0.097%

(1/2)^n = x
as "n" approaches infinity, x approaches 0

>Supposedly the more I toss the coin the more i'm EXPECTED to reach an average of 1/2.
Yes, though at any one particular stopping point it is still an expectation and will not necessarily be actually observed.
> There is no physical law that prevents me of getting 10 heads in a row.
>Or even 1,000,000 "heads" or 1,000,000,000 "heads" or infinite "heads" in a row.
That is true but that sweet young cocktail waitress serving you drinks at Toss number one will probably be watching her granddaughter be serving you drinks while you are still waiting for 1,000,000 heads in a row.

>So if I toss the coin 9 "heads" in a row my 10th toss my odds will still be 1/2
True. The coin has no memory and your memory of the past events has no effect on the coin. You could tell those gathered around that its your 9th toss or you could tell them its your third toss. Their knowledge of the past and emotional expectations for that next toss has no effect on that coin either.

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silly
Sally

Thanks everyone!!! Great answers!

hey mustangsally, how do you get this formula?

Quote: mustangsally

2^(n+1) - 2 (fair coin flip formula)

Im guessing "n" is my streak but how did you get the other numbers?

2^((3)+1) - 2 = 14

does that mean i should expect to see a streak of 3 in a row within 14 flips? sorry im not that great with math!

Thanks!

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silly

Sally

Awesome thanks!!!