Variance question

**FourLengthsClear** · 03-10-11, 07:59 PM

In short, yes.

On a roullette wheel (without a zero) the probability of seeing ten consecutive reds is identical to flipping ten consecutive heads.

**sapidoc** · 03-10-11, 08:47 PM

Since you asked about factoring in house advantage, I'll do a full breakdown that can help you better understand that effect on the number of 'reds' or 'heads or..etc... in a row:

The chance of a 50/50 coin going heads on a single flip is of course:

0.5 = 50%

For 2 heads in a row:

0.5 * 0.5 = 0.25 = 25%

For 3 heads in a row:

0.5 * 0.5 * 0.5 = (0.5)^3 = 0.125 = 12.5%

For 4 heads in a row, then:

(0.5)^4 = 0.0625 = 6.25%

So you can get the probability for in X heads in a row by:

(0.5)^X

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Or, even more generally, if the probability of heads is H and the number you would like to get in a row is X then:

H^X

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In Roulette now for example, if we say the chance of red is 48% then the formula is simply put:

(0.48)^X (where X = the number of reds you want in a row)

For 4 reds in a row it would then be:

(0.48)^4 = 0.05308416 = 5.31%

Which is slightly less than a 50/50 coin going heads 4 times in a row (6.25% from above)

So due to house edge on the 'coin flips' you will find that the chance is less than a real 50/50 shot, and to figure it out exactly you can use that formula.

**goucla** · 03-10-11, 09:50 PM

exactly i think the math theories in sports don't work, the only real way to play sharp is to play line movement and you might just get banned.

**Data** · 03-10-11, 11:10 PM

Originally posted by LegitBet

Will all 'events' that have an equal chance of one of two outcomes always have the same chance of having long one way runs?
For example, it's concievable that even though a coin flip is 50/50, one may see a run of 10 heads in a row. (this being the primary reason progressive systems don't work, I imagine).
Would that same chance of '10 in a row' be the same with every conceivable event where one of two outcomes are equally likely?
Would one expect to see the same crazy 10 in a row runs with the same frequency in blackjack (factoring house advantage), coin flips, roulette (again factoring 0 and 00)?
thanks in advance

So, there is your variance question?

**horsiehung** · 03-11-11, 12:37 AM

hey legit...you're too smart for me!

**Data** · 03-11-11, 01:08 AM

Originally posted by horsiehung

hey legit...you're too smart for me!

Ghost alert.

**LegitBet** · 03-11-11, 02:14 AM

I appreciate all the input and I'm not sure I was able to articulate my question clearly. I'll give it sme more thought and repost.
Thanks all

**MonkeyF0cker** · 03-11-11, 07:55 AM

Binomial distribution - Wikipedia

http://en.wikipedia.org/wiki/Binomial_distribution

**Blax0r** · 03-11-11, 11:01 AM

Originally posted by Data

Ghost alert.

Wow, that's definitely an interesting history; if it's not a ghost, horsiehung is legitbet's #1 fan.

**Vaioice** · 03-21-11, 03:48 AM

"The chance of a 50/50 coin going heads on a single flip is of course:

0.5 = 50%

For 2 heads in a row:

0.5 * 0.5 = 0.25 = 25%

For 3 heads in a row:

0.5 * 0.5 * 0.5 = (0.5)^3 = 0.125 = 12.5%

For 4 heads in a row, then:

(0.5)^4 = 0.0625 = 6.25%" i don't think this is correct. ur odds would still be 50/50,

-vai

**JustinBieber** · 03-21-11, 07:20 AM

lol ghost, ban him imo.

**Nookx** · 03-21-11, 10:49 AM

variance sucks yo

**thekid667** · 03-22-11, 12:01 AM

YOu gonna wait till .....fatboy....fatboy....wait till tommorrow

**Wackabrew** · 03-23-11, 02:55 PM

Originally posted by Vaioice

"The chance of a 50/50 coin going heads on a single flip is of course: 0.5 = 50% For 2 heads in a row: 0.5 * 0.5 = 0.25 = 25% For 3 heads in a row: 0.5 * 0.5 * 0.5 = (0.5)^3 = 0.125 = 12.5% For 4 heads in a row, then: (0.5)^4 = 0.0625 = 6.25%"

i don't think this is correct. ur odds would still be 50/50, -vai

The odds of the next flip being heads would be 50/50, as any single flip is an indpendent event. Sapidoc is calculating the probability of one individual event happening X times in a row.