How do you figure odds of flipping a coin in CONSECUTIVE heads? 1 heads = 50/50. 2 heads? 5 heads? Etc.?

Question

I know that a single flip of a coin is 50/50 odds of either heads or tails. How do you figure the mathmatical odds for CONSECUTIVE flips of the SAME outcome, either all heads OR all tails? I know that every SINGLE flip = 50/50, but what are the odds on groupings of flips producing ALL the same result? What do the odds reduce to on every flip past one of predicting the same consecutive outcome?

Answer 1

Ahh! True, 6 consecutive flips of a coin in which each lands heads has a (1/2)^6 probability of happening. But what are the odds that the seventh flip will be heads, too?

Answer 2

odds for a heads on flipping a coin = (1/2) / (1/2) = 1:1
odds for 2 consecutive heads = (1/4) / (3/4)= 1:3
odds for 3 consecutive heads = (1/8) / (7/8)= 1:7
odds for 4 consecutive heads = (1/16) / (15/16) = 1:15
odds for 5 consecutive heads = (1/32) / (31/32) = 1:32

Answer 3

To figure out the odds of head or tails consecutively, you multiply the odds of each flip together.

For example, what are the odds of landing on heads 3 times?

(1/2)(1/2)(1/2) = 1/8
.5 x .5 x .5 = .125

Odds on landing on Heads 2 times?

.5 x .5 = .25 or 1/4

Think of it this way: What are the odds that I will land on heads on the first flip? Then, what are the odds that I will land on heads on the 2nd flip? The 3rd? The 4th? You can see that the odds will remain the same for each SINGLE flip, because no matter what, even if you landed on heads or tails on the previous flip, your chances of landing on heads remain the same on each flip. To calculate the chance of landing on heads CONSECUTIVELY, you must multiply all the chances of each flip together.

You can also think of it this way:

(chance of landing on heads on a single flip)^# of flips

Take the chance of landing on heads and put it to the power of the # of flips.
.
.5^3 = .125 or 1/8
.5^2 = .25 or 1/4

Hope this makes sense!

Answer 4

Either heads or tails?

For x flips, the probability is 1 / (2 ^ (x-1) ).

Evan, you forgot to multiply by 2, because either heads or tails is okay.

Answer 5

(1/2)^n

1st flip = (1/2)
2nd flip = (1/4)
3rd flip = (1/8)
4th flip = (1/16)
5th flip = (1/32)
etc...

Answer 6

Do .5 to the x power, where x is the number of flips.

Answer 7

A simple answer to your question buddy: (0.5)^x, where x is the number of throws. hope this helps!! :)