
#1
Nov2812, 06:20 PM

P: 3

I hopefully have a rather simple question.... I am looking for the equation to calculate the probability when the total counts of heads and tails get off balance. To further explain, when I flip the coin once there is 50/50 chance getting one or the other.... If I flip the coin 10,000 times that probability will get closer and closer to .5. However, on the way to 10,000 flips.... Maybe after 100 flips, I might have 30 heads and 70 tails... Therefore my next flip should have have more of a chance of hitting heads (greater than .5 anyway).... How do I calculate that "improved" probability after the measuring x amount of data?
Thanks 



#2
Nov2812, 06:31 PM

Sci Advisor
P: 3,175

If you are flipping the coin in some way that you can change the probability of getting a head and make it more or less than 0.5 then you could look at how many total heads you have and try to "balance things out" by changing the probability of tossing a head. But if you assume you are always tossing a fair coin then you are tossing a fair coin, regardless of the past history of heads and tails. 



#3
Nov2812, 06:36 PM

P: 771





#4
Nov2812, 07:21 PM

HW Helper
P: 1,391

Coin toss probability when not 50/50
As the previous posters have mentioned, if the coin is fair the probability of heads is 0.5 no matter what the previous sequence of heads and tails is.
At the risk of going beyond what you have learned so far, Mikes1098, you could ask a different question that does depend on the past sequence of heads and tails: is this a fair coin or not? Using the Bayesian view of probability, you could calculate ##P(p_{heads} = 0.5  A_n)##  the probability that the probability of the coin coming up heads is 0.5, given a that you have observed a sequence, ##A_n##, of n previous coin toss outcomes. This probability will change after each new flip, as each new flip gives you another data point that helps you decide whether or not the coin is fair. You would need some model for how each new data point changes your estimate of ##P(p_{heads} = 0.5  A_n) \rightarrow P(p_{heads} = 0.5  A_{n+1})##. However, it is very important to note that this still has no bearing whatsoever on what the outcome of the next coin toss is! It is only an estimate of the probability that the coin is fair! If this estimate is telling you that the probability that ##p_{heads} = 0.5## is small, however, then you may wish to start betting more on tails (however, ##P(p_{heads} = 0.5  A_n)## doesn't tell you what the actual heads:tails ratio is). 



#5
Nov2812, 07:28 PM

P: 3

ok, maybe the 100 flip example was a bit exaggerated.... assuming it is a fair coin, after 10 flips it would not be shocking not to have perfect 5 heads and 5 tails. Eventually, which ever is behind will catch up after enough flips to equal the predicted .5... therefore doesn't the probability have to favor the one, that by chance, is lower? It seems there should be a way to take measured random data and alter the future probability have the remainder of the set.




#6
Nov2812, 07:35 PM

P: 3

Thanks Mute, that was exactly what I was looking for. I guess my definition of "fair coin" might have been off.




#7
Nov2812, 08:11 PM

HW Helper
P: 1,391




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