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What would the probability be for the 11th flip to be the same as 10th flip


New member
Feb 7, 2019
I'm wanting to know if there is a formula that can get the probability of then next flip, by taking the data/ averages of the last 10 flips.
So, if the last 10 flips were "H,T,T,H,H,T,T,H,H,H".
What would the probability be for the 11th flip to be the same as 10th flip?


Staff member
Feb 5, 2013
There are two possible outcomes for every toss so that probability would be 1/2. This is different than, say, "What is the probability of tossing two heads in a row?" which would be 1/2 * 1/2 = 1/4.

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
Assuming this is a fair coin, that the probability of "heads" or "tails" is always 1/2, then, pretty much by definition, the probability of heads or tails on the 11th flip is 1/2.

If we are not assuming that, but are considering the possibility that the coin is "weighted" so that one of "heads" or "tails" is more likely than the other, we could consider that "history", the first ten tosses, reflects which is more likely. Here the first ten tosses are "H,T,T,H,H,T,T,H,H,H", 6 heads, and 4 tails, then we would estimate the probability of heads on any one toss, and in particular, the eleventh toss, is 0.6 and the probability of tails is 0.4.

Notice that, in this scenario, the fact that there are more heads than tails in the 10 tosses means that another head is more likely than another tail. Future tosses will \left(\right)not "make up" for previous tosses.


New member
Feb 7, 2019