[AaronBrown]

While using the observed frequency as an estimate of future probability is convenient, it can't be right here. Clearly you are not 100% certain that something will happen in the future, based on 10 out of 10 past successes.

Bayesians begin with a prior distribution. For the binomial, this typically takes the form of a Beta distribution with parameters A and B. The estimate of probability is (A + successes)/(A + B + trials). A/(A + B) is your estimate of the probability before you saw any trials. If you're pretty sure of the probability, make A and B big; if you have almost no information, make A and B small.

In this case, I might say I think the probability of check raising is 50%, and I'm moderately confident of that, so I'll make A = B = 5. Even after seeing 10 out of 10, I still think there is only (5 + 10)/(5 + 5 + 10) = 75% probability of the next check raise being strong.

The classical approach is to set a confidence level on the probability. If the probability is 0.7411 or higher, there is at least a 5% chance of getting 10 out of 10. So the 95% confidence interval on the probability is 0.7411 to 1. The midpoint of that interval is 0.8706.

Your 10/11 estimate comes from a non-parametric argument. Assume that the check raiser gets some kind of signal that determines when he check raises. This could come from some logical system, or his mood, or his dog could tell him how to play. It doesn't matter. The odds that the next signal is weaker than all the ones that came before it is 1/11, as long as the signals are independent. So there's 10/11 chance that his next signal will be high enough to check raise.

All of these assume independence of the outcomes. In real Poker, he's going to decide based on what he thinks you think. He'll check raise if he thinks you still haven't caught on to the strategy, not if you don't. Unless his dog is telling him how to play.