Chances Question

[GTSamIAm]

Someone has check/raised with very strong hands 10 out of 10 times he has shown his hand. He now check/raises you in a pot, what are the chances he has a very strong hand, using your observations so far? Is it about 91%, i.e. 10/11? What is your best estimate, given the information you already have?

[uuDevil]

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Someone has check/raised with very strong hands 10 out of 10 times he has shown his hand. He now check/raises you in a pot, what are the chances he has a very strong hand, using your observations so far? Is it about 91%, i.e. 10/11? What is your best estimate, given the information you already have?

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Without more information, I think the best estimate of the probability he has a strong hand is the proportion of successes in the sample, which is 1. So fold unless you have a monster.

If Harrington's Law of Bluffing applies, or you've seen him bluff/semibluff in the past, that changes things.

[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.

[mosdef]

Aaron - I am partial (in theory, perhaps not in practice) is a simplified credibility theory approach, where you combine the observed data (10 for 10) with a full-credibility criteria and a "default" rate of a check-raise meaning strength.

For example, you may guess that the average opponent has the goods 75% of the time when he CR. However, you acknowledge that it actually varies by opponent. You estimate that if you were to see someone CR 100 times, you would say that that observed player has an actual rate equal to his observed rate. For a player with < 100 obseravations, you would combine the default rate with the observed rate. For the example above:

Estimated rate (given you've seen 10 for 10) = 100% x Z + 75% x (1 - Z),

Z = (10/100)^0.5

(This square-root rule is typical).

The problem with my suggestion is that you need a default rate and a full-credibility threshold. The problem with your approach is that you often end up with ranges too wide to be useful. It's a matter of taste as to which short-comings you'd rather live with.

As an aside, I think that credibility theory is the most underappreciated tool for evaluating poker statistics. Perhaps because it's less commonly understood that confidence intervals, which are first-year stats course material.

I'd be interested in your comments. I've read your posts and value your opinions (I'm sure that makes you feel like you've really accomplished something in life /images/graemlins/wink.gif )

[GTSamIAm]

So taking in no information other than he has check/raised 10 out of 10 times with a monster we have seen, our best mathematical approximation is 87%?

[hukilai]

You cannot calculate your chances here, as all terms you used are unquantifiable.

If you mean, say, what his chances of getting dealt AA again if he alreay got them ten times in a row, then the answer is that it is still 0.45%.