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View Full Version : Quick calculation

illunious
11-22-2004, 06:23 PM
There's 3 players to my left with preflop raise stats of 10%, 10%, and 10%, how do I calculate the chance of it getting raised? 30% just doesn't seem right.

Lost Wages
11-22-2004, 06:30 PM
10% of the time the first player will raise. Of the remaining 90% of the time the second player will raise 10% of the time, etc. 27.1%

Lost Wages

illunious
11-22-2004, 06:31 PM
Thanks, it was on the tip of my brain, just couldn't get it out /images/graemlins/smile.gif

schroedy
11-22-2004, 07:44 PM
Easy way to do these "multiple miss" questions = 1-(probability of missing).

Here each of your opponenets is a 90% non-raiser, so your odds of being raised at least once by at least one of them is 1 minus (.9)*(.9)*(.9) = 1 minus .729 = 27.1%.

illunious
11-22-2004, 08:03 PM
[ QUOTE ]
Here each of your opponenets is a 90% non-raiser, so your odds of being raised at least once by at least one of them is 1 minus (.9)*(.9)*(.9) = 1 minus .729 = 27.1%.

[/ QUOTE ]

This is a great solution, thank you. I'll just remove the "1 minus" part and use that number for the chance it -won't- be raised.

I find myself trying to make this type of calculation on the fly more often lately with the advent of PlayerView and GameTime+.

mannika
11-22-2004, 08:25 PM
The only problem with this calculation is that it assumes that player B's choice to raise or not is independant of player A's choice. In reality, this assumption is grossly inaccurate. If player A raises, player B will probably only call A's raise with about half of the hands that he would have raised with in the first place, and re-raise the other 50%.

This makes the problem more complex because we now need conditional probabilities.

If similar to the previous example, we take each players probability of a preflop raise given no raise as 12% p(PFR|no raise) = 0.12
and likewise p(PFR|raise) = 0.06

(the above numbers are fictional, but would probably result in an overall PFR% of 10% for each of the players)

If we take this as being true, then the probability of having a raise is 1 - (1 - 0.12)^3, or 31.9%.

In all cases with rational players (those who re-raise with less hands than they would raise with), the actual probability of having a raise will ALWAYS be greater than using the equation above.

astrodon
11-22-2004, 09:04 PM
Jeeze Loius! Either way you have about a 30% +/- indication of raise which was the original intuitive conclusion . Is it really important to be within 2% +/- on the fly?