Re: How many callers does Buzz need?
Hey Buzz,
I'm still not comfortable here.
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Thus we have a total of about 8 outs. And we have two chances.
Here’s the math set-up for the next step:
(8/44) + (36/44)*(8/43)
Then solving (8/44) + (6.7/44) = (14.7/44) = about one third.
Thus we expect to win about one third of the time and lose two thirds of the time. We lose one chip on each of the two times we lose. Therefore we need to win two chips each time we win.
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Let's look at the 36/44 number. The 36 represents the missed clubs. But what if the missed clubs pair the board? If we subtract the 9 non-club board pairers, we get 27/44, which lowers our chances.
On the other hand, 20 of the 36 provide the first leg of runner-runner low. This might be good and it might be bad: good if it gives us extra outs; bad if we are playing for half the pot. The 3 non-club 2s might be worth another 1.5 outs, giving us 9.5/44 + 27/44 * 8/43 = ~.3.
Strangely, however I massage these numbers, I end up ~.3. [img]/images/graemlins/smile.gif[/img]
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