X-post from MTT, HOH2, strange math on page 166

[KSakuraba]

I posted this to MTT-forum also but realized afterwards this might be more correct place to put it...

I dont know if this has been discussed here already or if i am just tired so please correct me if im way out here, but in the inflection point -chapter, where Harrington is calculating the odds of not getting an all-in called he uses percents as follows:

Player A calls: 2.8%
Player B calls: 5.3%
Player C calls: 11.1%
Player D calls: 2.8%
No-one calls: 78.0%

Shouldn't the real math go NOT like 100%-(A calls+ B calls+ C calls+ D calls) but (A doesn't call)*(B doesn't call)...

So the real number would be (1-0.028)*(1-0.053)*(1-0.111)*(1-0.028)~79,5%, right??

Just got me confused...

Thanks guys,

Saku

[madnak]

Yes, I think you're right. If I follow you.

If player A has a 50% chance to call and player B has a 50% chance to call, there is a 75% (not 100%) probability that one of them will call.

[Borodog]

Yeah, that first formula is spectacularly wrong. If A, B, C, and D each call 50% of the time, no-one calls -100% of the time?