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	I'm honestly not capable of putting him on any kind of range since we know absolutely nothing about him (60 hands/no reads). If villain would be myself, then I'm never holding anything but the nuts here (distant second possibility 7hxh). However, a braindead ape might hold any two cards, so let's try to figure this out by thinking that he could be one:
		
			
			
				
					  Originally Posted by nutsinho 
	
		
			
			
				
					  Originally Posted by pocketfours 
	
		
			
			
				
					  Originally Posted by nutsinho what does he have if u fold Why, you callin? no i am not. but i think its close and i wonder what range youd put him on 
 Let's simplify the problem and assume that we have exactly two possibilities:
 
 1. Villain is a tight nit like myself, or:
 2. Villain is a braindead ape, holding any two cards
 
 Then let's find out what the likelihood for villain to be an ape P(ape) has to be for the call to be profitable. Required equity: 330/760 = 43%
 
 Equity against tight nit:
 Hand 0: 	09.390% { 6d5d }
 Hand 1: 	90.610% { Ah7h, 9h7h, 87s, 7h6h, 87o }
 
 Equity against ape:
 Hand 0: 	60.871% { 6d5d }
 Hand 1: 	39.129% { random }
 
 Simple algebra:
 
 P(ape)*EV(ape) + P(nit)*EV(nit) = 43%
 P(ape)*EV(ape) + (1-P(ape))*EV(nit) = 43%
 P(ape)*EV(ape) + EV(nit) - P(ape)*EV(nit) = 43%
 P(ape)*EV(ape) - P(ape)*EV(nit) = 43% - EV(nit)
 P(ape)*(EV(ape) - EV(nit)) = 43% - EV(nit)
 P(ape) = (43% - EV(nit)) / (EV(ape) - EV(nit))
 P(ape) = (0.43 - 0.09) / (0.61 - 0.09) = 0.34 / 0.52 = 0.65
 
 65%, does that sound reasonable?
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