Re: Doyle Brunson is wrong about straights and free cards
Mathematically , it works out like this .
Suppose the flop contains an ace . We know that the other 2 cards cannot be {2-5} or {10-k) since it's always possible for your opponent to make a straight by the turn . This means that the other 2 cards must belong to {6-9} . But this means that your opponent may still be able to hit a straight by the turn .
So with 100% certainty , there is a greater than 0% chance that your opponent may hit a straight if the flop contains an ace .
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