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Old March 23rd, 2009, 04:55 PM
stophon stophon is offline
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Join Date: Feb 2009
Posts: 88

Originally Posted by Nazgul View Post
You didn't include the probability that the card underneath the ace has a 3 card clump.
I think I included the probabilities that it will appear in either a 2 or 3 card clump.
-------O (20%)



I did assume only one rifle, however this data can continue to be replicated for as many rifles as you so choose (just requires more work, I'm working on a formula. From what I can tell it's a pretty intense quadratic). The formula for the chance your ace does not get lost due to stripping is:
Where s is the number of times the deck is cut + the number of chunks that the deck is divided into as it is stripped. In a real casino you can improve on this number if you know approximately where your key card is. However if you just want objective math probability, that's the formula.

Suppose the deck is riffled once, stripped down once into five pieces and cut at the end. You are playing heads up against the dealer and you do not spread to multiple hands.

5 strips + 1 cut = 6
1-6/52 = (46/52) chance the key card and the ace are not seperated

Mathematical EV
Event Player Adv Probability Expected Value
Player Ace +.51------ .549-------- .2799
Dealer Ace -.34------ .335-------- -.1139
Ace lost___-.01----- .115-------- -.00115
Total .16485 = 16.5% Advantage

I assumed that if the ace was lost due to stripping the chances of it being in that clump of cards was now 0%. I know this isn't true but the odds of it showing back up shouldn't be mathematically significant. I use the hi-lo count, and at a count of -1 (ace lost) I have a 1% disadvantage. Again, not a perfect number (though hi-lo has a good betting correlation) but its fine for what we're doing.

You could also calculate how your EV changes for spreading to two hands. I'll work on getting the data for a more realistic number of riffles.
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