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Old 07-01-2008, 12:56 PM   #199
st.cronin
General Manager
 
Join Date: Oct 2004
Location: New Mexico
Here's how I work it out:

Assuming A 9 held by player, dealer shows 5. There are 3 As out of 49 cards left, giving 6.12% odds for a score of 21. Here's the table:

Code:
A 3 - 21 - 6.12 2 4 - 12 - 8.16 3 4 - 13 - 8.16 4 4 - 14 - 8.16 5 3 - 15 - 6.12 6 4 - 16 - 8.16 7 4 - 17 - 8.16 8 4 - 18 - 8.16 9 3 - 19 - 6.12 10 16 - 20 - 32.65

Adding the under 17 hands, we get:

Code:
21 - 6.12 less than 17 - 38.78 17 - 8.16 18 - 8.16 19 - 6.12 20 - 32.65

Dealer will bust about 40% of the time, giving all of those scores wins. In other cases (leaving pushes out of it): 20 will beat the dealer 75%. 19 will beat the dealer 50%. 18 will beat the dealer 25%. 17 and under will lose or push 100%. So add those to the table and we get:

Code:
21 - 6.12 (6.12) less than 17 - 38.78 (15.51) 17 - 8.16 (3.26) 18 - 8.16 (3.26 + 1.25 = 4.51) 19 - 6.12 (2.45 + 1.84 = 4.29) 20 - 32.65 (13.06 + 14.69 = 27.75)

Adding up all the win expectancies I get 61.41, making this a sure double situation. Where is my error?
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