I am a math student at the University of Utah and have studied the probability associated with blackjack. However, I am wondering about the basic strategy case of 9 vs 7 in a single deck black jack game. The numbers I'm getting says they are good enough that you should double (if allowed) rather than just take a hit. Here are the statistics:
Case: (2,7) vs 7:
Expected Occurence (%) = 96/1326 = 0.0724
HitWinProb = 50.32 %
HitPushProb = 7.36 %
HitLoseProb = 42.32 %
Case: (3,6) vs 7:
Expected Occurence (%) = 128/1326 = 0.0965
HitWinProb = 50.69 %
HitPushProb = 7.40 %
HitLoseProb = 41.53 %
Case: (4,5) vs 7:
Expected Occurence (%) = 128/1326 = 0.0965
HitWinProb = 51.05 %
HitPushProb = 7.42 %
HitLoseProb = 41.53 %
Thus, when you take each Hit-Win Prob, Hit Push Prob, and Hit-Lose Prob and weight them all by their expected occurence, you end up with the following:
Case: 9 vs 7:
Expected Occurence (%) = 352/1326 = 0.2655
HitWinProb = 50.72 %
HitPushProb = 7.40 %
HitLoseProb = 41.88 %
which is kinda like an average value but weighted slightly by their occurences in the ratio of 3:4:4 respectively or (3/11): (4/11): (4/11) percentage-wise.
Now, the advantage (money-wise) by hitting the nine is simply the expected win probability subtract the expected loss probability of simply:
50.72 - 41.88 = 8.84 %
This means that out of one hundred hands played, you would be up (over and above what you wagered) of 8.84 units
However, if you were to double this advantage by doubling your bet, you would now be up 8.84 x 2 = 17.68 units. So my question is this? Why are you telling people to only hit this when they could be taking advantage of the double down option?
To verify these results in actual play, I ran my own version of a blackjack simulation machine through 8000 reshuffles of a single deck game using only basic strategy choices (but with the 9 vs 7 choice changed to a double down instead of simply hitting). The simulation played 63,152 hands. Of those hands I hit 9 vs 7 190 times. (Expected = 190/63152 = 0.30 %) which is within a normal variation of the expected value of 0.2655 %. Of the 190 times I hit this I would always double down on my bet. I won 92/190 = 48.4 %; I pushed 18/190 = 9.47 % and lost 80/190 = 42.1% which is very close to the expected values. Thus, money-wise I gained 2*(92)-2*(80) = 24 units whereas had I only hit without doubling I would have gained only 12 units.
So why hit? Why not double? You're missing a huge advantage here!
Case: (2,7) vs 7:
Expected Occurence (%) = 96/1326 = 0.0724
HitWinProb = 50.32 %
HitPushProb = 7.36 %
HitLoseProb = 42.32 %
Case: (3,6) vs 7:
Expected Occurence (%) = 128/1326 = 0.0965
HitWinProb = 50.69 %
HitPushProb = 7.40 %
HitLoseProb = 41.53 %
Case: (4,5) vs 7:
Expected Occurence (%) = 128/1326 = 0.0965
HitWinProb = 51.05 %
HitPushProb = 7.42 %
HitLoseProb = 41.53 %
Thus, when you take each Hit-Win Prob, Hit Push Prob, and Hit-Lose Prob and weight them all by their expected occurence, you end up with the following:
Case: 9 vs 7:
Expected Occurence (%) = 352/1326 = 0.2655
HitWinProb = 50.72 %
HitPushProb = 7.40 %
HitLoseProb = 41.88 %
which is kinda like an average value but weighted slightly by their occurences in the ratio of 3:4:4 respectively or (3/11): (4/11): (4/11) percentage-wise.
Now, the advantage (money-wise) by hitting the nine is simply the expected win probability subtract the expected loss probability of simply:
50.72 - 41.88 = 8.84 %
This means that out of one hundred hands played, you would be up (over and above what you wagered) of 8.84 units
However, if you were to double this advantage by doubling your bet, you would now be up 8.84 x 2 = 17.68 units. So my question is this? Why are you telling people to only hit this when they could be taking advantage of the double down option?
To verify these results in actual play, I ran my own version of a blackjack simulation machine through 8000 reshuffles of a single deck game using only basic strategy choices (but with the 9 vs 7 choice changed to a double down instead of simply hitting). The simulation played 63,152 hands. Of those hands I hit 9 vs 7 190 times. (Expected = 190/63152 = 0.30 %) which is within a normal variation of the expected value of 0.2655 %. Of the 190 times I hit this I would always double down on my bet. I won 92/190 = 48.4 %; I pushed 18/190 = 9.47 % and lost 80/190 = 42.1% which is very close to the expected values. Thus, money-wise I gained 2*(92)-2*(80) = 24 units whereas had I only hit without doubling I would have gained only 12 units.
So why hit? Why not double? You're missing a huge advantage here!
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