Without doubling, you'll be lowering risk and variance but also adding to the house edge. In the short run, your money may last longer, but in the long run, it will run out sooner. Perhaps a compromise where you at least do the real good doubles like 11 vs. 6 and skip the marginal ones like A7 vs. 2?
Maybe someone here will volunteer to rank each doubling situation by its risk/reward ratio.
Peter Griffin, who wrote "Theory of Blackjack", has a section titled 'World's Worst Blackjack Player' where he quantifies the penalty a player will face for certain ill-begotten plays. The penalty for never doubling down is 1.6%. So if you are begin as a perfect basic strategy player where the casino typically has a .5% edge over you, then by never doubling down, you have just quadrupled the casino's edge against you!
If you are counting cards, there are counts where it is correct to do as you say, to hit rather than double. But without this knowledge, your best chances are to follow Basic to the letter.
Sometimes doubling down and only taking 1 small card will give the dealer the next card to bust.. when if instead of doubling you take the 2nd card that made you bust instead of the dealer..
I personally will not double if my wager is already on the higher end
No, actually that's not the correct answer. The advantage of doubling versus not doubling must be taken into account. For example, here's a chart:
chassss.jpg
"I don't think outside the box; I think of what I can do with the box." - Henri Matisse
No, for doubling down it is not sufficient that the chance of winning the hand is more than 50 percent.
You also have to compare the winning chance when doubling to the winning chance without doubling.
In the end, the expectation value (EV) decides whether to double or not.
An example: Suppose the chances for doubling in certain situation (such as 11 vs. dealer 6) are
win: 70 percent, push: 10 percent, loss: 20 percent
If you bet 100 dollars per hand and this situation occurs 10 times, and you double down every time,
then
you win 7 times 200 dollars = 1400 dollars
you lose 2 times 200 dollars = 400 dollars
you push 1 time (= 0 dollars, nothing changes)
so in the end you gain 1400 minus 400 = 1000 dollars.
Suppose the chances for NOT doubling (just hitting) in the same situation are
win: 80 percent, push: 10 percent, loss: 10 percent
If you again bet 100 dollars per hand and the situation occurs 10 times,
and you hit every time, then
you win 8 times 100 dollars = 800 dollars
you lose 1 time 100 dollars = 100 dollars
you push 1 time (nothing changes)
so in the end you gain 800 minus 100 = 700 dollars.
For both decisions, the chance of winning is greater than 50 percent.
But despite winning fewer hands when doubling (70 compared to 80 percent),
doubling down earns more money because every win is worth twice as much.
Now suppose the chance of winning when doubling is only 60 percent (not 70),
the push is again 10 percent, and the chance of losing is 30 percent (not 10).
Then
you win 6 times 200 dollars = 1200 dollars
you lose 3 times 200 dollars = 600 dollars
you push 1 time (nothing changes)
so in the end you gain 1200 minus 600 dollars = 600 dollars,
which is less than the 700 dollars you would have gained when hitting.
Again, for both decisions, the chance of winning is greater than 50 percent.
But in this case, hitting would be better than doubling down,
because it generates more net profit.
You have not learned basic strategy if you prefer not to double. Your edge is dependant on taking the best long term view, which means doubling when appropriate, which you aren’t. Don’t bother with indices until you get the blinders off.
Pay attention to post 3 and Schlesinger’s comments.
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