Thanks for the clarification! Here's my analysis attached below:
ev & var.jpg
Thanks for the clarification! Here's my analysis attached below:
ev & var.jpg
Dealers make mistake all the time. There are some discussions and a chapter in some books.
Now we are talking about the difference between 1 hand and 5 hands.
Yes, new dealer makes more mistakes, but a casual dealer makes even more.
Most of the dealer errors happen at the payoff and the hand total, sometimes colors changing.
If playing 1 to 2 hands, usually the dealer remembers the two hand totals and the dealer total, so less errors made.
If playing 5 hands, many dealers need to recount their hand and each player's hand total, after paying a hand, sometimes they make mistake from their wrong memory, such as paying 4 hands in a row, then she thinks it is a bust on the dealer...
Usually we know there is a mistake from us or from the dealer, and we can correct it by our choice, even after the cards are collected.
Some dealers don't know, or even choose to not collect it even they know there was a mistake, because some casinos will give a bad mark to the dealer by that.
My experience: 10K usd was made only by dealer errors in one year, BS and flat bets.
So the question is: whether more mistakes you make? or the dealers?
Dealer errs are indeed important, particularly on the new variations and side bets and hands with many cards. CV tests you for the following:
Insurance payoffs
Blackjack payoffs
No payoff on win
Bust on 21 or less
Stand on 16
Dealer should have busted
Lose on a push
No payoff on bonus or side bet
"I don't think outside the box; I think of what I can do with the box." - Henri Matisse
Your analysis for the variance of the five $1 bets isn't correct as it doesn't consider the covariance among the five hands. Do you have BJA3? The correct formula is at the bottom of page 20.
Of course, your conclusion is, nonetheless, correct. If you aren't counting, it is surely better to bet five $1 hands than one $5 hand.
Don
Email: [email protected]
Thanks for your response Mr. Schlesinger! Big fan btw, thank you sincerely for your invaluable contributions to the blackjack community. My earlier analysis took into account that the OP assumed that the dealer shuffles after every hand, so that the hands' outcomes were independent and therefore uncorrelated. If the OP had not said so, I would have applied your formula from BJA3 assuming no shuffling between hands.
I may be misunderstanding, but I don't see why shuffling after every hand would change the calculation for the variance of playing five $1 hands at the same time. My comment was that, in calculating that variance, you can't just add the variances of the five individual hands; you have to include in the calculation the covariance.
Am I missing something?
Don
I assumed that the player was playing five $1 hands one after the other rather than concurrently. So play and finish one $1 hand, dealer shuffles all used and unused cards together back into the shoe to restart the next one, then play the next $1 hand with the freshly-shuffled shoe, and repeat for five total hands across five total rounds, each with a fresh well-shuffled shoe. Yours seems to consider five $1 hands at once spread across the table played against the dealer all at once, in one round. OP was playing online where cards are not set aside from future hands but reshuffled back into the shoe right away, so I opted for the former in my interpretation of "playing five $1 hands". If it's the latter which is the case, then I agree that the covariances between hands have to be considered, since cards dealt in previous hands will be unavailable for future hands during the round. However, I fail to see why covariance has to be considered for the former because that is equivalent to playing one $1 hand once each against five separate dealers each with fresh shoes, so treated as five independent trials.
It seems determining variance for 2 or more simultaneous hands is dependent upon covariance. In the case of splitting A-A once there are 2 identical simultaneous hands. I did a little research and found the the covariance of 2 identical variables is simply equal to variance. I'm not too familiar with statistical math logic though.
As a simple example: single deck, A-A versus dealer up card of 9, split once, 1 card to split ace -
1. There are exactly 2 hands
2. The variance of each split hand taken independently is about .9022
3. The total variance of the 2 split hands taken as an aggregate is about 2.32262
4. The total variance can be computed from the following data (a - e):
a. probability of winning 2 units by splitting = .366139
b. probability of winning 1 unit by splitting = .0475381
c. probability of breaking even by splitting = .298499
d. probability of losing 1 unit by splitting = .0856004
e. probability of losing 2 units by splitting = .202223
My question is, "Can the total variance of a pair split in blackjack be determined by combinatorial analysis by using covariance?" Can the true value of total variance of the 2 split hands of 2.32262 in the above example be reconciled using covariance?
k_c
Last edited by k_c; 02-04-2024 at 04:38 PM. Reason: Fix typo in computed total variance value
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