Blackjack betting 'Strategy'

DouglasFromVictoria

New member
Joined
Sep 5, 2018
Messages
1
Hello everyone

I literally joined this forum to ask this question. I am very interested to get an answer from people much smarter than me.

I just recently started dealing cards at a casino near my place (primarily blackjack). From here, I have been able to gather some anecdotal insight into how people think and behave surrounding gambling, based on their verbal communication.

There is no doubt that blackjack favors the house over the long term, but I wanted to discuss a progressive betting strategy and the numerical pitfalls of this strategy.

So for some context - I had a player at my table and he was talking to another player about his betting strategy, basically he would double his bet each time he lost. This player knows that losing six consecutive hands results in his strategy failing (the basis of this will be discussed below). But he was saying ''How often do you lose six straight hands?'' This statement is what go me thinking of the whole summary below.

So lets assume that these players are NOT counting cards (as I am sure they are not), and even if they were, the dealer is using an automatic shuffler, so the dealer can reload cards after each hand, therefore never having any discards and therefore not allowing players to establish a real true count (or the denominator of their true count will always be six, the total number of cards in the shuffler).

Second, now that card counting is not occurring, the house is numerically favored. I have heard different percentages for how much the house is favored on a per-hand basis, but lets be relatively generous to the player and assume that a player has a 49.5% chance of winning any one hand (50.5% chance of losing). I guess the dealer would have to be reloading the discards into the automatic shuffler on each hand to keep this the same, but that is easily doable as a dealer has the free choice to do this.

The table minimum at my casino is $10 and the maximum is $500 per bet.

So under this strategy, if a player loses their $10 bet, their next bet $20, if they lose that bet they bet $40, then $80, then $160, then $320. Of course if they lose all of these 6 straight hands, they cannot keep doubling. The penalty for losing these six straight bets is ($10+$20+$40+$80+$160+$320)= $630. And even if they won their next bet at $500, they would still be at a loss of ($630-$500)=$130 for those seven hands. While these are the penalties for NOT winning six or seven straight hands, the reward for breaking that chain and winning one of these hands (be it hand number one or hand number six) is to stave off the penalty of losing six straight hands. This results in you winning $10, basically starting over.

So now for the important question of how often do you lose six straight hands

The probability is simple - you have a 0.505 chance of losing your first hand, and with the dealer eliminating counting by reloading the discards each hand, each hand is independent of the last hand. Therefore, you have a 0.505 chance of losing your second hand, and your third hand, and so on.

So, the odds of losing six consecutive hands of blackjack are: (0.505*0.505*0.505*0.505*0.505*0.505) =0.01658 or about 1.658% of the time.

Some of these players play A LOT. It would be a conservative estimate to say that they play ten hours a week. You can deal a hand of blackjack in 20 seconds at times, but for the sake of accommodating a dealer reloading the discards into the automatic shuffler each time and allowing the discards to be reset, as well as accommodating other variables of time, lets say that it takes two minutes to deal a hand of blackjack.

So if a player is playing 10 hours a week, with 52 weeks in a year, that amounts to 520 hours in one year, or (60 minutes*520 hours)=31,200 minutes per year. At 2 minutes per hand, this amounts to 15,600 hands per year.

Over the course of these 15,600 hands per year, you should lose six straight hands 1.658% of the time, as proved above. This results in (15,600*0.01658)=258.648 occurrences of losing six straight hands, call it 258.

Therefore, if you play 10 hours a week, at one hand every two minutes, you should expect to lose six straight hands 258 times. Again, the penalty for losing six straight hands is $630, so you can expect to lose (258 * $630) = $162,540 over the year.

This number seems high to me just on an anecdotal level. And one can argue about winning a seventh hand at $500 and therefore limiting your losses to ($630 - $500)=$130. So lets consider this.

We already know that the chances of losing six straight hands is 1.658% (or 0.01658), and with each hand being independent, the odds of losing the next hand are still 0.505. Therefore, the chances of losing the seventh hand, and making it seven straight lost hands, is 0.00837, or 0.837% of the time, slightly less than one percent of the time.

Of course the penalty for losing the $500 bet on the 7th straight hand means that you have doubled your bet up to the $320 bet (starting from $10). As we said, losing six straight hands results in a cumulative loss of $630. Losing this 7th hand results in a loss of ($500+$630)= $1,130. Going back to our 10 hours a week at 1 hand every two minutes basis - you can expect to lose seven straight hands 0.837 of 1% of the time (0.00837), with a penalty for this occurrence of $1,130. So with 15,600 hands taking place over the year, there should be seven straight loses on (0.00837*15,600)=130.572 different occurrences (call it 130).

Therefore, with losing seven straight hands 130 different times over the year, you should lose (130*$1,130) = $146,900 - not much better than the six straight loses from above ($162,540).

Am I on the right path with this? I mean everyone knows that the house always wins, and we can make different assumptions for the number of hours a player plays or the number of hands per hour, but (with counting eliminated and always using the full six decks in the shuffler) the chances of losing a hand is always 50.5%, and therefore the chances of losing six and seven straight hands are 0.01658 and 0.00837, respectively.

So it just seems like the only way to ever win over the long term at blackjack (without counting) is to make your losing hands smaller bets, and your winning hands bigger bets. But of course there is no way to determine when those hands will occur (since we are not counting). So even if someone happens to win 4 straight hands (which will happen (.495*.495*.495*.495) = 0.06 or about 6% of the time), their chances of winning the next hand remain at 49.5% and losing the next hand remain at 50.5%.

So if you do happen to win over the long term, you can chalk it up to straight guessing, and there is an almost infinitely small probabilistic chance of this occurring.

Sorry for the long, and wordy question, but I want to be well prepared for addressing this kind of thing.

Thanks
 
Hello everyone

I literally joined this forum to ask this question. I am very interested to get an answer from people much smarter than me.

I just recently started dealing cards at a casino near my place (primarily blackjack). From here, I have been able to gather some anecdotal insight into how people think and behave surrounding gambling, based on their verbal communication.

There is no doubt that blackjack favors the house over the long term, but I wanted to discuss a progressive betting strategy and the numerical pitfalls of this strategy.

So for some context - I had a player at my table and he was talking to another player about his betting strategy, basically he would double his bet each time he lost. This player knows that losing six consecutive hands results in his strategy failing (the basis of this will be discussed below). But he was saying ''How often do you lose six straight hands?'' This statement is what go me thinking of the whole summary below.

So lets assume that these players are NOT counting cards (as I am sure they are not), and even if they were, the dealer is using an automatic shuffler, so the dealer can reload cards after each hand, therefore never having any discards and therefore not allowing players to establish a real true count (or the denominator of their true count will always be six, the total number of cards in the shuffler).

Second, now that card counting is not occurring, the house is numerically favored. I have heard different percentages for how much the house is favored on a per-hand basis, but lets be relatively generous to the player and assume that a player has a 49.5% chance of winning any one hand (50.5% chance of losing). I guess the dealer would have to be reloading the discards into the automatic shuffler on each hand to keep this the same, but that is easily doable as a dealer has the free choice to do this.

The table minimum at my casino is $10 and the maximum is $500 per bet.

So under this strategy, if a player loses their $10 bet, their next bet $20, if they lose that bet they bet $40, then $80, then $160, then $320. Of course if they lose all of these 6 straight hands, they cannot keep doubling. The penalty for losing these six straight bets is ($10+$20+$40+$80+$160+$320)= $630. And even if they won their next bet at $500, they would still be at a loss of ($630-$500)=$130 for those seven hands. While these are the penalties for NOT winning six or seven straight hands, the reward for breaking that chain and winning one of these hands (be it hand number one or hand number six) is to stave off the penalty of losing six straight hands. This results in you winning $10, basically starting over.

So now for the important question of how often do you lose six straight hands

The probability is simple - you have a 0.505 chance of losing your first hand, and with the dealer eliminating counting by reloading the discards each hand, each hand is independent of the last hand. Therefore, you have a 0.505 chance of losing your second hand, and your third hand, and so on.

So, the odds of losing six consecutive hands of blackjack are: (0.505*0.505*0.505*0.505*0.505*0.505) =0.01658 or about 1.658% of the time.

Some of these players play A LOT. It would be a conservative estimate to say that they play ten hours a week. You can deal a hand of blackjack in 20 seconds at times, but for the sake of accommodating a dealer reloading the discards into the automatic shuffler each time and allowing the discards to be reset, as well as accommodating other variables of time, lets say that it takes two minutes to deal a hand of blackjack.

So if a player is playing 10 hours a week, with 52 weeks in a year, that amounts to 520 hours in one year, or (60 minutes*520 hours)=31,200 minutes per year. At 2 minutes per hand, this amounts to 15,600 hands per year.

Over the course of these 15,600 hands per year, you should lose six straight hands 1.658% of the time, as proved above. This results in (15,600*0.01658)=258.648 occurrences of losing six straight hands, call it 258.

Therefore, if you play 10 hours a week, at one hand every two minutes, you should expect to lose six straight hands 258 times. Again, the penalty for losing six straight hands is $630, so you can expect to lose (258 * $630) = $162,540 over the year.

This number seems high to me just on an anecdotal level. And one can argue about winning a seventh hand at $500 and therefore limiting your losses to ($630 - $500)=$130. So lets consider this.

We already know that the chances of losing six straight hands is 1.658% (or 0.01658), and with each hand being independent, the odds of losing the next hand are still 0.505. Therefore, the chances of losing the seventh hand, and making it seven straight lost hands, is 0.00837, or 0.837% of the time, slightly less than one percent of the time.

Of course the penalty for losing the $500 bet on the 7th straight hand means that you have doubled your bet up to the $320 bet (starting from $10). As we said, losing six straight hands results in a cumulative loss of $630. Losing this 7th hand results in a loss of ($500+$630)= $1,130. Going back to our 10 hours a week at 1 hand every two minutes basis - you can expect to lose seven straight hands 0.837 of 1% of the time (0.00837), with a penalty for this occurrence of $1,130. So with 15,600 hands taking place over the year, there should be seven straight loses on (0.00837*15,600)=130.572 different occurrences (call it 130).

Therefore, with losing seven straight hands 130 different times over the year, you should lose (130*$1,130) = $146,900 - not much better than the six straight loses from above ($162,540).

Am I on the right path with this? I mean everyone knows that the house always wins, and we can make different assumptions for the number of hours a player plays or the number of hands per hour, but (with counting eliminated and always using the full six decks in the shuffler) the chances of losing a hand is always 50.5%, and therefore the chances of losing six and seven straight hands are 0.01658 and 0.00837, respectively.

So it just seems like the only way to ever win over the long term at blackjack (without counting) is to make your losing hands smaller bets, and your winning hands bigger bets. But of course there is no way to determine when those hands will occur (since we are not counting). So even if someone happens to win 4 straight hands (which will happen (.495*.495*.495*.495) = 0.06 or about 6% of the time), their chances of winning the next hand remain at 49.5% and losing the next hand remain at 50.5%.

So if you do happen to win over the long term, you can chalk it up to straight guessing, and there is an almost infinitely small probabilistic chance of this occurring.

Sorry for the long, and wordy question, but I want to be well prepared for addressing this kind of thing.

Thanks

If you reload the cards and reshuffle after each hand, the process has no memory and there is no such thing as an advantageous betting strategy based solely on some previous hand's outcome. Said another way, each hand has NO IDEA whether you won or lost the last hand. Said another way, no hand has ANY influence on any subsequent hand. It is a common misconception - and many have made money selling this kind of silliness in various lotteries - that processes have memories.
 
If you reload the cards and reshuffle after each hand, the process has no memory and there is no such thing as an advantageous betting strategy based solely on some previous hand's outcome. Said another way, each hand has NO IDEA whether you won or lost the last hand. Said another way, no hand has ANY influence on any subsequent hand. It is a common misconception - and many have made money selling this kind of silliness in various lotteries - that processes have memories.
I agree with what you said but will offer the following. If you are the last player in the table and are card counting you will have a slight advantage (over 49.5%) since you did see some cards-even if the dealer shuffles after each hand. Will that knowledge bring your advantage over 50%, certainly not. However in the end you will be a better player seeing some cards. For example if you were dealt a 2 and a 9 and every one else, including the dealer's up card, was a 10 (or picture card) you will have a lower expectation on doubling down on your 11. I forget the cut-off number for this situation but possibly you should NOT even double down.

I remember the good ole days when casinos dealt down to the last card and used only one deck. Who remembers Beat The Dealer by Ed Thorpe?
 
I agree with what you said but will offer the following. If you are the last player in the table and are card counting you will have a slight advantage (over 49.5%) since you did see some cards-even if the dealer shuffles after each hand. Will that knowledge bring your advantage over 50%, certainly not. However in the end you will be a better player seeing some cards. For example if you were dealt a 2 and a 9 and every one else, including the dealer's up card, was a 10 (or picture card) you will have a lower expectation on doubling down on your 11. I forget the cut-off number for this situation but possibly you should NOT even double down.

I remember the good ole days when casinos dealt down to the last card and used only one deck. Who remembers Beat The Dealer by Ed Thorpe?

Absolutely agreed. There ARE strategies, just not of the variety suggested. :)
 
Top