What is the max times in a row you can lose at one deck blackjack?
September 14th, 2021 at 2:11:29 PM
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Game rules
You have a deck of cards
Every round you shuffle and you deal
You can shuffle and deal for ever
You play basic strategy for 1 deck
What is the max number of losses in a row?
You can lose a few times but
is it possible to lose 1000 times, 10000, 1 million times in a row
and never win a hand?
You have a deck of cards
Every round you shuffle and you deal
You can shuffle and deal for ever
You play basic strategy for 1 deck
What is the max number of losses in a row?
You can lose a few times but
is it possible to lose 1000 times, 10000, 1 million times in a row
and never win a hand?
September 14th, 2021 at 2:18:42 PM
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Your next hand is unaffected by your last hand if it is a new shuffle.
The older I get, the better I recall things that never happened
September 14th, 2021 at 2:25:26 PM
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This sort of like asking the maximum number of consecutive coin flips you can lose. The answer is infinity
It’s all about making that GTA
September 14th, 2021 at 2:42:11 PM
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You can't lose infinite times.
There is a distribution between win and losses.
If you only lose and not win means that
probability of winning is 0 and probability of losing is 100%
but at blackjack probability of win hands > 0
that means that eventually after a losing streak you will win.
There is a distribution between win and losses.
If you only lose and not win means that
probability of winning is 0 and probability of losing is 100%
but at blackjack probability of win hands > 0
that means that eventually after a losing streak you will win.
September 14th, 2021 at 2:49:39 PM
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Quote: MrSevenBoxYou can't lose infinite times.
In theory, you can. It's pretty much impossible, but it "can" happen.
You cannot say that there is a maximum number of consecutive losses. Otherwise, the probability of winning the next hand after that many losses would be 100%, but, to put it in a variation of your words, "at blackjack the probability of losing hands > 0."
Remember, the probability of a coin toss coming up heads 1000 times in a row is the same as the same coin coming up heads 999 times in a row and then the 1000th toss is tails.
September 14th, 2021 at 2:49:57 PM
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You don't have an infinite bankroll to play an infinite amount of hands.
The older I get, the better I recall things that never happened
September 14th, 2021 at 3:12:22 PM
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Quote: billryanYou don't have an infinite bankroll to play an infinite amount of hands.
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I am not asking how much bankroll you need to have to survive infinite losses.
September 14th, 2021 at 3:19:21 PM
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Quote: MrSevenBoxQuote: billryanYou don't have an infinite bankroll to play an infinite amount of hands.
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I am not asking how much bankroll you need to have to survive infinite losses.link to original post
Play for funsies. It won't change anything.
The older I get, the better I recall things that never happened
September 14th, 2021 at 3:44:07 PM
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Quote: MrSevenBoxQuote: billryanYou don't have an infinite bankroll to play an infinite amount of hands.
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I am not asking how much bankroll you need to have to survive infinite losses.link to original post
You did ask, however:
"You can lose a few times but is it possible to lose 1000 times, 10000, 1 million times in a row and never win a hand?"
and the answer is, yes. It's about as likely as tossing a coin 1000, 10,000, 1 million times and having it come up heads each time, but that doesn't mean it "can't" happen.
September 14th, 2021 at 4:12:34 PM
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I was really excited about this question. I thought it would be the most hands you could lose in one deck you versus a dealer playing BS. My guess is 8.
September 14th, 2021 at 4:25:11 PM
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Quote: FinsRuleI was really excited about this question. I thought it would be the most hands you could lose in one deck you versus a dealer playing BS. My guess is 8.
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That’s what I thought it was going to be also!
Can we make it that question and figure out?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 14th, 2021 at 4:29:02 PM
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What is the maximum amount of hands you could get out of a deck. Wouldn't that be the same answer, but a different question.
The older I get, the better I recall things that never happened
September 14th, 2021 at 4:30:45 PM
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What dealer rules? Does surrender count as a loss?
Have you tried 22 tonight? I said 22.
September 14th, 2021 at 4:53:22 PM
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Quote: billryanWhat is the maximum amount of hands you could get out of a deck. Wouldn't that be the same answer, but a different question.
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Might be right. Not sure it’s necessarily true since BS different than dealer strategy there will be hands that work one way and not the other.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 14th, 2021 at 4:56:01 PM
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That's true. H17 or S17 would be a factor. This is way above my abilities.
The older I get, the better I recall things that never happened
September 14th, 2021 at 5:25:15 PM
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What about all those pushes? How are those considered?
September 14th, 2021 at 5:40:38 PM
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Quote: BoSoxWhat about all those pushes? How are those considered?
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I think that if you didn't get paid, you didn't win.
May the cards fall in your favor.
September 14th, 2021 at 5:54:29 PM
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I get 11, but I can't get 12.
2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
4-4-9 Q-8
4-4-9 Q-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
5-6-6 J-J
9-J J-T
Leftover = 556T
2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
4-4-9 Q-8
4-4-9 Q-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
5-6-6 J-J
9-J J-T
Leftover = 556T
Climate Casino: https://climatecasino.net/climate-casino/
September 14th, 2021 at 5:56:54 PM
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Quote: billryanWhat is the maximum amount of hands you could get out of a deck. Wouldn't that be the same answer, but a different question.
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If there is no restriction as to the player's strategy, the "obvious" answer is, 13 hands, assuming the player always stands on the two dealt cards.
I do know that you can get 18 21s out of a single deck - I figured it out once after playing Turbo 21 on Pogo.
If you count all of the Aces as 11, the sum of the values in a single deck is 4 x (2 + 3 + ... + 9 + 10 + 10 + 10 + 10 + 11) = 380 = 21 x 18 + 2.
Set the 2 of spades aside
Hands 1-4: Ace & 10 of each suit
Hands 5-8: K, 8, 3 of each suit
Hands 9-12: Q, 7, 4 of each suit
Hands 13-15: J, 9, 2 of hearts, clubs, and diamonds
This leaves the four 5s, the four 6s, and the 9 and Jack of spades
Hand 16: 6 of hearts, 5 of hearts, 5 of clubs, 5 of diamonds
Hand 17: Jack, 6, 5 of spades
Hand 18: 9 of spades, 6 of clubs, 6 of diamonds
September 14th, 2021 at 6:01:47 PM
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Assuming BS and liberal rules...
Dealer (up-card first) followed by player (draw card last).
K-A > Q-J
K-A > Q-J
K-A > Q-J
K-A > Q-J
(Here, assume player split all four 8s then doubles.)
7-10 > 8-2-3
.......... 8-2-3
.......... 8-2-3
.......... 8-2-3
(Here, Player doubles.)
7-10 > 6-5-4
7-10 > 6-5-4
7-10 > 6-5-4
9-9 > 9-4-9 (bust)
5, 6 leftover
Dealer (up-card first) followed by player (draw card last).
K-A > Q-J
K-A > Q-J
K-A > Q-J
K-A > Q-J
(Here, assume player split all four 8s then doubles.)
7-10 > 8-2-3
.......... 8-2-3
.......... 8-2-3
.......... 8-2-3
(Here, Player doubles.)
7-10 > 6-5-4
7-10 > 6-5-4
7-10 > 6-5-4
9-9 > 9-4-9 (bust)
5, 6 leftover
Have you tried 22 tonight? I said 22.
September 14th, 2021 at 6:35:46 PM
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Quote: ThatDonGuyIn theory, you can. It's pretty much impossible, but it "can" happen.
I'm having trouble understanding this. Are there really streaks that may or may not happen after an infinite amount of hands? If a streak "can" happen, but has not yet happened, shouldn't that mean there was not an infinite amount of hands played.
September 14th, 2021 at 7:15:11 PM
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Quote: TomGI'm having trouble understanding this. Are there really streaks that may or may not happen after an infinite amount of hands? If a streak "can" happen, but has not yet happened, shouldn't that mean there was not an infinite amount of hands played.
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The problem is that infinity is not a well defined concept.
Infinite consecutive wins
Infinite consecutive losses
Infinite alternating wins and losses
Can all three of those things happen in an infinite series?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 14th, 2021 at 7:34:32 PM
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EDIT:******This is so horribly wrong.......completely forgot dealer has to take cards.....please IGNORE.......**********
I think you can lose all 13 hands (though I admit to not being 100% on BS, so just relying on Wizard's charts)
Player - Dealer (up card first).....player always stands (Edit: per BS chart)
1) 87 - 6 10...Stand
2) 87 - 6 10...Stand
3) 87 - 6 10...Stand
4) 87 - 6 10...Stand
5) J2 - 4 Q....Stand
6) J2 - 4 Q....Stand
7) J2 - 4 Q....Stand
8) J2 - 4 Q....Stand
9) 93 - 5 A....Stand
10) 93 - 5 A....Stand
11) 93 - 5 A....Stand
12) 93 - 5 K....Stand
13) KK - K A....Stand.......I don't see a line for Splitting KK in chart, nor a surrender listing for 20.....so I think that is Stand (that's how I play it)
I think you can lose all 13 hands (though I admit to not being 100% on BS, so just relying on Wizard's charts)
Player - Dealer (up card first).....player always stands (Edit: per BS chart)
1) 87 - 6 10...Stand
2) 87 - 6 10...Stand
3) 87 - 6 10...Stand
4) 87 - 6 10...Stand
5) J2 - 4 Q....Stand
6) J2 - 4 Q....Stand
7) J2 - 4 Q....Stand
8) J2 - 4 Q....Stand
9) 93 - 5 A....Stand
10) 93 - 5 A....Stand
11) 93 - 5 A....Stand
12) 93 - 5 K....Stand
13) KK - K A....Stand.......I don't see a line for Splitting KK in chart, nor a surrender listing for 20.....so I think that is Stand (that's how I play it)
September 14th, 2021 at 7:41:45 PM
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Quote: teliotI get 11, but I can't get 12.
link to original post
2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
4-4-9 Q-8
4-4-9 Q-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
5-6-6 J-J
9-J J-T
Leftover = 556T
I know this is what the thank you button is for, but this is amazing. I now wish I would have spent time thinking about this before just handing it off.
I think the Wizard would like a trivia question like this.
September 14th, 2021 at 9:08:23 PM
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Quote: teliotI get 11, but I can't get 12.
link to original post
2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
4-4-9 Q-8
4-4-9 Q-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
5-6-6 J-J
9-J J-T
Leftover = 556T
Since my first attempt was so horribly, seriously, terribly, awfully, embarrassingly wrong, let me start with a proven commodity courtesy of teliot and ask can I get a 12th one......possibly by making dealer have to draw in some instances.
** are my modifications **
2-2 A-K
2-2 A-K
3-3 A-K
3-3 A-K
** Q-4 5-6-6
** Q-4 5-4-8
T-7 Q-8
T-7 Q-8
6-7-9 7-5
** 4-8 6-5-J
9-J J-T
** 9-9 J-T
Anyway, My guess is I am missing something....a split or something.
September 14th, 2021 at 9:16:19 PM
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One player can lose 8 bets on one hand.
The dealer can win 15-20+ times in a row.
The dealer can win 15-20+ times in a row.
September 15th, 2021 at 1:40:35 AM
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I get 12 using the following
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T
September 15th, 2021 at 1:55:26 AM
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Quote: charliepatrickI get 12 using the following
link to original post
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T
Is standing on 12 a basic strategy play?
May the cards fall in your favor.
September 15th, 2021 at 2:06:18 AM
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If the dealer has an up card of 6 or 4 then yes. btw you can always change the 9 8 vs Q Q to Q7 vs Q8 leaving 9 5 vs 6 4 T to get the same result.Quote: DieterQuote: charliepatrickI get 12 using the following
link to original post
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T
Is standing on 12 a basic strategy play?link to original post
September 15th, 2021 at 2:12:18 AM
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Quote: charliepatrickIf the dealer has an up card of 6 or 4 then yes. btw you can always change the 9 8 vs Q Q to Q7 vs Q8 leaving 9 5 vs 6 4 T to get the same result.link to original postQuote: DieterQuote: charliepatrickI get 12 using the following
link to original post
2x 2 2 < A K
2x 3 3 < A K
2x 9 8 < Q Q
2x 9 8 < J J
4x 7 5 < 6 4 T
Is standing on 12 a basic strategy play?link to original post
Got it. Just looked a little funny for some reason.
Even if the player hits, they bust, so I suppose the hands per deck stays the same, with 1 player plus a dealer.
Nice solve!
May the cards fall in your favor.
September 15th, 2021 at 3:04:13 AM
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An easier solution (which loses more) is
4x 3 2 < A K
4x Q 8 < J 9
4x 6 5 (double) 4 < 7 T (7 is dealer's up-card, so correct to double 11)
four split (and doubled) hands of 8 3 (Q) > 7 T
four split (and doubled) hands of 9 2 (K) > 7 T
four split (and doubled) hands of 6 A (5) > 4 T T
This leaves JJJJ77444 to create
two split (and doubled) hands of 7 4 (J) > 4 J JI suspect you can use similar ideas to have the player lose more in a single shoe!
4x 3 2 < A K
4x Q 8 < J 9
4x 6 5 (double) 4 < 7 T (7 is dealer's up-card, so correct to double 11)
I don't know but this gives the 20 units profit for the player in a single deck
4x A K > 2 3 (wins 1.5 units each)
4x 7 4 (Q) > 8 J (doubled 11 vs 8, so wins 2 units each)
1x 5 6 (9) > 9 T (doubled 11 vs 9)
1x 5 6 (T) > 9 T (doubled 11 vs 9)
1x 5 6 (T) > 9 5 6 (doubled 11 vs 9)
Total win = 6+8+6 = 20
4x A K > 2 3 (wins 1.5 units each)
4x 7 4 (Q) > 8 J (doubled 11 vs 8, so wins 2 units each)
1x 5 6 (9) > 9 T (doubled 11 vs 9)
1x 5 6 (T) > 9 T (doubled 11 vs 9)
1x 5 6 (T) > 9 5 6 (doubled 11 vs 9)
Total win = 6+8+6 = 20
four split (and doubled) hands of 8 3 (Q) > 7 T
four split (and doubled) hands of 9 2 (K) > 7 T
four split (and doubled) hands of 6 A (5) > 4 T T
This leaves JJJJ77444 to create
two split (and doubled) hands of 7 4 (J) > 4 J J
September 15th, 2021 at 4:36:40 AM
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Quote: charliepatrickAn easier solution (which loses more) is
4x 3 2 < A K
4x Q 8 < J 9
4x 6 5 (double) 4 < 7 T (7 is dealer's up-card, so correct to double 11)I don't know but this gives the 20 units profit for the player in a single deck
4x A K > 2 3 (wins 1.5 units each)
4x 7 4 (Q) > 8 J (doubled 11 vs 8, so wins 2 units each)
1x 5 6 (9) > 9 T (doubled 11 vs 9)
1x 5 6 (T) > 9 T (doubled 11 vs 9)
1x 5 6 (T) > 9 5 6 (doubled 11 vs 9)
Total win = 6+8+6 = 20I suspect you can use similar ideas to have the player lose more in a single shoe!link to original post
four split (and doubled) hands of 8 3 (Q) > 7 T
four split (and doubled) hands of 9 2 (K) > 7 T
four split (and doubled) hands of 6 A (5) > 4 T T
This leaves JJJJ77444 to create
two split (and doubled) hands of 7 4 (J) > 4 J J
This is great. Thank you.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 15th, 2021 at 12:17:28 PM
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Quote: TomGI'm having trouble understanding this. Are there really streaks that may or may not happen after an infinite amount of hands? If a streak "can" happen, but has not yet happened, shouldn't that mean there was not an infinite amount of hands played.
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The numbers 1, 2, 3, ... are an infinite set of numbers. Let each number represent a hand.
Choose any positive integer N.
Regardless of what number N is, it is possible to lose hand number N.
September 15th, 2021 at 9:49:21 PM
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I think I dislike arguments about infinity more than I dislike betting system arguments.
I wish we could ban them on this forum.
I wish we could ban them on this forum.
September 17th, 2021 at 1:54:04 PM
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Quote: FinsRuleI think I dislike arguments about infinity more than I dislike betting system arguments.
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I wish we could ban them on this forum.
You are just jealous because you don't have an infinite bankroll. Were you to have one, you might look at the discussions with a different view.
The older I get, the better I recall things that never happened
