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?

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.

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.

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.

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.

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.

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?

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.

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.

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

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

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

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.

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?

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)

Quote:teliotI 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 = 556Tlink to original post

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.

Quote:teliotI 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 = 556Tlink to original post

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.

The dealer can win 15-20+ times in a row.

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

Quote:charliepatrickI 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 Tlink to original post

Is standing on 12 a basic strategy play?

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

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 Tlink to original post

Is standing on 12 a basic strategy play?link to original post

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.Quote:DieterQuote:charliepatrickI 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 Tlink to original post

Is standing on 12 a basic strategy play?link to original post 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!

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)

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

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!

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 Jlink to original post

This is great. Thank you.

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.

I wish we could ban them on this forum.

Quote:FinsRuleI think I dislike arguments about infinity more than I dislike betting system arguments.

I wish we could ban them on this forum.link to original post

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.