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!

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.

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.

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.