charliepatrick
Joined: Jun 17, 2011
• Posts: 2577
Thanks for this post from:
September 15th, 2021 at 3:04:13 AM permalink
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)

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

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
I suspect you can use similar ideas to have the player lose more in a single shoe!
unJon

Joined: Jul 1, 2018
• Posts: 3400
September 15th, 2021 at 4:36:40 AM permalink
Quote: charliepatrick

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)

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

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
I suspect you can use similar ideas to have the player lose more in a single shoe!

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.
ThatDonGuy

Joined: Jun 22, 2011
• Posts: 5368
Thanks for this post from:
September 15th, 2021 at 12:17:28 PM permalink
Quote: TomG

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.

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.
FinsRule
Joined: Dec 23, 2009
• Posts: 3793
September 15th, 2021 at 9:49:21 PM permalink
I think I dislike arguments about infinity more than I dislike betting system arguments.

I wish we could ban them on this forum.
billryan
Joined: Nov 2, 2009
• Posts: 13290
September 17th, 2021 at 1:54:04 PM permalink
Quote: FinsRule

I think I dislike arguments about infinity more than I dislike betting system arguments.

I wish we could ban them on this forum.