Hello everyone,
Let's assume I can correctly predict a 10 roughly 50% of the time (as opposed to being able to predict a 10 only about 31% of the time solely by chance) in a double deck, dealer hits soft 17 shoe game.
Can someone please calculate by how much this improves my chances of winning the various "Lucky Lady" sidebets (assuming the following payoff schedule: 4:1, 10:1, 25:1, 200:1, and 1000:1)?
Thank you!
Overkill,
Below is the Return table for Lucky Ladies DD with the 1000/200/25/10/4 paytable. The "Nomal Prob" and "Return" columns are for the normal game.
Your amazing ability to guess a First Card X correctly with a probability of 50% has the effect of multiplying the return for each winning hand by 1.625 (that's (1/2)/(4/13) = 13/8): these returns are shown in the "50% First X" column. By the way, I assumed you had no knowledge of the identity of the FCX, and that your second card prob is the normal result for a DD missing one random X.
As you can see, your ability raises the sidebet's EV from the normal -24.93% to a lofty +21.97%, and also means you'll win the SB just a shade under 17% of the time.
Hint: BET THE MAX ON THE SIDEBET ;-)
Hope this helps!
Dog Hand
Hand Pays Normal Prob Return 50% First X.
Qh-Qh & BJ 1000 8.69920E-06 0.0086992 0.0141362
Qh-Qh 200 1.78007E-04 0.0356015 0.0578524
Matched 20 25 2.80060E-03 0.0700149 0.1137743
Suited 20 10 2.09111E-02 0.2091113 0.3398058
Unsuited 20 4 8.06572E-02 0.3226288 0.5242718
Loser -1 8.95444E-01 -0.8954444 -0.8300971
Total -0.2493887 0.2197434
OK, thanks so much Dog Hand. I really appreciate it. I didn't think it would be that strong. And yes, your assumptions were right on. By the way, what is "FCX?" (When you refer to the "identity of the FCX," am I correct in assuming you mean the rank and suit of the first card?)
1) Could you keep everything in the original post the same but add in that whenever I place the Lucky Lady side bet, I not only can predict that my first card will be any kind of 10 with 50% accuracy, but I can predict that my second card will be any kind of 10 with 40% accuracy?
2)Could you figure the EV for the data in my original post for the 21 + 3 side bet (assuming all wins pay 9:1) instead of for Lucky Ladies?
3)Finally, could you tell me my EV for the 21 + 3 side bet if I can predict with 50% accuracy that my first card will be any kind of 10, and I can predict with 40% accuracy that my second card will be any kind of 10?
Thank you!
Johnny Bravo is indeed correct for Lucky Ladies ;-)
Hand Pays Probability 50/40 Return.
Qh-Qh & BJ 1000 3.35105E-05 0.0335105
Qh-Qh 200 3.69715E-04 0.0739431
Matched 20 25 6.04839E-03 0.1512097
Suited 20 10 4.51613E-02 0.4516129
Unsuited 20 4 1.74194E-01 0.6967742
Loser -1 7.74194E-01 -0.7741935
Total 0.6328568
OK, and thank you yet again. So, in sum, would you conjecture to say that the double deck 21 +3 off the top scenario would be slightly worse than the already poor EV for 21 + 3? And would you say that for the 50/40 scenario, it's too close to call?
Finally, how does using 6 decks (with "common" rules) change the EV for these 2 scenarios? Worse? Better?
Thank you!
Dog Hand,I think our 'wires' got crossed. What I meant is that I only will bet the Lucky Ladies sidebet when I predict my first card will be a 10, and please assume I can do so with 50% accuracy. Did your sim reflect that I could correctly predict the rank of ALL initial cards dealt to me (not just 10s) with 50 % accuracy?
Overkill,
I assumed that on some (as yet unspecified) fraction of the rounds you can predict a FCX with 50% accuracy, and that you will play LL only on those rounds.
As an example, for a DD game let's calculate the probability of the jackpot for "normal" play and for "FCX" play.
Normal:
1st card Qh: 2/104
2nd card Qh: (2-1)/(104-1)
Dealer BJ: 2*(8/(104-2))*((32-2)/(104-3)) (note: the "2" is because her BJ can be A/X or X/A with equal probability)
P(jackpot) = (2/104)*((2-1)/(104-1))*(2*(8/(104-2))*((32-2)/(104-3))) = 8.699196...x10^(-6)
FCX:
1st card Qh: (1/2)*(1/16) (note: the 1/2 is the FCX probability, and 1/16th of the time it'll be a Heart Q)
2nd card Qh: (2-1)/(104-1) (note: same as for "normal", since we aren't assuming any knowledge of your 2nd card here)
Dealer BJ: 2*8/(104-2)*(32-2)/(104-3) (note: same as for "normal")
P(jackpot) = ((1/2)*(1/16))*((2-1)/(104-1))*(2*8/(104-2)*(32-2)/(104-3)) = 1.413619...x10^(-5)
Hope this helps!
Dog Hand
what the "X" stood for in your "FCX" nomenclature. At first, I thought the "X" represented a variable, but now I see that it simply means a card that is some kind of 10: either a 10, J, Q, or K, correct? By the way, did you choose "X" because it is the Roman numeral equivalent of ten?
