Lucky Ladies BJ Side Bet
Trev
https://wizardofodds.com/games/blackjack/side-bets/lucky-ladies/
and from the return it looks like it doesn't.....i think
Quote: TrevorSimple question. Does A9 pay on suited/any 20 on the Lucky Ladies side bet?
Trev
Yes, it does. It pays just like any other 20 or suited 20.
ZCore13
Quote: TrevorSimple question. Does A9 pay on suited/any 20 on the Lucky Ladies side bet?
Trev
Yes it does.
Take Suited 20 - 864 combinations:
Without including A9 i get - (COMBIN(6 (decks) * 4 (10 cards), 2 (drawn)) * 4 (suits)) - (15 (QH) + 225 (Match 20)) = 864
Quote: TrevorIs that table (table D) correct then?
Take Suited 20 - 864 combinations:
Without including A9 i get - (COMBIN(6 (decks) * 4 (10 cards), 2 (drawn)) * 4 (suits)) - (15 (QH) + 225 (Match 20)) = 864
Looks like Wizard forgot about soft 20's. Here is an updated pay table:
| Event | Pays | Ways | Probability | Return |
|---|---|---|---|---|
| Pair of queens | 100 | 15 | 0.000309176 | 0.030917635 |
| Ranked 20 | 30 | 225 | 0.004637645 | 0.139129359 |
| Suited 20 | 10 | 1008 | 0.020776651 | 0.20776651 |
| Any 20 | 3 | 3888 | 0.080138511 | 0.240415533 |
| One Queen | 1 | 5184 | 0.106851348 | 0.106851348 |
| Loser | -1 | 38196 | 0.787286668 | -0.787286668 |
| Total | 48516 | 1 | -0.062206282 |
I was looking at another Lucky Ladies side bet that pays for 2QH + Dealer has blackjack and was curious as to how this is calculated.
Trev
Quote: TrevorJust out of interest, how do you calculate the number of permutations for a blackjack hand.
I was looking at another Lucky Ladies side bet that pays for 2QH + Dealer has blackjack and was curious as to how this is calculated.
Trev
d = number of decks
Total combinations combin(52*d,2)*combin(52*d-2,2)
2QH + BJ combin(d,2)*(4*d)*(16*d-2)
2QH + other combin(d,2)*(combin(52*d-2,2)-(4*d)*(16*d-2))
Matched 20 15*combin(d,2)*combin(52*d-2,2)
Suited 20 4*(combin(4,2)*d*d+d*d)*combin(52*d-2,2)
Unsuited 20 (combin(4,2)*(16*d*d)+4*d*3*d)*combin(52*d-2,2)
Wizard used permutations, so multiply these by 4 to match his numbers.
Unsuited 20 Looks weird. Might want to double check that one
My combinations are: =(COMBIN(4,2)+1)*PERMUT(4,2)*decks^2
In simple English, this is the ways to choose 2 ranks out of 4 for a 20 composed of two different face cards. Then add one for the A/9 20. Then choose two suits out of four. You use permut, because there are two ways to assign two different suits to two different ranks. So now you have two separate ranks and suits. Then multiply both by the number of decks.
Quote: WizardThanks for the correction. Indeed, I forgot about the A/9 twenties in version D. However, I still get a different number of unsuited 20's than you, Miplet.
My combinations are: =(COMBIN(4,2)+1)*PERMUT(4,2)*decks^2
In simple English, this is the ways to choose 2 ranks out of 4 for a 20 composed of two different face cards. Then add one for the A/9 20. Then choose two suits out of four. You use permut, because there are two ways to assign two different suits to two different ranks. So now you have two separate ranks and suits. Then multiply both by the number of decks.
Don’t forget that they can both be the same rank, but different suits.
combin(4,2)*4*d*4*d +4*d*3*d
Pick 2 of 4 suits
There are 4 ranks for each of the suits and d decks of those ranks. Then add in the A9. 4 suits from d decks for the A and 3 suits from d decks for the 9.
