Math Help

[roliin]

If a blackjack is dealt 1 in every 21 hands, does that mean I can expect a blackjack with a queen of hearts in it 1 in every 336 hands (tens,jacks, queens, kings =16 x 21)?

Also, on wizard of odds I see the probability of getting 2 queen of hearts (6 decks) is .000295. Is that the same as .0295% or 1 in 2950?

TY

[aceside]

Everything else looks good, but 0.0295% is 1 in 3390.

[DSchles]

If a blackjack is dealt 1 in every 21 hands, does that mean I can expect a blackjack with a queen of hearts in it 1 in every 336 hands (tens,jacks, queens, kings =16 x 21)?

Also, on wizard of odds I see the probability of getting 2 queen of hearts (6 decks) is .000295. Is that the same as .0295% or 1 in 2950?

TY

For the first one, you don't mention how many decks you're playing against. For a six-deck game, the first one is 2 x 24/312 x 6/311 = 0.002968 = 1 in 336.9
For the second: 6/312 x 5/311 = 0.0003091 = 1 in 3,235. Since Mike's answer is different, he obviously wasn't doing it for six decks.

Don

[Dog Hand]

For the first one, you don't mention how many decks you're playing against. For a six-deck game, the first one is 2 x 24/312 x 6/311 = 0.002968 = 1 in 336.9
For the second: 6/312 x 5/311 = 0.0003091 = 1 in 3,235. Since Mike's answer is different, he obviously wasn't doing it for six decks.

Don

Don,

The Wiz was indeed calculating the probability for 6D, but he has a separate entry for two heart Q's with a dealer BJ; see "Analysis of Pay Table A":

https://wizardofodds.com/games/black.../lucky-ladies/

Hope this helps!

Dog Hand

[roliin]

For the first one, you don't mention how many decks you're playing against. For a six-deck game, the first one is 2 x 24/312 x 6/311 = 0.002968 = 1 in 336.9
For the second: 6/312 x 5/311 = 0.0003091 = 1 in 3,235. Since Mike's answer is different, he obviously wasn't doing it for six decks.

Don

Curious as to why the discrepency? Mike's table says for 6 decks (Table A), probability of getting 2 queens is .000295. Table C however, also 6 decks, agrees with you (.000309). Why the difference in probability if they're both 6 decks? Shouldn't the "pay table" be irrelivent?

[DSchles]

Did you not see Dog Hand's post above? If the dealer has a BJ AND there are two Queen of Hearts for the player, then the math is different. You have to remove the dealer's cards.

Don

[roliin]

Did you not see Dog Hand's post above? If the dealer has a BJ AND there are two Queen of Hearts for the player, then the math is different. You have to remove the dealer's cards.

Don

I got it, I think. The tables are just a little confusing. Table A lists the probability of getting two queens with a dealer BJ and then, seperately, the probability of getting two queens:

Lucky Ladies Pay Table A — 6 decks

HAND	PERMUTATIONS	PROBABILITY	PAYS	RETURN
Q of hearts pair & dealer has BJ	135360	0.000015	1000 to 1	0.014563
Q of hearts pair	2738340	0.000295	125 to 1	0.036827

Table D excludes dealer BJ with two queens and just lists the probability of getting two queens:

Lucky Ladies Pay Table D — 6 decks

HAND	PAYS	COMBINATIONS	PROBABILITY	RETURN
Queen of hearts pair	100	15	0.000309	0.030918

I assume that Table A is giving the probabilitiy of two queen of hearts (below two queen hearts w dealer Bj) with dealers two cards and Table D is giving just the straight probability of getting two queen of hearts without consideration that the dealer will have two cards?

Sorry in advance if I'm beating a dead horse.

[DSchles]

Everything you wrote above is exactly correct. No need to apologize.

Don