What are the odds

In a Blackjack game what are the odds of the 1st, 2nd, 12th and 13th cards being an Aces with 6 decks in a 5 player table? Also, what are the odds of this happening twice in one night in the same table?

Quote: Said

In a Blackjack game what are the odds of the 1st, 2nd, 12th and 13th cards being an Aces with 6 decks in a 5 player table? Also, what are the odds of this happening twice in one night in the same table?

I can't answer the second question because I do not know how many rounds you played at that table over the course of the night. The answer to your first question is very simple since you want them to be specific cards. There are 24 Aces of 312 cards, so:

(24/312) * (23/311) * (288/310) * (287/309) * (286/308) * (285/307) * (284/306) * (283/305) * (282/304) * (281/303) * (280/302) * (22/301) * (21/300) =

0.00001487058 or 1/0.00001487058 = 67,246.87 (Rounded)

That may seem odd, but you are highly specifying the position of Four Aces in thirteen cards, so that's what changes the likelihood a bit. If the four Aces could be anywhere in the first thirteen cards dealt, then:

nCr(24,4)*nCr(288,9)/nCr(312,13) = 0.010632470547348 or 1/0.010632470547348 = 1 in 94.05 to get exactly four of the Aces dealing out thirteen cards from a six deck shoe.

As you can see, if you do not demand that the Aces be in a specific position, this doesn't even have the faintest appearance of anything unusual.

Right, thanks a lot! Actually I was playing switch blackjack last night (so each of the 5 players had 2 games) and making a side bet that payed 40 to 1 in four of a kind. One guy got 2 pairs of aces (four of a kind) and won the bet...only about 30 rounds later I got exactly the same thing. Lets say thy were in total 100 rounds. What are the odds of getting any four of a kind? Also, is 40 to 1 a "fair" payout for such odds (considering any four of a kind and the specific one of Aces)

I'm still not entirely sure what you are asking, but I can definitely say that the payout is not fair or the casino would not offer the bet, at least, not in most cases.

Are you saying you get paid if your first four cards result in a 4OaK, the odds of that are not good:

(1 * 23/311 * 22/310 * 21/309) = 0.00035668861 or 1/0.00035668861 = 1 in 2803.57

I'm assuming there is more to the side bet than that.

Quote: Mission146

I can't answer the second question because I do not know how many rounds you played at that table over the course of the night. The answer to your first question is very simple since you want them to be specific cards. There are 24 Aces of 312 cards, so:

(24/312) * (23/311) * (288/310) * (287/309) * (286/308) * (285/307) * (284/306) * (283/305) * (282/304) * (281/303) * (280/302) * (22/301) * (21/300) =

0.00001487058 or 1/0.00001487058 = 67,246.87 (Rounded)

That may seem odd, but you are highly specifying the position of Four Aces in thirteen cards, so that's what changes the likelihood a bit. If the four Aces could be anywhere in the first thirteen cards dealt, then:

nCr(24,4)*nCr(288,9)/nCr(312,13) = 0.010632470547348 or 1/0.010632470547348 = 1 in 94.05 to get exactly four of the Aces dealing out thirteen cards from a six deck shoe.

As you can see, if you do not demand that the Aces be in a specific position, this doesn't even have the faintest appearance of anything unusual.

Looks like your math is "Cards #1, 2, 12, 13 are Aces, while 3-11 are specifically non-Aces". Don't think the original question had the non-aces part.

I'd say it's:

24/312 * 23/311 * 22/310 * 21/309 = 0.00002743 = 1 in 1/0.00002743 = 1 in 36,446.

Quote: RS

Looks like your math is "Cards #1, 2, 12, 13 are Aces, while 3-11 are specifically non-Aces". Don't think the original question had the non-aces part.

I'd say it's:

24/312 * 23/311 * 22/310 * 21/309 = 0.00002743 = 1 in 1/0.00002743 = 1 in 36,446.

;)

You just made an ass out of me, but rightly so, LOL, I shouldn't have assumed that the other nine cards absolutely could not be Aces.

Well, OP, now you have the Math for those cards all being Aces and for ONLY those cards all being Aces.

Thanks, RS!

Quote: Said

Right, thanks a lot! Actually I was playing switch blackjack last night (so each of the 5 players had 2 games) and making a side bet that payed 40 to 1 in four of a kind. One guy got 2 pairs of aces (four of a kind) and won the bet...only about 30 rounds later I got exactly the same thing. Lets say thy were in total 100 rounds. What are the odds of getting any four of a kind? Also, is 40 to 1 a "fair" payout for such odds (considering any four of a kind and the specific one of Aces)

Hard to say whether that's a "fair" payout, because it's probably not the only pay on the sidebet (or nobody would play it). It's a component of some bet where the top payout is 4OAK, right? Though perhaps there's a higher one for 4OAK suited or colored? In isolation, no, it's probably not "fair". But if the total sidebet paytable adds up to a HE of maybe 4% or less, it's one of the "fairer" sidebets out there IMO. And 40:1 makes it more attractive to a lot of casinos than 500:1 or some huge payout like that. Too much exposure. 40:1 would also make it possible to pay either better odds on the lower wins, or pay more frequently with more lower hands possible.

Say they paid 500:1 odds. That might make it impossible to pay even money on a pair; instead they'd have to push it or something. Or if they pay 10:1 on 3OAK, they can now only afford to pay 5:1 or 3:1 because the paytable's so top-heavy. I don't know what sidebet they're using, but you have to look at the whole paytable to have some idea whether it's a good bet (given that the Wizard says all sidebets are sucker bets).