There is a video blackjack machine that offers a bonus for having 4 blackjacks at the same time (dealer blackjack excluded).
What are the chances of getting 4 blackjacks?
A single deck is used.
We get to play on 7 hands per game.

I don't know the odds, but this has to be one of the worst bonuses I've heard of.

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Hi dlarcher10, welcome to the forums.
I'd encourage you to 'sound this out' and try to solve it yourself, even though I'm providing the info below. Problems like this sound really complicated, but really aren't when you 'sound it out' and take the probably one piece at a time.
In statistics when you need multiple events to happen (frequently when you use the word 'AND') the resulting probability is multiplicative... When any one of multiple events could happen (frequently when you use the word 'OR') the resulting probability is additive. Thus:
P(4 blackjacks) = P(1st blackjack) * P(2nd blackjack given 1st) * P(3rd blackjack given 1st 2) * P(4th blackjack given 1st 3)
P(1st blackjack) = P(1st ace BJ) OR P(1st 10 BJ) = P(1st ace 2nd 10) + P(1st 10 2nd ace) = [(4/52)*(16/51)] + [(16/52)*(4/51)] = (.0769*.3137) + (.3077*.0784) = .0241 + .0241 = .0482... or ~4.8% chance of getting dealt a blackjack (~1 in 21 hands, as shown previously by the Wiz).
P(2nd blackjack given 1st) = (same as above just remove one ace, 10, and 2 total cards from the deck)... [(3/50)*(15/49)]*2 = (.06*.3061)*2 = .0368, or ~3.7%
P(3rd blackjack given 1st 2) = (same as above with more removals)... [(2/48)*(14/47)]*2 = .0248, or ~2.5%
P(4th blackjack given 1st 3) = (same as above with more removals)... [(1/46)*(13/45)]*2 = .0126, or ~ 1.3%
Thus, P(4 blackjacks) = .0482 * .0368 * .0248 * .0126 = .0000005543, or ~ .00005543%... which is ~1 in 1.9 million.
I'd expect the payout to be a million bucks, and then they'd still be shorting you. So more than likely a terrible bet.

2 * (16/52 * 4/51) = 128/2652
128/2652 * Number of hands = 128/2652 * 7 = 896/2652 ~ 1/3
Did you take into account that any of the 7 hands can make 4 blackjacks?

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Assuming there are 7 players:
There are (7)C(4) = 35 groups of 4 players that can have the blackjacks
The first player can have any of 4 aces and any of 16 10-cards, or 64 possible hands
The second player can have any of the 3 remaining aces and any of the 15 remaining 10-cards, or 45 hands
The third player can have either of the 2 remaining aces and any of the 14 remaining 10-cards, or 28 hands
The fourth player can have the remaining ace and any of the 13 remaining 10-cards, or 13 hands
The first of the other three players can have any of the (44)C(2) remaining hands, the second any of the (42)C(2) remaining hands, and the third any of the (40)C(2) remaining hands.
Divide this product by (52)C(2) x (50)C(2) x (48)C(2) x (46)C(2) x (44)C(2) x (42)C(2) x (40)C(2), and you get about 1 / 51,685.
Simulation seems to confirm this calculation.

7 x 2 x 16/52 x 4/51 = 896/2652
6 x 2 x 15/50 x 3/49 = 540/2450
5 x 2 x 14/48 x 2/47 = 280/2256
4 x 2 x 13/46 x 1/45 = 104/2070
896/2652 x 540/2450 x 280/2256 x 104/2070 = 1/2153
I'd like to know if I am wrong please.
Thank you for your answers

Playing it correctly means you've already won.

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So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.