It uses a 32-card deck, 7s through Aces inclusive. The cards are shuffled and cut, and exactly 13 cards are dealt into a pile, and then this pile is turned over and the cards are spread. If there are any Aces, they are removed and set aside. That whole process counts as one "deal." All of the cards (except any removed Aces) are then collected up and shuffled, and two more deals take place. The object of the game is to get as many Aces in as few deals as possible.

So, what would be the odds of getting:

One Ace in one deal

Two Aces in one deal

Three Aces in one deal

Four Aces in one deal

One Ace in two deals

Two Aces in two deals

Three Aces in two deals

Four Aces in two deals

One Ace in three deals

Two Aces in three deals

Three Aces in three deals

Four Aces in three deals

Quote:DieterThe discards are reshuffled into the stub?

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Yes, the only thing not reshuffled are any Aces removed from the 13 card stock dealt out.

Quote:TigerWuI came across this game in an old book on playing cards. It wasn't specifically described as a gambling game, but it seems like you could easily make it one, with different levels/rounds of betting...

It uses a 32-card deck, 7s through Aces inclusive. The cards are shuffled and cut, and exactly 13 cards are dealt into a pile, and then this pile is turned over and the cards are spread. If there are any Aces, they are removed and set aside. That whole process counts as one "deal." All of the cards (except any removed Aces) are then collected up and shuffled, and two more deals take place. The object of the game is to get as many Aces in as few deals as possible.

So, what would be the odds of getting:

One Ace in one deal

Two Aces in one deal

Three Aces in one deal

Four Aces in one deal

One Ace in two deals

Two Aces in two deals

Three Aces in two deals

Four Aces in two deals

One Ace in three deals

Two Aces in three deals

Three Aces in three deals

Four Aces in three deals

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I have your probabilities, but my grandson just woke up, so I’ll have to post them later. It may not be until late tonight/early morning depending on how my evening goes…

ETA: For those who might want to give it a try in the meantime, I used the hypergeometric distribution, with standard probabilities on the first draw and conditional probabilities thereafter.

Quote:DieterI'm also curious about the probability for zero aces after 3 rounds.

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That's an easy one.

There are C(32,13) ways to draw 13 cards from the 32-card deck, of which C(28,13) hace no aces.

If this happens, all 32 cards are back in the deck for the second, and again for the third, deals.

The probability of zero aces in three deals is (C(28,13) / C(32,13))^3, or about 1 in 798.56.

On the other hand, the probability of four aces in just one deal is C(28,9) / C(32,13), or "only" about 1 in 50.29.

Quote:camaplETA: For those who might want to give it a try in the meantime, I used the hypergeometric distribution, with standard probabilities on the first draw and conditional probabilities thereafter.

Sounds more like a brute force problem to me.

On the first draw, there are C(32,13) ways to draw the cards.

If you have four aces, there are C(28,9) ways to draw the nine non-aces.

If you have three, there are four sets of three aces and C(28,10) ways to draw the ten non-aces, but you now have to take into account there are only 29 cards in the deck for the second draw.

If you have two, there are six pairs of aces and C(28,11) ways to draw the eleven non-aces, but you now have to take into account there are only 30 cards in the deck for the second draw.

If you have one, there are four choices for the ace and C(28,12) ways to draw the twelve non-aces, but you now have to take into account there are only 31 cards in the deck for the second draw.

If you have none, there are C(28,13) ways to draw the thirteen non-aces; the deck will have all 32 cards for the second draw.

4 in 1: 0.0198832

3 or more in 1: 0.1709956

2 or more in 1: 0.5419077

1 or more in 1: 0.8922136

4 in 1-2: 0.1785634

3 or more in 1-2: 0.5760810

2 or more in 1-2: 0.8870311

1 or more in 1-2: 0.9883821

4 in 1-3: 0.4191971

3 or more in 1-3: 0.8299064

2 or more in 1-3: 0.9762792

1 or more in 1-3: 0.9987477

Total number of Aces after…

1st Draw Prob

0 10.7786%

1 35.0306%

2 37.0912%

3 15.1112%

4 1.9883%

Total 100.0000%

2nd Draw Prob

0 1.1618%

1 10.1351%

2 31.0950%

3 39.7518%

4 15.8680%

Prior 4 1.9883%

Total 100.0000%

3rd Draw Prob

0 0.1252%

1 2.2469%

2 14.6373%

3 41.0709%

4 24.0634%

Prior 4 17.8563%

Total 100.0000%

After the first draw, the probability that one has already found the four Aces must be accounted for in order for each total to add to 100%. Think of this as, “all four Aces have already been found, so the current draw isn’t necessary”.

The reason that TDG’s and my figures agree is that the “brute force” method he describes IS the hypergeometric distribution… Whodathunkit?

Round 1 | Probability | R 2 (given 3) | R 2 (given 2) | R 2 (given 1) | R 2 (given 0) | Round 2 |

4 | 0.019 883 204 | 0.067 740 018 | 0.066 508 381 | 0.022 288 651 | 0.002 143 140 | 0.178 563 393 |

3 | 0.151 112 347 | 0.083 372 329 | 0.188 440 413 | 0.109 417 014 | 0.016 287 860 | 0.397 517 616 |

2 | 0.370 912 125 | 0.115 963 331 | 0.155 007 436 | 0.039 979 294 | 0.310 950 061 | |

1 | 0.350 305 895 | 0.063 592 794 | 0.037 758 222 | 0.101 351 016 | ||

0 | 0.107 786 429 | 0.011 617 914 | 0.011 617 914 | |||

Round 2 | R 3 (given 3) | R 3 (given 2) | R 3 (given 1) | R 3 (given 0) | Round 3 | |

4 | 0.178 563 393 | 0.178 197 552 | 0.055 756 563 | 0.006 448 585 | 0.000 231 001 | 0.419 197 094 |

3 | 0.397 517 616 | 0.219 320 064 | 0.157 976 927 | 0.031 656 691 | 0.001 755 610 | 0.410 709 293 |

2 | 0.310 950 061 | 0.097 216 571 | 0.044 846 979 | 0.004 309 225 | 0.146 372 775 | |

1 | 0.101 351 016 | 0.018 398 761 | 0.004 069 824 | 0.022 468 585 | ||

0 | 0.011 617 914 | 0.001 252 254 | 0.001 252 254 |