TigerWu
TigerWu
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September 2nd, 2022 at 9:38:05 AM permalink
I 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
Dieter
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Dieter
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September 2nd, 2022 at 10:10:10 AM permalink
The discards are reshuffled into the stub?
May the cards fall in your favor.
TigerWu
TigerWu
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September 2nd, 2022 at 10:12:30 AM permalink
Quote: Dieter

The 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.
camapl
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September 2nd, 2022 at 1:33:02 PM permalink
Quote: TigerWu

I 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
link to original post



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.
Expectation is the root of all heartache.
Dieter
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Dieter
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September 2nd, 2022 at 1:52:15 PM permalink
I'm also curious about the probability for zero aces after 3 rounds.
May the cards fall in your favor.
ThatDonGuy
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Dieter
September 2nd, 2022 at 4:08:55 PM permalink
Quote: Dieter

I'm also curious about the probability for zero aces after 3 rounds.
link to original post


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: camapl

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.


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.
ThatDonGuy
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camaplTigerWu
September 2nd, 2022 at 6:06:46 PM permalink
Here is what I get:

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
camapl
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TigerWu
September 3rd, 2022 at 6:24:25 AM permalink
Our figures agree, as far as I can tell. TDG’s are cumulative, while mine are exact… As both can be useful, here are mine:

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?
Expectation is the root of all heartache.
TigerWu
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September 3rd, 2022 at 8:06:58 AM permalink
Cool, guys, thanks! I thought it was an interesting little game.
charliepatrick
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September 3rd, 2022 at 8:20:56 AM permalink
I get very similar figures using N(C,R) etc for when you're looking for Aces when there are 4,3,2,1,0 left..
Round 1ProbabilityR 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 2R 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
jafdevera004
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September 6th, 2022 at 6:59:44 PM permalink
Yes thankyou ! I've been looking for someone to send these graphs. I really appreciate it !
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