ksdjdj
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March 9th, 2021 at 3:36:30 AM permalink
Hi,

Below are some questions about 3 card rummy I would like answered, if possible:

1. The chance of the player and dealer getting a tie on 0 points ?

2. The overall chance of the player getting 0 points (whether the dealer has qualified or not) ?

3. Using the information from the questions above, how you would get a probability of 0.002919 as the answer for "Player beats dealer with 0 points " ?

Note : see "Ante Bet Return Table" in the link here>>> https://wizardofodds.com/games/vegas-three-card-rummy/ <<< , for where I got the probability figure for question 3.

Thanks for any help in advance.

PS if you want to know why I am asking these questions , then I may tell you via PM.
Last edited by: ksdjdj on Mar 9, 2021
gordonm888
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March 9th, 2021 at 9:11:52 AM permalink
Question: is a suited AK worth zero points, the same as a suited A2?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
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March 9th, 2021 at 9:26:44 AM permalink
Quote: gordonm888

Question: is a suited AK worth zero points, the same as a suited A2?

According to the rules, no - ace is low, and king is high.

Quote: ksdjdj

Below are some questions about 3 card rummy I would like answered, if possible:

1. The chance of the player and dealer getting a tie on 0 points ?

2. The overall chance of the player getting 0 points (whether the dealer has qualified or not) ?

3. Using the information from the questions above, how you would get a probability of 0.002919 as the answer for "Player beats dealer with 0 points*** " ?

You can't use the numbers from 1 and 2 to calculate 3, as 3 does not include when the dealer does not qualify while the player hand has 0 points.
ksdjdj
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March 9th, 2021 at 12:53:46 PM permalink
Quote: gordonm888

Question: is a suited AK worth zero points, the same as a suited A2?

No it is not, see post made by ThatDonGuy above.

Quote: ThatDonGuy

(snip)
You can't use the numbers from 1 and 2 to calculate 3, as 3 does not include when the dealer does not qualify while the player hand has 0 points.

Ok, what about if you use the "Dealer doesn't qualify" probability figures from "Ante Bet Return Table"(see link here > Link ) can you then answer "question 3."?

Here are my attempts at answering questions1. 2. and 3. from my OP.

1. The chance of the player and dealer getting a tie on 0 points ?

Note: this is just a guess, since I don't really know how to work this one out (i just used 96/22100^2 to get this figure).

2. The overall chance of the player getting 0 points (whether the dealer has qualified or not) ?
Ans: 96/22100 = ~ 0.004344

3. Using the information from the questions above, how you would get a probability of 0.002919 as the answer for "Player beats dealer with 0 points "

I would attempt to get the answer of 0.002919 by doing this:

i) Start with the "overall chance of the player getting 0 points" = ~ 0.004344
ii) Work out the "chance of the player and dealer getting a tie on 0 points" (I will use 1/53,000, until someone provides a more accurate figure).
iii) Work out / find the chance of the "dealer not qualifying". (I will use the 0.223879 from the link above, near the top of this post)
iv) Add ii) and iii) together to get ~ 0.223897...
v) Take the figure from iv) away from 1 (i get about 0.776102...)
vi) Use figure from i) and multiply it by the figure from v) (I get ~ 0.00337...)

0.00337 does not = or roughly = 0.002919, so I don't know what I am missing.
Last edited by: ksdjdj on Mar 9, 2021
gordonm888
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March 9th, 2021 at 1:11:09 PM permalink
I'll do an exact calculation. The chance of dealer getting a zero is different for the cases when player has an AAA vs 555 vs QJTs vs A23s because of the effect of removing cards form the deck..
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
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March 9th, 2021 at 1:11:35 PM permalink
Quote: ksdjdj

Here are my attempts at answering questions1. 2. and 3. from my OP.

1. The chance of the player and dealer getting a tie on 0 points ?

Note: this is just a guess, since I don't really know how to work this one out (i just used 96/22100^2 to get this figure).

OOPS - I read the rules wrong. I thought any suited pair meant the whole hand was worth zero. Only straight flushes and threes of a kind are worth zero.
There are 44 straight flushes (11 values for the lowest card x 4 suits) and 52 threes of a kind (13 values x 4 sets of 3), which is, er, 96.
Let me get back to you on this one...
gordonm888
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March 9th, 2021 at 3:01:22 PM permalink
EDITED: Found an error in my spreadsheet. The results below have been corrected.

If I did this right:

The first column is player hands that make a Zero, 2nd hand is the probability that dealer does not Qualify when player has that hand, and the third column is the probability of tie-ing with a dealer hand that is worth 0. Obviously, the rest of the probability is a win for player.

Player 3 of a kinds
AAA 0.377225358 0.022742076
222 0.367781155 0.023827616
333 0.357359965 0.022362136
444 0.344984802 0.022362136
555 0.334563613 0.022362136
666 0.336409032 0.022362136
777 0.323871038 0.022362136
888 0.311984368 0.022362136
999 0.301563178 0.022362136
TTT 0.288536691 0.022362136
JJJ 0.289188016 0.022362136
QQQ 0.28929657 0.021819366
KKK 0.282946157 0.023284846

Player's 3-card straight flushes
KQJs 0.289405124 0.02236210
QJTs 0.291196266 0.022036474
JT9s 0.295104212 0.022253582
T98s 0.302485888 0.022253582
987s 0.313829787 0.022253582
876s 0.325282241 0.022253582
765s 0.332718194 0.022253582
654s 0.338688667 0.022253582
543s 0.345636127 0.022253582
432s 0.356708641 0.022742076
32As 0.367238385 0.002333912

The differences in some of the above numbers is due to differences in the number of dealer straight flushes that the player hand blocks.
Last edited by: gordonm888 on Mar 9, 2021
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
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March 9th, 2021 at 3:27:48 PM permalink
I'm coming in late to this but I'm guessing you want to count how many ways the Dealer can get either Trips or Straight Flush given certain Player hands. I see that Suited 32A pays a bonus for the sidebet, but for the main game only "0 points" is considered. So my guess is a Trips ties a Straight Flush.
If the Player has AsAhAd then the dealer can get
(i) 222-KKK = 12 x 4 = 48
(ii) KQJ-432 (s/h/d/c) = 10 x 4 = 40
(iii) 32A (c) = 1 (i.e. 32A (s/h/d) are impossible)
Total = 89 (48+(44-3))
Perms = 49 x 48 x 47 / 6 = 18424
So chance = 89/18424.
(I think you may have only looked at Trips vs Trips = 48/18424.)

If the Player has 5s5h5d then this rules out more Straight Flushes (765,654,543 s/h/d = 9 rather than 3 for AAA.) So the total would be 48+(44-9) = 83.

Similarly KKK QQQ and 222 are 83, 86 and 86.

(NB: I haven't looked at your edited figures.)
charliepatrick
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March 9th, 2021 at 3:57:11 PM permalink
Quote: ksdjdj

...3. Using the information from the questions above, how you would get a probability of 0.002919 as the answer for "Player beats dealer with 0 points " ?...

Just an estimate using very rough figures and estimates.

(A) Chances of Player getting 0 points is 52 Trips and 44 SFs = 96 out of 22100 perms (52x51x50/6) = 0.004 343 891.

(A 2 Chances of Dealer getting 0 points is approximately the same; chances of both happening = 0.000 018 869 (if squaring above)
(4388+3698)/22100/18424 if working it out accurately (I think) = 0.000 019 859.)
Anyway whichever it is, the value is quite small.

(B) Chances of Dealer not qualifying given an unknown Player hand is probably just less than 1/3 (otherwise you would always play, and that would be bad game design). So the chances of a Dealer qualifying is marginally over 2/3. (Given the Player mimics the Dealer and the strategy could have been 19 or 21 points instead, then it's not too much more than 2/3.)

So the chances of Player having 0 and the Dealer qualifying is only slightly greater than A * 2/3 = 0.002 895 928.

Since the dealer will qualify slightly more than 2/3, and the factor for tying is negligible, 0.002919 seems possible.
gordonm888
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March 9th, 2021 at 7:40:40 PM permalink
I'll try this once agian, with another glitch fixed, and with better formatting.

This is a composition-dependent calculation of the Dealer Not Qualifying and of a Player/Dealer Tie for each and every player hand that is zero points.

Player Hand (Zero Points)
Dealer Prob of Not Qualifying
Prob of Tieing
AAA
0.377225358
0.004342162
222
0.367781155
0.00493921
333
0.357359965
0.004776379
444
0.344984802
0.004776379
555
0.334563613
0.004776379
666
0.336409032
0.004776379
777
0.323871038
0.004776379
888
0.311984368
0.004776379
999
0.301563178
0.004776379
TTT
0.288536691
0.004776379
JJJ
0.289188016
0.004776379
QQQ
0.28929657
0.004722102
KKK
0.282946157
0.004884933
KQJs
0.289405124
0.004722102
QJTs
0.291196266
0.004667825
JT9s
0.295104212
0.004722102
T98s
0.302485888
0.004722102
987s
0.313829787
0.004722102
876s
0.325282241
0.004722102
765s
0.332718194
0.004722102
654s
0.338688667
0.004722102
543s
0.345636127
0.004722102
432s
0.356708641
0.004776379
32As
0.367238385
0.004287885
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
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March 10th, 2021 at 8:20:34 AM permalink
I have been trading PMs with the OP, but let me add something here:

If the probability of an event happening is P, the expected number of events before it happens N times in a row is 1/P + (1/P)^2 + (1/P)^3 + ... + (1/P)^N.
This equals (1 - P^N) / (P (1 - P)).
Er, no, it doesn't; it equals (1 - P^N) / (P^N (1 - P)), doesn't it?
Thanks to @charliepatrick for pointing that out
Last edited by: ThatDonGuy on Mar 10, 2021
gordonm888
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March 10th, 2021 at 10:42:57 AM permalink
Deleted, never mind, found an error. The posted strategy of Raise 20, Fold 21 appears to be correct.
Last edited by: gordonm888 on Mar 10, 2021
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
charliepatrick
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March 10th, 2021 at 11:04:35 AM permalink
Quote: ThatDonGuy

...This equals (1 - P^N) / (P (1 - P)).

I'm slightly confused as for small P (i.e. 1-P or 1-PN)~=1 this would equate to 1/P.

I ran some trials using N=2 and P=1/100 to get averages of 10112, 10076, 10116, 10115, 10075 - which suggests the number is nearer 10000 than 100. This sorts of makes sense as if you roll a 100-sided die (which is how I'm simulating it) on average 1 in 100 rolls will be a 1, and then 1 in 100 of those will tend to have the follow-up roll of a 1.
gordonm888
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March 10th, 2021 at 12:19:11 PM permalink
Deleted. Never mind, found an error.
Last edited by: gordonm888 on Mar 10, 2021
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
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March 10th, 2021 at 1:19:26 PM permalink
Quote: charliepatrick

I ran some trials using N=2 and P=1/100 to get averages of 10112, 10076, 10116, 10115, 10075 - which suggests the number is nearer 10000 than 100.

The expected value is 1/p + 1/p^2 = 1 / (1/100) + 1 / (1/100)^2 = 100 + 10,000 = 10,100.

((1/100)^2 - 1) / ((1/100)^2 x (1/100 - 1)) = (-9999/10,000) / (1/10,000 x -99/100)) = (-999,900/1,000,000) / (-99/1,000,000)) = 10,100.
charliepatrick
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March 10th, 2021 at 2:21:37 PM permalink
I agree with Wizard's House Edge and the Player strategy.
Best 20 = 965 EV = -0.906698
Worst 20 = K9A EV = -0.920104
Best 21 = 8s7h6s EV = -1.017152 (although it's more likely the Dealer won't qualify, you can't win any hands)
Worst 30 = KsJhTs/KsJhTd = -1.119409
The coding traps are AKQ isn't a straight, looking for two card flush straights and, in some cases, checking for Pairs as well.

The number of times the Player has Zero points and then...
Win 1188508 (0.002 919)
Tie 8020 (0.000 020)
Lose 0
DNQ 572176 (0.001 405)
Total perms = 52*51*50/6*49*48*47/6 = 407170400.
Darenbug