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.

Quote:gordonm888Question: 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.

Quote:gordonm888Question: 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 ?

Answer (attempt): About 1/53,000

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.

Quote:ksdjdjHere 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 ?

Answer (attempt): About 1/53,000

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).

The actual answer is...

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...

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.

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.)

Just an estimate using very rough figures and estimates.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 " ?...

(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.

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 |

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

I'm slightly confused as for small P (i.e. 1-P or 1-PQuote:ThatDonGuy...This equals (1 - P^N) / (P (1 - P)).

^{N})~=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.

Quote:charliepatrickI 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.

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.