JB78NJ
JB78NJ
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April 12th, 2020 at 7:56:04 PM permalink
Hello All,

This is my first post here. I've been so inspired by people's generosity on this site with their time and knowledge that I decided to post the problem I am trying to work through. I've done what I think may be correct work, but I am an amateur with an English degree so not super confident. Any advise or insight would be greatly appreciated. I'm working on a concept where the following can occur:

2 cards are dealt from a 6-deck shoe no jokers and if those two cards are a pair of As, Ks, Qs or Js then two more cards can be dealt (4 total). If any of those two additional cards are also A-J cards, then the hand value / payout is higher.

Because you need a pair first and then other cards are dealt, I am using the method of taking the number of possible cards for the situation divided by the remaining cards in the deck for each card (ex. drawing 4 aces in a row in standard deck would be 4/52 * 3/51 * 2/50 * 1/49). I'm nervous that this doesn't work because I am not using the x choose y formulas that most people seem to use for calculating probabilities in cards, but if my method here doesn't work I'd love to know why.

Of these 4-card hands with pairs for the first two cards, do these probabilities make sense? It's more important that I'm OK conceptually, so no need to double check the calculations (excel did that), but in terms of concept please help me understand if I am doing it wrong. Thanks in advance for any guidance.

Notes for convenience:
[A-J] = Ace, King, Queen, or Jack
[non A-J] = 2 thorugh 10
There are 312 total cards in 6-deck shoe no jokers
There are 96 total [A-J] cards (4 ranks * 4 suits * 6 decks)
There are 24 of any single rank (4 suits * 6 decks)
There are 216 total [non A-J] cards (312 total - 96 A-J)

Four of a Kind [A-J]:
96[A-J]/312 * (23 of same Rank)/311 * (22 of same Rank)/310 * (21 of same Rank)/309
96*23*22*21 / 312*310*309*308 =
1,020,096 / 9,294,695,280 = 0.010975%

Two Pair [A-J]:
96[A-J]/312 * (23 of 1st Rank)/311 * (72 other rank [A-J])/310 * (23 of 2nd Rank)/309
96*23*72*23 / 312*311*310*309 =
3,656,448 / 9,294,695,280 = 0.039339%

3 of a Kind [A-J] + 1 other rank [A-J]:
96[A-J]/312 * (23 of 1st Rank)/311 * (22 of 1st Rank)/310 * (72 of other rank [A-J])/309
96*23*22*72 / 312*311*310*309
3,497,472 / 9,294,695,280 = 0.037629%

3 of a Kind [A-J] + 1[non A-J]:
96[A-J]/312 * (23 of Rank)/311 * (22 of rank)/310 * (216 other [non A-J])/309
96*23*22*216 / 312*311*310*309
10,492,416 / 9,294,695,280 = 0.112886%

Pair [A-J] + 2 other ranks but not another pair(A-J):
96[A-J]/312 * (23 of 1st Rank)/311 * (72 of 2nd rank [A-J])/310 * (48 of 3rd rank [A-J])/309
96*23*72*48 / 312*311*310*309
7,630,848 / 9,294,695,280 = 0.082099%

Pair[A-J] + 1 other rank [A-J] + 1 [non A-J]:
96[A-J]/312 * (23 of 1st Rank)/311 * (72 of 2nd rank [A-J])/310 * (216 [non A-J])/309
96*23*72*216 / 312*311*310*309
34,338,816 / 9,294,695,280 = 0.369445%

Pair [A-J] + 2 [non A-J]:
96[A-J]/312 * (23 of Rank)/311 * (216 other [non A-J]/310 * (215 other [non A-J])/309
96*23*216*215 / 312*311*310*309
102,539,520 / 9,294,695,280 = 1.103205%

Lose (Do not draw [A-J] pair in first 2 cards):
100% - sum of % above =
100% - 1.755640% = 98.244360%

Thank you so much for any help and please forgive me for a long first post. I hope anyone who understands this stuff really well will be able to look relatively quickly and say if I'm on the wrong track without actually having to look through all of it in detail. If I did do it all wrong, then at least it's a good learning experience.

All the best and I wish you all good health.

Best,

John
Ace2
Ace2
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JB78NJ
April 12th, 2020 at 8:11:48 PM permalink
A quick and easy reasonableness test is to run the numbers for infinite deck (no suits). So, for instance, the probability of four of a kind for A-J would be 4 / 13^4 = 0.014 %. Your number is 0.011% so that seems reasonable
Itís all about making that GTA
ksdjdj
ksdjdj
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JB78NJ
April 12th, 2020 at 8:56:48 PM permalink
Hi and welcome to WoV,

Quote: JB78NJ

(snip)

Of these 4-card hands with pairs for the first two cards, do these probabilities make sense? It's more important that I'm OK conceptually, so no need to double check the calculations (excel did that), but in terms of concept please help me understand if I am doing it wrong. Thanks in advance for any guidance.

Notes for convenience:
[A-J] = Ace, King, Queen, or Jack
[non A-J] = 2 thorugh 10
There are 312 total cards in 6-deck shoe no jokers
There are 96 total [A-J] cards (4 ranks * 4 suits * 6 decks)
There are 24 of any single rank (4 suits * 6 decks)
There are 216 total [non A-J] cards (312 total - 96 A-J)

Four of a Kind [A-J]:
96[A-J]/312 * (23 of same Rank)/311 * (22 of same Rank)/310 * (21 of same Rank)/309
96*23*22*21 / 312*310*309*308 =
1,020,096 / 9,294,695,280 = 0.010975%


I think the way you are doing it is an acceptable way, and I would probably do it that way too.

The first line and answer for " Four of a Kind [A-J]: " looks good to me, you made a typo on the 2nd line, it should be "96*23*22*21 / 312*311*310*309 ="


Quote:


Lose (Do not draw [A-J] pair in first 2 cards):
100% - sum of % above =
100% - 1.755640% = 98.244360%

Thank you so much for any help and please forgive me for a long first post. I hope anyone who understands this stuff really well will be able to look relatively quickly and say if I'm on the wrong track without actually having to look through all of it in detail. If I did do it all wrong, then at least it's a good learning experience.

All the best and I wish you all good health.

Best,

John


For "Lose (Do not draw [A-J] pair in first 2 cards):", i get 97.724462...% ***

*** Proof:

Chance of drawing an "A-J pair" in first 2 cards = 96 x 23 / (312 x 311) = 0.02275537966856294830571357902548
Chance of not drawing "A-J Pair" in first 2 cards= 1 - 0.02275537966856294830571357902548
= 0.97724462033143705169428642097452
Therefore the chance of not drawing "A-J Pair" in first 2 cards = 97.724462...%

note: Since I don't know the thing you are testing, my answer and proof could still be "wrong" when applied to what you are doing, but going by the way you have worded it as " Lose (Do not draw [A-J] pair in first 2 cards)", then I think the above answer i did is correct.

Lastly, I haven't checked the math thoroughly for the other parts, but as a "concept" the way you are doing is seems good to me.
DogHand
DogHand
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April 12th, 2020 at 9:21:19 PM permalink
Quote: JB78NJ

<snip>2 cards are dealt from a 6-deck shoe no jokers and if those two cards are a pair of As, Ks, Qs or Js then two more cards can be dealt (4 total). If any of those two additional cards are also A-J cards, then the hand value / payout is higher.

Because you need a pair first and then other cards are dealt, I am using the method of taking the number of possible cards for the situation divided by the remaining cards in the deck for each card (ex. drawing 4 aces in a row in standard deck would be 4/52 * 3/51 * 2/50 * 1/49). I'm nervous that this doesn't work because I am not using the x choose y formulas that most people seem to use for calculating probabilities in cards, but if my method here doesn't work I'd love to know why.

Of these 4-card hands with pairs for the first two cards, do these probabilities make sense? It's more important that I'm OK conceptually, so no need to double check the calculations (excel did that), but in terms of concept please help me understand if I am doing it wrong. Thanks in advance for any guidance.
<snip> 3 of a Kind [A-J] + 1 other rank [A-J]:
96[A-J]/312 * (23 of 1st Rank)/311 * (22 of 1st Rank)/310 * (72 of other rank [A-J])/309
96*23*22*72 / 312*311*310*309
3,497,472 / 9,294,695,280 = 0.037629%<snip>

,

JB78NJ,

I see an error in this calculation. You have correctly calculated the case of first drawing three of a kind in honors, followed by a fourth non-match honor: for example, AAAJ. But what about drawing them in the order AAJA: wouldn't that also qualify?

According to your rules, the first two have to be matched honors, so the beginning of your calculation is correct: 96[A-J]/312 * (23 of 1st Rank)/311

The problem now is that the next two cards can be drawn in two ways: first the match, then the non-match (that's what you calculated as (22 of 1st Rank)/310 * (72 of other rank [A-J])/309), or first the non-match, then the match, which would be this:

(72 of other rank [A-J])/310 * (22 of 1st Rank)/309.

Of course, you can see by inspection that both of these probabilities for the last two cards yield the same result (because multiplication of scalars is commutative: 3*5 = 5*3), so we can combine them into one term simply by multiplying by 2.

Thus, the probability of 3 of a Kind [A-J] + 1 other rank [A-J] is this:

96[A-J]/312 * (23 of 1st Rank)/311 *2*[ (22 of 1st Rank)/310 * (72 of other rank [A-J])/309]
96*23*2*[22*72] / 312*311*310*309
2*3,497,472 / 9,294,695,280 = 2*0.037629% = 0.075258%.

You have similar counting errors in the calculations for these:
3 of a Kind [A-J] + 1[non A-J]
Pair[A-J] + 1 other rank [A-J] + 1 [non A-J]

Hope this helps!

Dog Hand
JB78NJ
JB78NJ
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Joined: Mar 31, 2020
April 13th, 2020 at 1:45:53 PM permalink
Quote: DogHand


I see an error in this calculation. You have correctly calculated the case of first drawing three of a kind in honors, followed by a fourth non-match honor: for example, AAAJ. But what about drawing them in the order AAJA: wouldn't that also qualify?



Seems so obvious now when you point it out but didn't even think about that while I was putting this together. Very helpful thanks!
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