OT - Card Game Probability

[Mustang]

Ok, had an odd situation come up playing cards tonight. Not sure if everyone is familiar with 7/27 so, I'll explain below.

The game was 7/27. The idea is get either a total of 7 or 27. Face cards are worth 1/2, 2-10 is worth face value and an Ace is worth 1 or 11. The deal starts with 1 card up and 1 card down. Starting to the left of the dealer, each player has a choice to take another card or stop. Once it goes around the table completely with no one taking a card, the game stops and the closest to 7 without going over wins half the pot and the closest to 27 without going over wins half the pot. You can play both high and low without declaring. (Obviously.. A/A/5 is the best, it is both 7 and 27).

Ok, on the deal we had Ax, Ax,3x,4x.. No one took a card, everyone bet and the game ended. The Cards were flipped up A6, A6, 34, 43.. all 7's. None of us has ever seen this and we are curious. What are the odds of 4 players having 7 total on the deal???

Is it just a simple 24/52*23/51*22/50*21/49*4/48*4/47*3/48*3/45??

Since on the first card, the chances of getting a card to make seven out of 2 cards is limited to A->6. So, the odds of getting one of those cards is 24/52. (Little under 50%). Then, the odds of getting the card to match on the second card (with 4 people) is roughly 4/48, then 4/47, etc...

Am I roughly on target here or???

[QuikSand]

Originally posted by Mustang
What are the odds of 4 players having 7 total on the deal???


Well, there are 52 x 51 = 2,652 combinations of cards that you could be dealt to start off with. Of those, 24 x 23 = 552 of them add to seven in this game (which I detest, by the way).

Seems to me that the calculation for all four people being dealt a sum of seven would be:

552/2652 x 551/2651 x 550/2650 x 549/2649

...which seems to come out at about 536-1.

[QuikSand]

Originally posted by Mustang
Is it just a simple 24/52*23/51*22/50*21/49*4/48*4/47*3/48*3/45??

Since on the first card, the chances of getting a card to make seven out of 2 cards is limited to A->6. So, the odds of getting one of those cards is 24/52. (Little under 50%). Then, the odds of getting the card to match on the second card (with 4 people) is roughly 4/48, then 4/47, etc...

Looking quicly, I think you're close - but you failed to delete the missing low cards from the deck after the first round.

For example - if A gets dealt a 3 initially, that fits your pattern so far. B then has to be dealt a card from A-6. It's possible that that card is a 4 -- and that's one of the cards that A needs to get his sum of seven. This can't be ignored when you then come back to calculate how likely it is that A will get the card he needs -- calling it 4/48 there overloks teh fact that the A-6 group has been depleted, and that it's a little more likely that one of the cards that are needed to sum to seven are already taken elsewhere.

[albionmoonlight]

Not to stir up a row here, but why do you detest this game (of which I had not heard until this post), Quik?

[dixieflatline]

Quicksand,

I think you may be double conting. The hands that make up 7 are Ax6x, 2x5x, and 3x4x where x can be any suit. So there are 16 possiblities that you get a A6 starting hand. Order doesn't matter so multiply by two because 6A is as good as A6. That's 32 hands for A6. The same goes for 25, and 34 which makes 92 chances. The rest of the math looks right to me though.

I also detest this game as it is very random and usually ends up with huge pots won with little skill. I detest 3-13-27 more htough if anyone has played that variant.

[Mustang]

On a side note, I do detest this game too.

[QuikSand]

Originally posted by dixieflatline
I think you may be double conting. The hands that make up 7 are Ax6x, 2x5x, and 3x4x where x can be any suit. So there are 16 possiblities that you get a A6 starting hand. Order doesn't matter so multiply by two because 6A is as good as A6. That's 32 hands for A6. The same goes for 25, and 34 which makes 92 chances. The rest of the math looks right to me though.


Am I double counting? I think I am, essentially, double counting everything in a manner of speaking, so it should wash out, right?

In my count of 2,652 combinations, I am counting Ah-6d as one combination, and then 6d-Ah as another one. I do the same when counting the 552 combinations that add to seven. Seems to me it just washes out in the probabilities (I could have divided both the numerator and denominatr of each component by two, but it would result in the same probability)

Am I missing something? (Certainly possible)

[QuikSand]

Originally posted by albionmoonlight
Not to stir up a row here, but why do you detest this game (of which I had not heard until this post), Quik?

Like the above comments - I think the game is just too much pure luck. Also, as implemented by the people I know who call it at dealer's choice games, they allow continued rounds of betting even after some people are "done" (presumably having hit on or near seven) and others are still taking more cards (presumably trying to get to 27). When this happens, the person who hit seven can keep pushing the betting with a pat hand, while others contine to chase (with no chance to upset the person who has the pat hand). I think this makes for a flawed betting structure - not a good, balanced one that works well in many other poker games.

[Radii]

Originally posted by QuikSand
Also, as implemented by the people I know who call it at dealer's choice games, they allow continued rounds of betting even after some people are "done" (presumably having hit on or near seven) and others are still taking more cards (presumably trying to get to 27). When this happens, the person who hit seven can keep pushing the betting with a pat hand, while others contine to chase (with no chance to upset the person who has the pat hand). I think this makes for a flawed betting structure - not a good, balanced one that works well in many other poker games.

The way we play it in the home game I attend, when you "pass" and do not get a card, you put a chip on your cards(or some other notation to indicate you've passed). On your first two passes, you may bet and raise like normal. After your third pass, you no longer have the option to draw, and you may no longer bet or raise, you may only fold.

I don't know if this is standard or not, but we also play that you may draw after you have passed in a previous round. If I get dealt a 5, I will pass in the first round and see what others are doing. There have been rare occasions where I win the low by default when everyone else goes high with something like a 5 or a 6(or 9, etc).

Like most of the dealers choice games I've played, its one that is fun to play w/ friends when it gets called, but is not one I expect to win a lot of money at. I suspect that the proper strategy from a profit perspective involves folding immediately if you a) don't get dealt a 7, or b) don't have a single face card draw to a 7.

[Radii]

Originally posted by Radii
After your third pass, you no longer have the option to draw, and you may no longer bet or raise, you may only fold.



dola... I meant to say, you may no longer bet or raise, you may only call or fold. I keep leaving out key words in my posts today, maybe I didn't get enough sleep last night, sheesh.

[dixieflatline]

Originally posted by Quicksand
In my count of 2,652 combinations, I am counting Ah-6d as one combination, and then 6d-Ah as another one. I do the same when counting the 552 combinations that add to seven. Seems to me it just washes out in the probabilities (I could have divided both the numerator and denominatr of each component by two, but it would result in the same probability)


I think your denominator is perfect but I don't think there are 552 combinations for the numerator. If that were the case %20 percent of the time you would have a 7 dealt to you which seems a little high to me. I'm going to do a list to get to 96(what I think the numerator should be) so please let me know what I am leaving out

A6 - 16 of these for all suit combinations
25 - 16 more
34 - 16 more
43 - 16 more
52 - 16 more
6A - 16 more

I think that is all of them which would give you about a 4% chance of getting dealt a 7 to start.

[QuikSand]

You're right - I booted the calculation completely. It's not actual double-counting, but rather an eggregious overcounting of the seven combos (I actually just counted all the combinations of everything involving two cards A through 6). (D'oh!)

[QuikSand]

on further reflection...

Seems to me that the calculation for all four people being dealt a sum of seven would be:

96/2652 x 95/2651 x 94/2650 x 93/2649

...which seems to come out at about 619,000 to 1.

[Mustang]

Quicksand,

Wouldn't those numbers be a little different because, you don't get 2 cards at one time?

I had the first player at 24/52 * 1/12. A little higher odds than yours.

I came up with about 926,000 - 1...

[QuikSand]

Might be a big brain cramp on my part. Your 24/52 x 1/12 isn't right, though... you cannot simply assume that the second card is a random draw from the deck (whic is what your 4/48 assumes). When you assume that the other players have themselves gotten cards from the A-6 subset, that skews the probabilities.

[QuikSand]

The chance that the first person gets a seven, absent any information about the other players' cards, is pretty clearly:

24/52 (chance that he gets something A-6 with first card)
x
4/51 (remaining chance that he gets one of the cards he needs to add to seven)

...which calculates to 96/2652 (as above).

I think step by step through the others works just as well, though the calculations get a bit more complicated, since you have to calculate two scenarios (to look at it this way)-- what if the first card matches one of those already used by the first player, and what if it doesn't. But I'm pretty sure that my math from above (3:19) is correct...