[bonanova]

It was poker night at Morty's, and true to form, Alex

arrived early to start out at his lucky table. As the night

wore on, the number of players dwindled. And when it

finally came down to Alex against Doyle the new kid in

town, Alex lay down his cards and issued a challenge.

I'll bet my stack of chips here against yours, right now,

that you can't answer one of the simplest poker questions:

a puzzle that any poker player worth his spit should be

able to solve.

These weren't the first words Alex had spoken. All thru

the tournament he had been verbally sizing up the new kid,

looking for a goad that would make this moment a reality.

Doyle, as it happened, was a sharp player. Better perhaps

in raw talent than even Alex himself. But he had a weakness.

Doyle couldn't let a pi**ing contest go by without getting

involved. And when Alex suggested that he couldn't answer

even a basic question about poker, he jumped at the bait.

You're on! he shouted, laying his cards down as well. What's

this question you think I can't answer?

Easy now, Alex began to milk the moment, let's not get all

excited about it. Like I said, it's a simple question. All

poker players know the precedence of suits and hands.

All I'm asking you is to tell me the best Full House you

could possibly be dealt.

Doyle thought for no more than a moment before he answered.

And Alex then began to explain simply and clearly why he now

owned all of Doyle's chips, and had become the tourney champion.

What answer would you have given?

[dms172]

three aces and two kings

[Guest]

I'm inclined to say

Ace of Hearts, Ace of Spades, Ace of Diamonds, Ten of Hearts, Ten of Clubs

You're not just trying to get the highest value, but reduce the chances of your opponent getting dealt a higher hand. Adding tens instead of kings cuts down on the availability of straight flushes to other players. Might not be ideal, but I think I'm on the right track.

[Prime]

I like Alex. He is a hustler.

Are there Jokers in the deck by chance?

[bonanova]

I like Alex. He is a hustler.

Are there Jokers in the deck by chance?

Nope, No jokers.

No pun intended by saying nothing is out of the box here - except the cards.

I won't comment on the two posted answers yet, except to say they aren't [quite] correct.

By the way, Doyle gave dms's answer.

[Guest]

I'm inclined to say

Ace of Hearts, Ace of Spades, Ace of Diamonds, Ten of Hearts, Ten of Clubs

You're not just trying to get the highest value, but reduce the chances of your opponent getting dealt a higher hand. Adding tens instead of kings cuts down on the availability of straight flushes to other players. Might not be ideal, but I think I'm on the right track.

Any straight flush would beat a full house, so I don't think we specifically need to be aiming to reduce the number of royal flushes available - just reducing the number of straight flushes. In that case, the 10 of hearts is slightly redundant, as we have already wiped out the possibility of a 10 J Q K A Hearts flush by claiming the ace.

So... I think aces and nines would be marginally better.

Thinking further... we want to reduce the total number of hands that could beat us... would tens over fours be an even better combo? that adds in a few full houses that can beat us, but does it remove more straight flushes?

And should we be taking the relative probability of FHs and SFs into account?

More work to be done, methinks.

[Guest]

are you talking about 5-card draw or texas hold 'em poker? cos then the probabilities would be slightly different since they share some of the cards. just a thought

if playing 5 card draw then i would say tens over fours probably. but if playing texas told 'em then it would depend on what the flop, turn and river were.

[Guest]

Based on my very basic knowledge of poker, i'd have to guess at:-

3 aces and two kings. I'm not sure of the precedence of suits but just assume i've picked the highest precedence for both kings and aces.

[bonanova]

are you talking about 5-card draw or texas hold 'em poker? cos then the probabilities would be slightly different since they share some of the cards. just a thought

if playing 5 card draw then i would say tens over fours probably. but if playing texas told 'em then it would depend on what the flop, turn and river were.

The OP talks about being dealt a poker hand - that is, being dealt 5 cards.

Let's remove all the variables by making the game 5-card stud poker.

Each player gets 5 cards, all closed, no exchange.

There are [say] five opponents, and the hands come from a single deck.

[bonanova]

Based on my very basic knowledge of poker, i'd have to guess at:-

3 aces and two kings. I'm not sure of the precedence of suits but just assume i've picked the highest precedence for both kings and aces.

have to worry about suits. Only one hand can have three aces, no matter what suit they are. But Aces and Kings is not the strongest hand.

[Guest]

Any straight flush would beat a full house, so I don't think we specifically need to be aiming to reduce the number of royal flushes available - just reducing the number of straight flushes. In that case, the 10 of hearts is slightly redundant, as we have already wiped out the possibility of a 10 J Q K A Hearts flush by claiming the ace.

So... I think aces and nines would be marginally better.

Thinking further... we want to reduce the total number of hands that could beat us... would tens over fours be an even better combo? that adds in a few full houses that can beat us, but does it remove more straight flushes?

And should we be taking the relative probability of FHs and SFs into account?

More work to be done, methinks.

I can see where you're going with this,

in poker, you don't just want to have the best cards, you want people to bet against you so you win money.

If you try to reduce the probability of other people getting straight flushes to beat your full house, then you also reduce the probability of other people getting straight non-flushes. Someone with a straight is likely to lose a lot of money to someone with a full house, so you might want to increase that probability and take the tiny increased chance of losing.

I guess I'm saying that the term "best Full House" is a little ambiguous. You could eliminate my objection by saying that everyone is all-in blind before the deal.

[Guest]

"this one!" Alex says as he shows his hand. Since the new guy already showed his hand, Alex knew at that point that he had won. This moment obviously came at the last hand when his opponet was all-in. Alex read the player well.

Edit- to answer the direct question. I would have answered the question "the one you have in your hand right now." then took all of my chips with my losing hand.

Edited November 7, 2008 by Grayven

[HoustonHokie]

I can't imagine answering a bona question without probability and a proof, so here goes:

In 5 card stud, there are 311,875,200 possible combinations of 5 cards and 2,598,960 possible distinct hands (ignoring order of the cards), which appear with equal probability in any hand of stud poker. There are 3,744 possible full houses, 624 four of a kinds, and 40 possible straight flushes, which should be the only hands we really care about (ignoring the valid question about betting raised by Chuck Rampart). So, the question is, if you hold one of the 3,744 full houses, is there one of those which is better at removing the potential 644 better hands from the deck?

Let's deal with the 624 four of a kinds first (it's easier). Any full house will eliminate 2/13 of those hands right off the bat because those two ranks are eliminated, so we're down to 528 four of a kinds. 528 = 11 * 48, which is indicative of the 11 remaining ranks which could still form the four of a kind and the 48 possible cards which could be the kicker in that hand. Our full house elminates 5 kickers from each of those hands, so we really have only 11 * 43 = 473 four of a kinds left. I don't see anything we can do in our selection of the "optimum" full house to affect this result, so the net effect on four of a kinds is somewhat inconsequential. However, I present it to give completeness to the solution.

What we really care about are the 40 straight flushes which remain. Can we eliminate more of them by our selection of an "optimum" full house? The answer is, undoubtedly, "yes".

Doyle's answer has been revealed as Aces full of Kings. How many of the 40 SF's does this combination eliminate? It depends somewhat on the suits, but let's assume that Doyle was at least smart enough to specify that one of the Kings was of the suit that is not represented in the Aces. By choosing Aces, you elminate three Ace-high and 3 Ace-low SFs, for a total of 6 SFs. The King which shares a suit with one of the Aces removes one other King-high SF, whereas the other King removes an Ace-high and a King-high SF, so a total of 3 more SFs are removed. So Doyle's answer removes a total of 9 SFs from contention. Aces full of Kings can therefore be beaten by 473 + 31 = 504 hands (0.01939% chance of losing).

Someone has suggested 10s full of 4s as a better combination, because it would remove a lot of SFs - better for that would be 10s full of 5s. Again, make the assumption that one of the 5s is a different suit than the three 10s (although it doesn't really matter because the 10s and 5s don't appear in the same SF the way Aces and Kings do). So, let's consider how this plays out (no pun intended). Every SF contains either a 10 or a 5, so the suit which is shared by our 10s and 5s is completely removed from contention, dropping 10 of the possible SFs. The remaining two 10s and one 5 all eliminate 5 more SFs each (the 10 removes 10-high thru A-high; the 5 removes 5-high thru 9-high), for an additional 15 SFs removed. Now, these gains are going to be more than offset by the fact that any higher full house will beat 10s over 5s - approximately 1100 hands. So the choice of 10s full of 5s removes a total of 25 SFs but adds 1100 FHs, and can therefore be beaten by 473 + 15 + 1100 = 1588 hands (0.06110% chance of losing) - much worse than Aces full of Kings.

I think it's now obvious that we have to have an Ace-high full house to remove the possibility of being beaten by another full house, which far outweighs the meager reductions in total numbers of hands which comes from removing a high percentage of SFs. I think the best combination then is Aces full of 9s (although it could be Aces full of 8s, 7s, or 6s). As noted before with Doyle's hand, the Aces remove 6 SFs from contention (Ace-high & Ace-low). The 9s remove 5 more SFs each (9-high thru K-high), for a total of 16 SFs removed. Aces full of 9s can then be beaten by only 431 + 24 = 455 hands (0.01751% chanse of losing), and it can be shown that the same is true for Aces full of 8s, 7s, or 6s.

I think that's the best possible solution, although I'm not sure that in practice I'd really be concerned about less than a 100th of a percent difference - unless I was in Morty's that night and all my chips depended on it!

edit:grammar

Edited November 7, 2008 by HoustonHokie

[Prime]

I can't imagine answering a bona question without probability and a proof, so here goes:

In 5 card stud, there are 311,875,200 possible combinations of 5 cards and 2,598,960 possible distinct hands (ignoring order of the cards), which appear with equal probability in any hand of stud poker. There are 3,744 possible full houses, 624 four of a kinds, and 40 possible straight flushes, which should be the only hands we really care about (ignoring the valid question about betting raised by Chuck Rampart). So, the question is, if you hold one of the 3,744 full houses, is there one of those which is better at removing the potential 644 better hands from the deck?

Let's deal with the 624 four of a kinds first (it's easier). Any full house will eliminate 2/13 of those hands right off the bat because those two ranks are eliminated, so we're down to 528 four of a kinds. 528 = 11 * 48, which is indicative of the 11 remaining ranks which could still form the four of a kind and the 48 possible cards which could be the kicker in that hand. Our full house elminates 5 kickers from each of those hands, so we really have only 11 * 43 = 473 four of a kinds left. I don't see anything we can do in our selection of the "optimum" full house to affect this result, so the net effect on four of a kinds is somewhat inconsequential. However, I present it to give completeness to the solution.

What we really care about are the 40 straight flushes which remain. Can we eliminate more of them by our selection of an "optimum" full house? The answer is, undoubtedly, "yes".

Doyle's answer has been revealed as Aces full of Kings. How many of the 40 SF's does this combination eliminate? It depends somewhat on the suits, but let's assume that Doyle was at least smart enough to specify that one of the Kings was of the suit that is not represented in the Aces. By choosing Aces, you elminate three Ace-high and 3 Ace-low SFs, for a total of 6 SFs. The King which shares a suit with one of the Aces removes one other King-high SF, whereas the other King removes an Ace-high and a King-high SF, so a total of 3 more SFs are removed. So Doyle's answer removes a total of 9 SFs from contention. Aces full of Kings can therefore be beaten by 473 + 31 = 504 hands (0.01939% chance of losing).

Someone has suggested 10s full of 4s as a better combination, because it would remove a lot of SFs - better for that would be 10s full of 5s. Again, make the assumption that one of the 5s is a different suit than the three 10s (although it doesn't really matter because the 10s and 5s don't appear in the same SF the way Aces and Kings do). So, let's consider how this plays out (no pun intended). Every SF contains either a 10 or a 5, so the suit which is shared by our 10s and 5s is completely removed from contention, dropping 10 of the possible SFs. The remaining two 10s and one 5 all eliminate 5 more SFs each (the 10 removes 10-high thru A-high; the 5 removes 5-high thru 9-high), for an additional 15 SFs removed. Now, these gains are going to be more than offset by the fact that any higher full house will beat 10s over 5s - approximately 1100 hands. So the choice of 10s full of 5s removes a total of 25 SFs but adds 1100 FHs, and can therefore be beaten by 473 + 15 + 1100 = 1588 hands (0.06110% chance of losing) - much worse than Aces full of Kings.

I think it's now obvious that we have to have an Ace-high full house to remove the possibility of being beaten by another full house, which far outweighs the meager reductions in total numbers of hands which comes from removing a high percentage of SFs. I think the best combination then is Aces full of 9s (although it could be Aces full of 8s, 7s, or 6s). As noted before with Doyle's hand, the Aces remove 6 SFs from contention (Ace-high & Ace-low). The 9s remove 5 more SFs each (9-high thru K-high), for a total of 16 SFs removed. Aces full of 9s can then be beaten by only 431 + 24 = 455 hands (0.01751% chanse of losing), and it can be shown that the same is true for Aces full of 8s, 7s, or 6s.

I think that's the best possible solution, although I'm not sure that in practice I'd really be concerned about less than a 100th of a percent difference - unless I was in Morty's that night and all my chips depended on it!

edit:grammar

Seems you made a typo in the end. You meant your end result to be 473(four of a kind) + 24(SF) = 497 hands beat aces full of nines. (Instead of 431+24=455).

[Guest]

Taking psychology out of the picture and just going by pure statistics, the most winning full house hand would be 10's over 5's.

Note: Suites would be irrelevant

Edited November 7, 2008 by RyanJ

[bonanova]

I can't imagine answering a bona question without probability and a proof, so here goes:

In 5 card stud, there are 311,875,200 possible combinations of 5 cards and 2,598,960 possible distinct hands (ignoring order of the cards), which appear with equal probability in any hand of stud poker. There are 3,744 possible full houses, 624 four of a kinds, and 40 possible straight flushes, which should be the only hands we really care about (ignoring the valid question about betting raised by Chuck Rampart). So, the question is, if you hold one of the 3,744 full houses, is there one of those which is better at removing the potential 644 better hands from the deck?

Let's deal with the 624 four of a kinds first (it's easier). Any full house will eliminate 2/13 of those hands right off the bat because those two ranks are eliminated, so we're down to 528 four of a kinds. 528 = 11 * 48, which is indicative of the 11 remaining ranks which could still form the four of a kind and the 48 possible cards which could be the kicker in that hand. Our full house elminates 5 kickers from each of those hands, so we really have only 11 * 43 = 473 four of a kinds left. I don't see anything we can do in our selection of the "optimum" full house to affect this result, so the net effect on four of a kinds is somewhat inconsequential. However, I present it to give completeness to the solution.

What we really care about are the 40 straight flushes which remain. Can we eliminate more of them by our selection of an "optimum" full house? The answer is, undoubtedly, "yes".

Doyle's answer has been revealed as Aces full of Kings. How many of the 40 SF's does this combination eliminate? It depends somewhat on the suits, but let's assume that Doyle was at least smart enough to specify that one of the Kings was of the suit that is not represented in the Aces. By choosing Aces, you elminate three Ace-high and 3 Ace-low SFs, for a total of 6 SFs. The King which shares a suit with one of the Aces removes one other King-high SF, whereas the other King removes an Ace-high and a King-high SF, so a total of 3 more SFs are removed. So Doyle's answer removes a total of 9 SFs from contention. Aces full of Kings can therefore be beaten by 473 + 31 = 504 hands (0.01939% chance of losing).

Someone has suggested 10s full of 4s as a better combination, because it would remove a lot of SFs - better for that would be 10s full of 5s. Again, make the assumption that one of the 5s is a different suit than the three 10s (although it doesn't really matter because the 10s and 5s don't appear in the same SF the way Aces and Kings do). So, let's consider how this plays out (no pun intended). Every SF contains either a 10 or a 5, so the suit which is shared by our 10s and 5s is completely removed from contention, dropping 10 of the possible SFs. The remaining two 10s and one 5 all eliminate 5 more SFs each (the 10 removes 10-high thru A-high; the 5 removes 5-high thru 9-high), for an additional 15 SFs removed. Now, these gains are going to be more than offset by the fact that any higher full house will beat 10s over 5s - approximately 1100 hands. So the choice of 10s full of 5s removes a total of 25 SFs but adds 1100 FHs, and can therefore be beaten by 473 + 15 + 1100 = 1588 hands (0.06110% chance of losing) - much worse than Aces full of Kings.

I think it's now obvious that we have to have an Ace-high full house to remove the possibility of being beaten by another full house, which far outweighs the meager reductions in total numbers of hands which comes from removing a high percentage of SFs. I think the best combination then is Aces full of 9s (although it could be Aces full of 8s, 7s, or 6s). As noted before with Doyle's hand, the Aces remove 6 SFs from contention (Ace-high & Ace-low). The 9s remove 5 more SFs each (9-high thru K-high), for a total of 16 SFs removed. Aces full of 9s can then be beaten by only 431 + 24 = 455 hands (0.01751% chanse of losing), and it can be shown that the same is true for Aces full of 8s, 7s, or 6s.

I think that's the best possible solution, although I'm not sure that in practice I'd really be concerned about less than a 100th of a percent difference - unless I was in Morty's that night and all my chips depended on it!

HH has it. Alex, watch out .. !

Nice job.

[Guest]

Ha, yeah I was way off. I was looking at straights and not straight-flushes; in which case in does work out, but apparently straights do not beat a full house... Nice job hokie.

[bonanova]

Any straight flush would beat a full house, so I don't think we specifically need to be aiming to reduce the number of royal flushes available - just reducing the number of straight flushes. In that case, the 10 of hearts is slightly redundant, as we have already wiped out the possibility of a 10 J Q K A Hearts flush by claiming the ace.

So... I think aces and nines would be marginally better.

Thinking further... we want to reduce the total number of hands that could beat us... would tens over fours be an even better combo? that adds in a few full houses that can beat us, but does it remove more straight flushes?

And should we be taking the relative probability of FHs and SFs into account?

More work to be done, methinks.

Tip of the hat to armcie for first suggestion of Aces and nines.

[Guest]

Aces full of 5s.

The 5s knock out as many SF as possible (5 each); being lower doesn't hurt the hand at all, but does tend to increase the the quality of any opponent's non-winning hands. This will help the player maximize their return on their full house.

Edited November 12, 2008 by xucam

[HoustonHokie]

Aces full of 5s.

The 5s knock out as many SF as possible (5 each); being lower doesn't hurt the hand at all, but does tend to increase the the quality of any opponent's non-winning hands. This will help the player maximize their return on their full house.

To be really picky about it, maybe Aces full of 6s would be better. Because the Ace appears in a 5-high SF, the 5s are only going to knock out 4 of the SFs in suits where you hold the Ace as well (either 1 or 2 of your Aces). Aces full of 6s knocks out 5 SFs, no matter what the suits.

All we can really do is postulate around this, though. The betting question depends on multiple factors: the number of players at the table, the hands they hold, and, most importantly, their BAC . I guess the ideal betting situation would be to knock out as many hands as possible where "rational" people wouldn't bet much (say pair of deuces) while leaving a lot of hands that still lose to your full house, but would encourage betting by the same "rational" people (say three Kings). The question is finding the point at which the threat of losing to a stronger hand is balanced by the probability that you will win a certain pot - and that's a lot more math than I'm willing to do right now .