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Waddauno
5th August 2006, 08:02 AM
There's a dice game played in the bars in my area called Horse. Two or more players wager a dollar (could be more, but for our purposes let's use a dollar), and roll 5 dice. The goal is to get the most dice of a kind while getting the highest kind. For example, three threes is not as good as four threes, and not as good as three fours. The best hand is five sixes. "Nonsuited" dice are ignored. If two players roll the same best hand the pot is pushed and everyone puts another buck in the kitty and rolls anew.

Ones are wild. To further complicate the question I have not yet posed, each player actually gets three rolls to compile the best hand. As an example, it's your roll. You roll a six, two ones, a two and a four. The ones are wild so you actually have 3 sixes. You may stay on three sixes, or pick up the two and the four and roll those dice again in order to strengthen your position. But if you take another roll everyone else automatically gets two rolls to try to beat your hand, though they are also limited to two rolls and cannot opt for the third unless you did.

I hope that's clear as to the rules, though I'm not sure you need all that. Happy to clarify if necessary.

So here's my odds question. Say you have three people playing a game of Horse. Two of the three are married, and are basically playing from the same bank. Thier winnings will be shared and so will thier losses. Is the third person at any statistical disadvantage in that circumstance?

The couple is putting up two dollars in order to win one. The solo roller is putting up one to win two. But he only gets one chance to establish the best hand, while the couple gets two. And the couple are playing against one another at a null outcome and thus only really playing against the solo guy, whereas the solo is playing against two people.

This topic is based on real life and has been the subject of much debate amongst folks who are in no way mathmagicians, so I'm hoping for some of your thoughts.

Just thinking
5th August 2006, 11:10 AM
It would seem to me the odds of the third player (non-married, player C) do not in any way change as to whether or not the others are playing as a 'team'. If players A and B decide to share their winnings or not affects player C in no way. He is always playing against two opponents in your scenario -- so the odds of his outcome beating their's stays the same. And the game you are describing sounds very much like Yahtzee (http://grail.sourceforge.net/demo/yahtzee/rules.html), although somewhat simpler.

Waddauno
5th August 2006, 11:18 AM
From the other side, the couple has more chances to win, but win less money if they do, so I think that balances it out.

Rasmus
5th August 2006, 11:20 AM
There's a dice game played in the bars in my area called Horse. Two or more players wager a dollar (could be more, but for our purposes let's use a dollar), and roll 5 dice. The goal is to get the most dice of a kind while getting the highest kind. For example, three threes is not as good as four threes, and not as good as three fours.

Are two fives better than one six? How do these kinds of results rank?

But if you take another roll everyone else automatically gets two rolls to try to beat your hand, though they are also limited to two rolls and cannot opt for the third unless you did.


I am not sure I understand this. Does this mean the first player of a round gets to decide how many rolls everyone can have maximum? Does he decide that before or after he sees whateveryone else has in the first round? (I assume player one goes first and completes his round, right?)

So here's my odds question. Say you have three people playing a game of Horse. Two of the three are married, and are basically playing from the same bank. Thier winnings will be shared and so will thier losses. Is the third person at any statistical disadvantage in that circumstance?

Did anyone hear a cricket?

I will think about this, though ...

Just thinking
5th August 2006, 11:28 AM
From the other side, the couple has more chances to win, but win less money if they do, so I think that balances it out.

I believe this is known as Risk Assessment (http://en.wikipedia.org/wiki/Risk_assessment). And it is the same for all 3 players.

Risk assessment is measuring two quantities of the risk R, the magnitude of the potential loss L, and the probability p that the loss will occur.

HappyCat
5th August 2006, 11:42 AM
If the player with the best hand is randomly determined (the game favors each player equally), then the unmarried player has a 1/3rd chance of winning 2 dollars, 2/3rds chance to lose 1 dollar. The married pair have a combined 2/3rds chance to win 1 dollar, 1/3rd chance to lose 2 dollars. The expected outcome of this for the unmarried person is (1/3 * 2) - (2/3 * 1) = 2/3 - 2/3 = 0. For the married couple, the expected out come is (2/3 * 1) - (1/3 * 2) = 2/3 - 2/3 = 0. So for both the unmarried person and the married couple, the expected outcome of any hand is to break even. Therefore, there is no advantage nor disadvantage to playing as a couple.

The odds will not change if the game favors one player more than the others. For example, the game gives the 3rd person to roll a 7/10 chance of winning, the second a 2/10 chance of winning, and the first a 1/10 chance of winning. If their seats are randomly determined, the couple would then have a 1/3 chance of having 3/10 chance of winning, a 1/3 chance of having 8/10 chance of winning, and a 1/3 chance of having 9/10 chance of winning. The couple then has an expected probability of winning of (1/3 * 9/10) + (1/3 * 8/10) + (1/3 * 3/10) = 3/10 + (2+2/3)/10 + 1/10 = (6 + 2/3)/10 = 2/3, which is the same probability they had in an equally weighted game.

If the seating arrangement is not random, then the game may or may not be in favor of the couple, depending on the rules used to determine seating

Waddauno
5th August 2006, 11:52 AM
Rasmus: Yes, 2 fives are better than one six. Four 2's are better than 3 sixes, and so on.

And yes, the first player decides how many rolls everyone else has because he decides how many he will take. For example, if on your first roll you had 4 sixes, you would probably stand and pass the cup. Because you only took one roll to get your 4 sixes, everyone else must try to beat 4 sixes with a single roll. If you rolled a straight, you'd scoop up all 5 dice and try again, allowing everyone else two rolls as well. If you decide to take all three tries, called "all week", everyone else has all week to try and beat whatever you came up with in three rolls. If the next player beats you using only one roll, he wins and gets the pot and a new hand ensues. If doesn't beat you on that first roll, he's got two more shots to do it.

I don't mean to call the crickets. Is this a dumb question? What if we increase the number of players to 5? Are they better positioned than any one other player? Then they are paying $2 to win $5 and getting two chances to do so, while everyone else is paying $1 to win five and getting one chance to do so. Given that they are playing as essentially one person, does their "one person" have any advantage or disadvantage compared to the other players?

Waddauno
5th August 2006, 12:03 PM
HappyCat, thanks, that was precisely the type of answer I was looking for. If I apply that to the 5 person game, the single player has a 1/5 chance to win $4, and a 4/5 chance to lose $1. The couple has a 2/5 chance to win $3, and a 3/5 chance to lose $2. Applying your formula, I get the same result of zip difference.

Thanks for laying it out for me.

HappyCat
5th August 2006, 12:08 PM
If you are playing with n players, the probability of the couple winning n-2 dollars is 2/n. The probability that they will lose 2 dollars is (n-2)/n. So their expected outcome is (2/n * (n-2)) - (2 * (n-2)/n) = (2*(n-2))/n - (2*(n-2))/n = 0. A single person's odds of winning n-1 dollars is 1/n, where their odds of losing 1 dollar is (n-1)/n. His expected outcome is (n-1 * 1/n) - (1 * (n-1)/n) = (n-1)/n - (n-1)/n = 0. Therefore, no matter how many people sit at the table, the couples odds of coming out ahead will be no different than a lone person's odds of coming out ahead.

Edit: No problem :)

GreedyAlgorithm
5th August 2006, 02:07 PM
Because of the structure of this game, the couple's expected winnings are at best 0. They can actually manage to do worse, following the strategy where each player tries to make the best roll he can individually. Why? Because the couples can counterfeit some of their wins. Suppose you roll 13345, and reroll the 45 to get 12333 for 4 3s. Now the first of the couple rolls twice and gets 4 4s. If the second of the couple tries to win, and accidentally ties, the whole pot gets pushed to another round and they feel very silly. So don't worry, you're doing at least as well and possibly better than they are.

fribble
5th August 2006, 02:47 PM
Because of the structure of this game, the couple's expected winnings are at best 0. They can actually manage to do worse, following the strategy where each player tries to make the best roll he can individually. Why? Because the couples can counterfeit some of their wins. Suppose you roll 13345, and reroll the 45 to get 12333 for 4 3s. Now the first of the couple rolls twice and gets 4 4s. If the second of the couple tries to win, and accidentally ties, the whole pot gets pushed to another round and they feel very silly. So don't worry, you're doing at least as well and possibly better than they are.

But surely this applies to the non-married competitors as well, thus isn't a disadvantage?

Rasmus
5th August 2006, 03:12 PM
But surely this applies to the non-married competitors as well, thus isn't a disadvantage?

No, if the first player of the couple wins, then they would already have the pot - between them. By scoring even, however, they would forfeit the win they already had.

So I guess they do have an advantage: Every few rounds, Partner B will play last in a game where Partner A is currently the winning player.

Player B can then secure the pot for the couple by losing on purpose. (In attempting to win, they might accidentally break even and lose the pot into the next round.)

fribble
5th August 2006, 04:00 PM
No, if the first player of the couple wins, then they would already have the pot - between them. By scoring even, however, they would forfeit the win they already had.

But, given the assumption that A and B are blindly out to maximise their own scores, the draw would happen regardless of whether the couple are married or not. It is equally likely between A and B as it is between B and C, or A and C.

So I suspect that with appropriate use of sacrifices, as you describe, a couple may be able to push the expected value slightly above 0.


edit: typo

mumchup
5th August 2006, 05:01 PM
Your forgetting something. As the female half of a couple, I can assert that the couple's wagers/winnings are split thusly:
Wagering-
"Ooh! I want to play too! Steve do you have another Dollar?
Winnings (2 outcomes)
"Look Steve you won! Great, because I need another drink!"
or
"Look I won! (puts money in pocket) Can you get me another drink Steve?"

rockoon
5th August 2006, 11:25 PM
Seems to me that there is a positional advantage in this game, similar to poker.

Of course in this game it looks like the first player has the advantage, rather than the last as it is in poker.

Just thinking
6th August 2006, 09:20 AM
Might not the final player in this game (Horse) have an edge? After all, if up to his turn the best hand was four 3's, he might break up an otherwise good hand in order to out do his former opponent, since he would surely lose with say only four 2's.

Plus, don't some forms of Poker expose no cards (except possibly an ace) until the very end, in which case no advantage is given to position?

Rasmus
6th August 2006, 09:38 AM
Might not the final player in this game (Horse) have an edge? After all, if up to his turn the best hand was four 3's, he might break up an otherwise good hand in order to out do his former opponent, since he would surely lose with say only four 2's.

Yes, I think so.

Plus, don't some forms of Poker expose no cards (except possibly an ace) until the very end, in which case no advantage is given to position?

As far as I understand it, you still have more information if you play later, because you know what those before you did - even if you might not know their cards.

Antiquehunter
6th August 2006, 10:32 AM
Just speculating - I haven't figured out the math. But...

I don't think the last position player has an advantage in this game. Sure - he knows what he has to beat, but there are very few varieties of hands in this game. Lets say I throw 4 5's in two rolls. Your first roll is 6,6,3,3,3. You need to either turf the 6,6 and hope to throw 3,3; 1,1; or 1,3 / 3,1. (You have to make 5 3's) Conversely you could turf the 3,3,3 and try for any combination of the three dies that includes at least a 6 and a 1 (or 2 sixes or 2 ones) in order to get at least 4 6's - you're better off chasing sixes.

Note - I haven't done the math - but...
If I roll first, and I know that in a 3 person game on a single roll that the median winning hand is say... 3 threes - then I can have a positional advantage by forcing the other players to make only one roll when I have the best of it - assuming my initial roll is the median or better. (In hold'em, if I can always get all my money in the middle as a 52% - 48% favorite, I'll be a long term winner) Likewise there is a stop point at roll 2 (any 4 of a kind for example) where you're better off to stop trying to improve. You'd need to figure out this chart for x number of players in the game - the greater the number of players, the better hand you need to stand a better than average chance of winning.

Since full houses and straights don't count in this version, only matching dies with aces wild, then I can't think of any rolls a late or last position player would choose to 'break up' to improve. If they have to beat 4 5's in 2 rolls and roll 1 was 6,6,3,3,3 its a no-brainer - you have to roll the 3 threes again and hope for 6's and 1's.

There is no concealed information from any player in this game - so no 'reading' skills, or determination based on the play of a hand given more information.

A more interesting version would be to add a 'raise' element - after your first roll you can choose to raise the bet IF you throw a second time.

I 'ante' $1 and on Roll 1 I throw 6,6,1,2,4 for 3 sixes. I could stand pat, or I could 'raise' to $2 and throw again in an attempt to improve further. Then player B would have to 'ante' $1 to throw the dice the first time and either 'call' my raise to $2 to throw a second time or 'fold' his $1. Another interesting twist would be to add a re-raise to $3. At least these twists would incorporate more understanding of the odds of improving your hand, and would also add a little more 'gamble' incentive rather than a straight up game of chance.

But in either variety (as described or with my modification) I believe there is a 'basic strategy' that would allow the first roller to have a small advantage in my opinion.

I'm not a statistician - this is just speculation.