Optimal strategy in two-player Guts
May 22nd, 2015 at 5:35:26 PM
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Let's assuming the following rules:
They keep playing until some wins an uncontested pot (the other guy is "out").
Let's assume that your opponent's strategy is to go "in" with a number of x or greater. Given that value of x, what should be your optimal strategy? Please answer for:
1. x<0.5
2. x=0.5
3. x>0.5
- Both players ante $1 into the pot.
- Each player is given a random number uniformly distributed from 0 to 1.
- Each player must simultaneously declare "in" or "out."
- If only one player declares "in," then he automatically wins the pot.
- If both players declare "in," then the player with the higher number wins the pot, the loser must match it, and they play again.
- If both players declare "out," then they play again.
They keep playing until some wins an uncontested pot (the other guy is "out").
Let's assume that your opponent's strategy is to go "in" with a number of x or greater. Given that value of x, what should be your optimal strategy? Please answer for:
1. x<0.5
2. x=0.5
3. x>0.5
Please put answers in spoiler tags.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 27th, 2015 at 7:20:43 PM
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In cases 5 and 6 when they play again, they don't ante again, do they? So the pot is always $2, right?
May 28th, 2015 at 3:48:45 AM
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you can only get a 0 or 1 as a number? the player sees his number before he declares in or out?Quote: WizardLet's assuming the following rules:
- Both players ante $1 into the pot.
- Each player is given a random number uniformly distributed from 0 to 1.
- Each player must simultaneously declare "in" or "out."
- If only one player declares "in," then he automatically wins the pot.
- If both players declare "in," then the player with the higher number wins the pot, the loser must match it, and they play again.
- If both players declare "out," then they play again.
They keep playing until some wins an uncontested pot (the other guy is "out").
Let's assume that your opponent's strategy is to go "in" with a number of x or greater. Given that value of x, what should be your optimal strategy? Please answer for:
1. x<0.5
2. x=0.5
3. x>0.5Please put answers in spoiler tags.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
May 28th, 2015 at 4:10:03 AM
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Quote: AxelWolf]you can only get a 0 or 1 as a number? the player sees his number before he declares in or out?
Including decimals, I believe.
Or rather, think of it as between 0 and 100. And he bets if < 50, = 50, or > 50 for 1,2,3 respectively.
Although, I am a bit confused. Does this mean ( < 0.5 ) that he ALWAYS goes in? Or he only goes "in" if he has less than 0.5?
May 28th, 2015 at 6:33:53 AM
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Quote: RSIncluding decimals, I believe.
Or rather, think of it as between 0 and 100. And he bets if < 50, = 50, or > 50 for 1,2,3 respectively.
Although, I am a bit confused. Does this mean ( < 0.5 ) that he ALWAYS goes in? Or he only goes "in" if he has less than 0.5?
Pretty sure it means the player has a fixed strategy where he goes in if he has some number greater then x and that number x is less then .5 . So for instance x could be .25 and he would go in if he got anything over that like .31 and would go out if he had anything lower then that like .21 .
May 28th, 2015 at 6:44:14 AM
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Quote: WizardLet's assuming the following rules:
- Both players ante $1 into the pot.
- Each player is given a random number uniformly distributed from 0 to 1.
- Each player must simultaneously declare "in" or "out."
- If only one player declares "in," then he automatically wins the pot.
- If both players declare "in," then the player with the higher number wins the pot, the loser must match it, and they play again.
- If both players declare "out," then they play again.
They keep playing until some wins an uncontested pot (the other guy is "out").
Let's assume that your opponent's strategy is to go "in" with a number of x or greater. Given that value of x, what should be your optimal strategy? Please answer for:
1. x<0.5
2. x=0.5
3. x>0.5Please put answers in spoiler tags.
My answer:
Whenever playing this with friends it's generally best to always declare in, because like in poker, you give yourself 2 ways to win. You can either 'bluff' your opponent and he'll say out, or you could end up randomly drawing a higher ranked card and winning anyways. The cards are kind of erroneous. If you both have the same chance of drawing a number from 0-1, then the difference maker would be who chickens out more in the long run. Thus, by always saying in you'll win the 50% of the time you're supposed to, and your opponent will lose when they chicken out in borderline decisions and say out.
Playing it correctly means you've already won.
May 28th, 2015 at 6:53:21 AM
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If your opponent's strategy is go to in with x or greater, then:
If x<0.5, then go in with 0.25+(x/2) or greater.
If x=0.5, then any strategy will be equally good.
if x<0.5, then go in x>0 (always go in).
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 28th, 2015 at 7:10:21 AM
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Quote: WizardLet's assuming the following rules:
4. If only one player declares "in," then he automatically wins the pot.
5. If both players declare "in," then the player with the higher number wins the pot, the loser must match it, and they play again.
Questions:
1. Is the pot always $2?
2. If both players are in, does the loser always put $2 into the pot and the winner nothing, or does the winner put in $1 and the loser put in what was in the pot?
For example, does it work like this:
1. A and B each put in $1 to start
2. Both are in; A wins the $2
3. B puts in $2, and A puts in $1
4. Both are in; B wins the $3
5. A puts in $3, and B puts in $1
6. A is in, but B is out; A wins the $4, and the game ends there
May 28th, 2015 at 7:37:37 AM
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Quote: ThatDonGuyQuestions:
1. Is the pot always $2?
2. If both players are in, does the loser always put $2 into the pot and the winner nothing, or does the winner put in $1 and the loser put in what was in the pot?
For example, does it work like this:
1. A and B each put in $1 to start
2. Both are in; A wins the $2
3. B puts in $2, and A puts in $1
4. Both are in; B wins the $3
5. A puts in $3, and B puts in $1
6. A is in, but B is out; A wins the $4, and the game ends there
My understanding of guts is that:
1) everyone puts an ante in on every round, and
2) whoever is 'in' and loses must match the pot as his 'penalty' for losing, and still must ante if he wants to be in the next round.
So...
1. A and B each put in $1 to start
2. Both are in; A wins the $2
3. B puts in $2, and A and B put in $1 (pot is now $4)
4. Both are in; B wins the $4
5. A puts in $4, and A and B put in $1 (pot is now $6)
6. A is in, but B is out; A wins the $6, and the game ends there
I played a little 3-card guts in high school for nickels. Occasionally, if enough people were playing, we would see a $20 pot (which was a lot for us!)
Wiz, you may want to check the inequality signs in your answer.Quote: Wizard
"Dealer has 'rock'... Pay 'paper!'"
May 28th, 2015 at 7:46:52 AM
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Quote: Joeman...2) whoever is 'in' and loses must match the pot as his 'penalty' for losing, and still must ante if he wants to be in the next round...
These are what we call "burn" games. We use them in a lot of home poker games. We usually cap the max burn at $3 or $4. Also in these poker games, you have to "beat the deck" where after all players see who won or lost, a hand is dealt off the top of the deck in which the winning player must beat also. That certain game keeps being played until only one person stays in and also beats the deck.
DUHHIIIIIIIII HEARD THAT!
May 28th, 2015 at 8:08:03 AM
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Oh. That would have been a good idea; we never thought of that! At the time, I would have had to mow 2 neighbors' lawns to come up with the funds to match a $20 pot!Quote: IbeatyouracesWe usually cap the max burn at $3 or $4.
We never did that, either. Out of curiosity, if you won, but couldn't beat the deck, would you get any of the pot? Would you have to match the pot?Quote:Also in these poker games, you have to "beat the deck"
"Dealer has 'rock'... Pay 'paper!'"
May 28th, 2015 at 8:19:50 AM
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Quote: Joeman....We never did that, either. Out of curiosity, if you won, but couldn't beat the deck, would you get any of the pot? Would you have to match the pot?
If you couldn't beat the deck, the pot stayed and another hand was dealt. You had to match the pot, but up to the max burn. My favorite burn game is Bloody 7's.
DUHHIIIIIIIII HEARD THAT!
May 29th, 2015 at 10:33:22 AM
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Wiz, I am still not entirely sure on the rules. Can you confirm this win / loss table:
(Hero) (Villain) (Hero outcome)
in out +1
out in -1
out out 0
in in +3 if h>v, - 3 if h<v
Thanks!
Sorry about the terrible format, looked better when I typed it...
(Hero) (Villain) (Hero outcome)
in out +1
out in -1
out out 0
in in +3 if h>v, - 3 if h<v
Thanks!
Sorry about the terrible format, looked better when I typed it...
May 29th, 2015 at 2:40:06 PM
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The Wizard has not explained exactly who puts how much into the pot and when, so it is hard to solve the problem in general.
However, I believe he has made a mistake in his analysis for the case where my opponent goes in whenever his number is 0.5 or greater).
The Wizard states that
"If your opponent's strategy is go to [sic] in with x or greater, then:...
If x=0.5, then any strategy will be equally good."
Suppose my strategy is to never go in but to always stay out. Then I obviously cannot win anything and must lose whatever I put into the pot, so my EV is negative. If, however I choose the same strategy as my opponent, then by symmetry my EV is zero. So it is incorrect that "any strategy will be equally good."
However, I believe he has made a mistake in his analysis for the case where my opponent goes in whenever his number is 0.5 or greater).
The Wizard states that
"If your opponent's strategy is go to [sic] in with x or greater, then:...
If x=0.5, then any strategy will be equally good."
Suppose my strategy is to never go in but to always stay out. Then I obviously cannot win anything and must lose whatever I put into the pot, so my EV is negative. If, however I choose the same strategy as my opponent, then by symmetry my EV is zero. So it is incorrect that "any strategy will be equally good."
