RealiseNothing
what is that?It is Cowpea
Re: 2015 permutation X2 marathon
Assume they wanna make as much money as possible for themself.
HSC 2015 MX2 Permutations & Combinations Marathon (archive) (2 Viewers)
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RealiseNothing
what is that?It is Cowpea
Re: 2015 permutation X2 marathon
Sorry should have mentioned this. They can't choose the same number.
Re: 2015 permutation X2 marathon
Also, is the following true?: B knows the number A chose, and A knows that B will know, and B knows A will know B will know, and ... .
What happens if the drawn number is directly between the numbers picked by A and B (though this may be irrelevant...but telling us that it's irrelevant may provide a clue that you don't want to provide lol)?
Also, is the following true?: B knows the number A chose, and A knows that B will know, and B knows A will know B will know, and ... .
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RealiseNothing
what is that?It is Cowpea
Re: 2015 permutation X2 marathon
Both players are together/next to each other whilst playing the game, so B hears A choosing a number, then A hears B choosing their number. So yes to your question pretty much.
Yep it's irrelevant (the optimal numbers for them to choose won't have a number that is the same distance from each). But for the sake of the question let's say the split the money 50/50.
Both players are together/next to each other whilst playing the game, so B hears A choosing a number, then A hears B choosing their number. So yes to your question pretty much.
Paradoxica
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Re: 2015 permutation X2 marathon
I smell a prisoner's dilemma
I smell a prisoner's dilemma
Re: 2015 permutation X2 marathon
The only prisoner's dilemma I'm aware of is whether or not he should pick up the soap.
Paradoxica
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glittergal96
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Re: 2015 permutation X2 marathon
In my interpretation, A chooses first, and B chooses second.
First observe that B might as well choose a number adjacent to A's. This is because if there is a gap between A and B (say B = A + 2), then B loses value on the numbers strictly between A and B, whilst still losing on all numbers on the other side of A.
For small A then, B's optimal response is to choose A+1, securing wins on all X > A. Similarly, the optimal response for large A is A-1.
The task is then to find the k=A that maximises the quantity
 =\min(2\sum_{j\leq k}j,30\cdot 31-2\sum_{k<j} j)\\ \\ =\min(k(k+1),930-k(k-1)).)
(f(k) is twice the expectated profit for A if he chooses the number k and B responds optimally.)
The first expression (in the pair that we are taking the min of) increases and the second one decreases and they intersect between 21 and 22. This means that to find the best k for A to choose, we simply need to compare f(21) and f(22).
Directly calculating gives f(22)>f(21) which means A's optimal choice is 22 and B's optimal response is 21.
Out of interest, A is slightly profiting in this game.
Note. The sums above aren't quite correct for k=1,30, but these are obviously far from optimal, and you can compute the EV of these strategies separately if you like.
In my interpretation, A chooses first, and B chooses second.
First observe that B might as well choose a number adjacent to A's. This is because if there is a gap between A and B (say B = A + 2), then B loses value on the numbers strictly between A and B, whilst still losing on all numbers on the other side of A.
For small A then, B's optimal response is to choose A+1, securing wins on all X > A. Similarly, the optimal response for large A is A-1.
The task is then to find the k=A that maximises the quantity
(f(k) is twice the expectated profit for A if he chooses the number k and B responds optimally.)
The first expression (in the pair that we are taking the min of) increases and the second one decreases and they intersect between 21 and 22. This means that to find the best k for A to choose, we simply need to compare f(21) and f(22).
Directly calculating gives f(22)>f(21) which means A's optimal choice is 22 and B's optimal response is 21.
Out of interest, A is slightly profiting in this game.
Note. The sums above aren't quite correct for k=1,30, but these are obviously far from optimal, and you can compute the EV of these strategies separately if you like.
Drsoccerball
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glittergal96
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Re: 2015 permutation X2 marathon
A two player variant of poker is played as follows.
There are three cards (A,K,Q) in the deck and the two players are each dealt one of these cards randomly. (They can see their card but not their opponents.)
Both players are forced to put $A into the pot (an ante). (You may assume A < 1.)
Player 1 is forced to check in this version of poker.
Player 2 then has the option of either checking or betting $1.
If Player 2 checked, then the cards are turned over and the player with the higher card wins the pot of $2A.
If Player 2 made a bet, then player $1 has the option of either calling the bet (putting in a matching $1 himself) or folding, which forfeits the hand.
A mixed strategy for this game consists of a set of probabilities determining how a player should act in certain situation. Eg, one strategy for player 2 might be to bet 90% of the time and call 10% of the time with an ace, bet 80% of the time and call 20% of the time with a king, and call 100% of the time with a queen.
An optimal strategy for a player is a mixed strategy that offers the player the best guaranteed outcome (on average). Of course if the opponent plays stupidly you will do much better than this guaranteed outcome, and if you only play a few hands you might have bad luck and do much worse. The guaranteed outcome is just an outcome that you will not do worse than in the long term.
a) Find the optimal strategies for both players in this game.
b) Which player has the advantage?
c) How much money does the player with the advantage win on average?
(Answers should be in terms of A.)
Note. This vastly simplified model actually gives some insight into the basics of actual poker strategy.
A two player variant of poker is played as follows.
There are three cards (A,K,Q) in the deck and the two players are each dealt one of these cards randomly. (They can see their card but not their opponents.)
Both players are forced to put $A into the pot (an ante). (You may assume A < 1.)
Player 1 is forced to check in this version of poker.
Player 2 then has the option of either checking or betting $1.
If Player 2 checked, then the cards are turned over and the player with the higher card wins the pot of $2A.
If Player 2 made a bet, then player $1 has the option of either calling the bet (putting in a matching $1 himself) or folding, which forfeits the hand.
A mixed strategy for this game consists of a set of probabilities determining how a player should act in certain situation. Eg, one strategy for player 2 might be to bet 90% of the time and call 10% of the time with an ace, bet 80% of the time and call 20% of the time with a king, and call 100% of the time with a queen.
An optimal strategy for a player is a mixed strategy that offers the player the best guaranteed outcome (on average). Of course if the opponent plays stupidly you will do much better than this guaranteed outcome, and if you only play a few hands you might have bad luck and do much worse. The guaranteed outcome is just an outcome that you will not do worse than in the long term.
a) Find the optimal strategies for both players in this game.
b) Which player has the advantage?
c) How much money does the player with the advantage win on average?
(Answers should be in terms of A.)
Note. This vastly simplified model actually gives some insight into the basics of actual poker strategy.
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