What is the reasonable betting strategy for this game?
I was bored the other day and thought up a kind of complex card game. Now I can't work out the reasonable betting strategy. I hope someone else will find this amusing to think about and maybe help me out.
Consider a two-player card game using a standard 52 card deck between a dealer and a player with a very large, but finite amount of money.
First Variation
First the dealer shuffles the deck and deals herself a hand of thirteen cards, face down, which she does not look at. The dealer’s hand is a “success” if and only if it contains the thirteen spades in order from Ace to Two. Otherwise, the dealer’s hand is a “failure”.
After the dealer deals herself a hand, the first round of the game begins with the dealer proposing the player a wager for some stakes at certain odds. If the player accepts those stakes, then the dealer plays a card from the deck, face up. If the card is a spade, then obviously the dealer’s hand isn’t successful, and the player keeps the stake. However, if the card drawn is not a spade, then the player must either choose to “stop” or “continue” play. If the dealer chooses “stop”, the dealer exposes her hand. If the dealer’s hand is a success, the player loses the wager, otherwise, the player wins the stakes. Either way, the game ends. On the other hand, if the player chooses to continue, the player withdraws her wager, and the game proceeds to the next round.
At the start of each subsequent round, the dealer offers the player a wager at certain odds (not necessarily the same odds as in previous rounds). If the player refuses the wager, the game is over. If the player accepts the wager, then the dealer plays another card from the deck and offers the player the choice to stop or continue, as above.
After 39 rounds, there are no more cards to play, so at the end of this round, so if the player chooses to stop the game is over, and the player withdraws her wager. But if the player chooses to continue, the dealer exposes her hand and declares whether the dealer or the player has won.
What are the correct odds for the dealer to offer the player at every round of the game to make the game “fair”, i.e. to make the expectation of each bet zero? Assuming the dealer is offering fair odds, what is the player’s best strategy? (It seems like the player should always choose ‘stop’ on the first round.)
Second Variation
The second variation is like the first, except that if the player chooses to stop, at any round, then the player forfeits half of her wager for that round.
How does this change the odds the dealer should offer to make the game fair? What is the optimal strategy for the player now? (It seems like the player should always choose ‘continue’ now.)
Third Variation
The third variation is like the second variation, except that the player cannot choose whether to stop or continue. Instead, she must flip a fair coin. If it lands “heads,” play continues; if it lands “tails,” play stops.
What odds should the dealer offer at every round now to make it worth the player’s time to play?
probability
asked Nov 28 at 17:36
shane
1184
add a comment |
I was bored the other day and thought up a kind of complex card game. Now I can't work out the reasonable betting strategy. I hope someone else will find this amusing to think about and maybe help me out.
Consider a two-player card game using a standard 52 card deck between a dealer and a player with a very large, but finite amount of money.
First Variation
First the dealer shuffles the deck and deals herself a hand of thirteen cards, face down, which she does not look at. The dealer’s hand is a “success” if and only if it contains the thirteen spades in order from Ace to Two. Otherwise, the dealer’s hand is a “failure”.
After the dealer deals herself a hand, the first round of the game begins with the dealer proposing the player a wager for some stakes at certain odds. If the player accepts those stakes, then the dealer plays a card from the deck, face up. If the card is a spade, then obviously the dealer’s hand isn’t successful, and the player keeps the stake. However, if the card drawn is not a spade, then the player must either choose to “stop” or “continue” play. If the dealer chooses “stop”, the dealer exposes her hand. If the dealer’s hand is a success, the player loses the wager, otherwise, the player wins the stakes. Either way, the game ends. On the other hand, if the player chooses to continue, the player withdraws her wager, and the game proceeds to the next round.
At the start of each subsequent round, the dealer offers the player a wager at certain odds (not necessarily the same odds as in previous rounds). If the player refuses the wager, the game is over. If the player accepts the wager, then the dealer plays another card from the deck and offers the player the choice to stop or continue, as above.
After 39 rounds, there are no more cards to play, so at the end of this round, so if the player chooses to stop the game is over, and the player withdraws her wager. But if the player chooses to continue, the dealer exposes her hand and declares whether the dealer or the player has won.
What are the correct odds for the dealer to offer the player at every round of the game to make the game “fair”, i.e. to make the expectation of each bet zero? Assuming the dealer is offering fair odds, what is the player’s best strategy? (It seems like the player should always choose ‘stop’ on the first round.)
Second Variation
The second variation is like the first, except that if the player chooses to stop, at any round, then the player forfeits half of her wager for that round.
How does this change the odds the dealer should offer to make the game fair? What is the optimal strategy for the player now? (It seems like the player should always choose ‘continue’ now.)
Third Variation
The third variation is like the second variation, except that the player cannot choose whether to stop or continue. Instead, she must flip a fair coin. If it lands “heads,” play continues; if it lands “tails,” play stops.
What odds should the dealer offer at every round now to make it worth the player’s time to play?
probability
asked Nov 28 at 17:36
shane
1184
Why make everything hinge on such an implausible success? The probability of being dealt all the spades is $ 1.57.. times 10^{-12}$. If you play one game a second for ten thousand years, the odds are you still won't see a success.
– lulu
Nov 28 at 17:47
Note: I was only looking at the unordered hand consisting of all the spades. If you want them in order, then "success" becomes significantly less likely.
– lulu
Nov 28 at 17:50
Probability of success should be 1/[52!/(52-13)!] which is approximately 10^-22. But the super low probability shouldn't matter, since I can still give you odds that make the expectation zero, even if the event is really unlikely. My real goal here was to to give us lots of different points to make a decision at (every time a card is flipped), not to deal with really big or small numbers per se.
– shane
Nov 28 at 18:56
For the first variant, I think the odds the dealer should give the player for the first round is 52!/(52-13)! to 1 and the odds the dealer should give on last round should be 13! to 1, because even if you know all the cards are spades, there are still 13! different orders those cards could be in.
– shane
Nov 28 at 19:03
So my conjecture would be that for the first variant, the odds for round n>0 should be: (52-[n-1])!/(52 - [n-1] -13)!, but I don't know how to approach the other two variants.
– shane
Nov 28 at 19:05
add a comment |
I was bored the other day and thought up a kind of complex card game. Now I can't work out the reasonable betting strategy. I hope someone else will find this amusing to think about and maybe help me out.
Consider a two-player card game using a standard 52 card deck between a dealer and a player with a very large, but finite amount of money.
First Variation
First the dealer shuffles the deck and deals herself a hand of thirteen cards, face down, which she does not look at. The dealer’s hand is a “success” if and only if it contains the thirteen spades in order from Ace to Two. Otherwise, the dealer’s hand is a “failure”.
After the dealer deals herself a hand, the first round of the game begins with the dealer proposing the player a wager for some stakes at certain odds. If the player accepts those stakes, then the dealer plays a card from the deck, face up. If the card is a spade, then obviously the dealer’s hand isn’t successful, and the player keeps the stake. However, if the card drawn is not a spade, then the player must either choose to “stop” or “continue” play. If the dealer chooses “stop”, the dealer exposes her hand. If the dealer’s hand is a success, the player loses the wager, otherwise, the player wins the stakes. Either way, the game ends. On the other hand, if the player chooses to continue, the player withdraws her wager, and the game proceeds to the next round.
At the start of each subsequent round, the dealer offers the player a wager at certain odds (not necessarily the same odds as in previous rounds). If the player refuses the wager, the game is over. If the player accepts the wager, then the dealer plays another card from the deck and offers the player the choice to stop or continue, as above.
After 39 rounds, there are no more cards to play, so at the end of this round, so if the player chooses to stop the game is over, and the player withdraws her wager. But if the player chooses to continue, the dealer exposes her hand and declares whether the dealer or the player has won.
What are the correct odds for the dealer to offer the player at every round of the game to make the game “fair”, i.e. to make the expectation of each bet zero? Assuming the dealer is offering fair odds, what is the player’s best strategy? (It seems like the player should always choose ‘stop’ on the first round.)
Second Variation
The second variation is like the first, except that if the player chooses to stop, at any round, then the player forfeits half of her wager for that round.
How does this change the odds the dealer should offer to make the game fair? What is the optimal strategy for the player now? (It seems like the player should always choose ‘continue’ now.)
Third Variation
The third variation is like the second variation, except that the player cannot choose whether to stop or continue. Instead, she must flip a fair coin. If it lands “heads,” play continues; if it lands “tails,” play stops.
What odds should the dealer offer at every round now to make it worth the player’s time to play?
probability
asked Nov 28 at 17:36
shane
1184
I was bored the other day and thought up a kind of complex card game. Now I can't work out the reasonable betting strategy. I hope someone else will find this amusing to think about and maybe help me out.
Consider a two-player card game using a standard 52 card deck between a dealer and a player with a very large, but finite amount of money.
First Variation
First the dealer shuffles the deck and deals herself a hand of thirteen cards, face down, which she does not look at. The dealer’s hand is a “success” if and only if it contains the thirteen spades in order from Ace to Two. Otherwise, the dealer’s hand is a “failure”.
After the dealer deals herself a hand, the first round of the game begins with the dealer proposing the player a wager for some stakes at certain odds. If the player accepts those stakes, then the dealer plays a card from the deck, face up. If the card is a spade, then obviously the dealer’s hand isn’t successful, and the player keeps the stake. However, if the card drawn is not a spade, then the player must either choose to “stop” or “continue” play. If the dealer chooses “stop”, the dealer exposes her hand. If the dealer’s hand is a success, the player loses the wager, otherwise, the player wins the stakes. Either way, the game ends. On the other hand, if the player chooses to continue, the player withdraws her wager, and the game proceeds to the next round.
At the start of each subsequent round, the dealer offers the player a wager at certain odds (not necessarily the same odds as in previous rounds). If the player refuses the wager, the game is over. If the player accepts the wager, then the dealer plays another card from the deck and offers the player the choice to stop or continue, as above.
After 39 rounds, there are no more cards to play, so at the end of this round, so if the player chooses to stop the game is over, and the player withdraws her wager. But if the player chooses to continue, the dealer exposes her hand and declares whether the dealer or the player has won.
What are the correct odds for the dealer to offer the player at every round of the game to make the game “fair”, i.e. to make the expectation of each bet zero? Assuming the dealer is offering fair odds, what is the player’s best strategy? (It seems like the player should always choose ‘stop’ on the first round.)
Second Variation
The second variation is like the first, except that if the player chooses to stop, at any round, then the player forfeits half of her wager for that round.
How does this change the odds the dealer should offer to make the game fair? What is the optimal strategy for the player now? (It seems like the player should always choose ‘continue’ now.)
Third Variation
The third variation is like the second variation, except that the player cannot choose whether to stop or continue. Instead, she must flip a fair coin. If it lands “heads,” play continues; if it lands “tails,” play stops.
What odds should the dealer offer at every round now to make it worth the player’s time to play?
probability
probability
asked Nov 28 at 17:36
shane
1184
asked Nov 28 at 17:36
shane
1184
asked Nov 28 at 17:36
shane
1184
asked Nov 28 at 17:36
shane
1184
asked Nov 28 at 17:36
shane
1184
1184
Why make everything hinge on such an implausible success? The probability of being dealt all the spades is $ 1.57.. times 10^{-12}$. If you play one game a second for ten thousand years, the odds are you still won't see a success.
– lulu
Nov 28 at 17:47
Note: I was only looking at the unordered hand consisting of all the spades. If you want them in order, then "success" becomes significantly less likely.
– lulu
Nov 28 at 17:50
Probability of success should be 1/[52!/(52-13)!] which is approximately 10^-22. But the super low probability shouldn't matter, since I can still give you odds that make the expectation zero, even if the event is really unlikely. My real goal here was to to give us lots of different points to make a decision at (every time a card is flipped), not to deal with really big or small numbers per se.
– shane
Nov 28 at 18:56
For the first variant, I think the odds the dealer should give the player for the first round is 52!/(52-13)! to 1 and the odds the dealer should give on last round should be 13! to 1, because even if you know all the cards are spades, there are still 13! different orders those cards could be in.
– shane
Nov 28 at 19:03
So my conjecture would be that for the first variant, the odds for round n>0 should be: (52-[n-1])!/(52 - [n-1] -13)!, but I don't know how to approach the other two variants.
– shane
Nov 28 at 19:05
add a comment |
Why make everything hinge on such an implausible success? The probability of being dealt all the spades is $ 1.57.. times 10^{-12}$. If you play one game a second for ten thousand years, the odds are you still won't see a success.
– lulu
Nov 28 at 17:47
Note: I was only looking at the unordered hand consisting of all the spades. If you want them in order, then "success" becomes significantly less likely.
– lulu
Nov 28 at 17:50
Probability of success should be 1/[52!/(52-13)!] which is approximately 10^-22. But the super low probability shouldn't matter, since I can still give you odds that make the expectation zero, even if the event is really unlikely. My real goal here was to to give us lots of different points to make a decision at (every time a card is flipped), not to deal with really big or small numbers per se.
– shane
Nov 28 at 18:56
For the first variant, I think the odds the dealer should give the player for the first round is 52!/(52-13)! to 1 and the odds the dealer should give on last round should be 13! to 1, because even if you know all the cards are spades, there are still 13! different orders those cards could be in.
– shane
Nov 28 at 19:03
So my conjecture would be that for the first variant, the odds for round n>0 should be: (52-[n-1])!/(52 - [n-1] -13)!, but I don't know how to approach the other two variants.
– shane
Nov 28 at 19:05
Why make everything hinge on such an implausible success? The probability of being dealt all the spades is $ 1.57.. times 10^{-12}$. If you play one game a second for ten thousand years, the odds are you still won't see a success.
– lulu
Nov 28 at 17:47
Why make everything hinge on such an implausible success? The probability of being dealt all the spades is $ 1.57.. times 10^{-12}$. If you play one game a second for ten thousand years, the odds are you still won't see a success.
– lulu
Nov 28 at 17:47
Note: I was only looking at the unordered hand consisting of all the spades. If you want them in order, then "success" becomes significantly less likely.
– lulu
Nov 28 at 17:50
Note: I was only looking at the unordered hand consisting of all the spades. If you want them in order, then "success" becomes significantly less likely.
– lulu
Nov 28 at 17:50
Probability of success should be 1/[52!/(52-13)!] which is approximately 10^-22. But the super low probability shouldn't matter, since I can still give you odds that make the expectation zero, even if the event is really unlikely. My real goal here was to to give us lots of different points to make a decision at (every time a card is flipped), not to deal with really big or small numbers per se.
– shane
Nov 28 at 18:56
Probability of success should be 1/[52!/(52-13)!] which is approximately 10^-22. But the super low probability shouldn't matter, since I can still give you odds that make the expectation zero, even if the event is really unlikely. My real goal here was to to give us lots of different points to make a decision at (every time a card is flipped), not to deal with really big or small numbers per se.
– shane
Nov 28 at 18:56
For the first variant, I think the odds the dealer should give the player for the first round is 52!/(52-13)! to 1 and the odds the dealer should give on last round should be 13! to 1, because even if you know all the cards are spades, there are still 13! different orders those cards could be in.
– shane
Nov 28 at 19:03
For the first variant, I think the odds the dealer should give the player for the first round is 52!/(52-13)! to 1 and the odds the dealer should give on last round should be 13! to 1, because even if you know all the cards are spades, there are still 13! different orders those cards could be in.
– shane
Nov 28 at 19:03
So my conjecture would be that for the first variant, the odds for round n>0 should be: (52-[n-1])!/(52 - [n-1] -13)!, but I don't know how to approach the other two variants.
– shane
Nov 28 at 19:05
So my conjecture would be that for the first variant, the odds for round n>0 should be: (52-[n-1])!/(52 - [n-1] -13)!, but I don't know how to approach the other two variants.
– shane
Nov 28 at 19:05
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Why make everything hinge on such an implausible success? The probability of being dealt all the spades is $ 1.57.. times 10^{-12}$. If you play one game a second for ten thousand years, the odds are you still won't see a success.
– lulu
Nov 28 at 17:47
Note: I was only looking at the unordered hand consisting of all the spades. If you want them in order, then "success" becomes significantly less likely.
– lulu
Nov 28 at 17:50
Probability of success should be 1/[52!/(52-13)!] which is approximately 10^-22. But the super low probability shouldn't matter, since I can still give you odds that make the expectation zero, even if the event is really unlikely. My real goal here was to to give us lots of different points to make a decision at (every time a card is flipped), not to deal with really big or small numbers per se.
– shane
Nov 28 at 18:56
For the first variant, I think the odds the dealer should give the player for the first round is 52!/(52-13)! to 1 and the odds the dealer should give on last round should be 13! to 1, because even if you know all the cards are spades, there are still 13! different orders those cards could be in.
– shane
Nov 28 at 19:03
So my conjecture would be that for the first variant, the odds for round n>0 should be: (52-[n-1])!/(52 - [n-1] -13)!, but I don't know how to approach the other two variants.
– shane
Nov 28 at 19:05