Card Expected Value with Option to Skip Cards
2
$begingroup$
In a normal deck of cards, you can either reveal the top card or guess whether that card is black. If you reveal the top card, you get to see what the card is and the game continues with one less card in the deck. If you were to make a guess, the game ends, and you get paid out $$100$ if the card is black and $$0$ if it's red.
What is the optimal strategy of this game and its expected value?
The lower bound has to be $0.5times $100 = $50$ since you can just guess on the first card.
expected-value card-games dynamic-programming
edited Jan 2 at 22:56
Henry
101k482169
asked Jan 2 at 22:54
tiger88tiger88
365
$endgroup$
add a comment |
2
$begingroup$
In a normal deck of cards, you can either reveal the top card or guess whether that card is black. If you reveal the top card, you get to see what the card is and the game continues with one less card in the deck. If you were to make a guess, the game ends, and you get paid out $$100$ if the card is black and $$0$ if it's red.
What is the optimal strategy of this game and its expected value?
The lower bound has to be $0.5times $100 = $50$ since you can just guess on the first card.
expected-value card-games dynamic-programming
edited Jan 2 at 22:56
Henry
101k482169
asked Jan 2 at 22:54
tiger88tiger88
365
$endgroup$
1
$begingroup$
It is difficult to see how the answer can be anything other than $$50$
$endgroup$
– Henry
Jan 2 at 22:58
$begingroup$
My best guess would have been to reveal cards until you revealed your first black card and then bet on the very next card. Coming up with the expression for expected value for that and plugging into wolfram alpha suggests that even that would just give an expectation of $50$.
$endgroup$
– JMoravitz
Jan 2 at 23:04
1
$begingroup$
You can argue recursively that "guessing now" is optimal, regardless of what stage you are at. Obviously at stage 52 (with only one card left) guessing now is optimal. But at stage 51 (when we have 2 cards left) we can either guess now (equivalent to betting the top card is black) or reveal (equivalent to betting the bottom is black). Both not-yet-seen cards have the same probability of being black (regardless of history of previous stages), so there is no reason to pick the second over the first, so we might as well guess now. Apply the same reasoning to stage 50, 49, 48, and so on.
$endgroup$
– Michael
Jan 2 at 23:40
add a comment |
2
2
2
$begingroup$
In a normal deck of cards, you can either reveal the top card or guess whether that card is black. If you reveal the top card, you get to see what the card is and the game continues with one less card in the deck. If you were to make a guess, the game ends, and you get paid out $$100$ if the card is black and $$0$ if it's red.
What is the optimal strategy of this game and its expected value?
The lower bound has to be $0.5times $100 = $50$ since you can just guess on the first card.
expected-value card-games dynamic-programming
edited Jan 2 at 22:56
Henry
101k482169
asked Jan 2 at 22:54
tiger88tiger88
365
$endgroup$
In a normal deck of cards, you can either reveal the top card or guess whether that card is black. If you reveal the top card, you get to see what the card is and the game continues with one less card in the deck. If you were to make a guess, the game ends, and you get paid out $$100$ if the card is black and $$0$ if it's red.
What is the optimal strategy of this game and its expected value?
The lower bound has to be $0.5times $100 = $50$ since you can just guess on the first card.
expected-value card-games dynamic-programming
expected-value card-games dynamic-programming
edited Jan 2 at 22:56
Henry
101k482169
asked Jan 2 at 22:54
tiger88tiger88
365
edited Jan 2 at 22:56
Henry
101k482169
asked Jan 2 at 22:54
tiger88tiger88
365
edited Jan 2 at 22:56
Henry
101k482169
edited Jan 2 at 22:56
Henry
101k482169
edited Jan 2 at 22:56
Henry
101k482169
101k482169
asked Jan 2 at 22:54
tiger88tiger88
365
asked Jan 2 at 22:54
tiger88tiger88
365
asked Jan 2 at 22:54
tiger88tiger88
365
365
1
$begingroup$
It is difficult to see how the answer can be anything other than $$50$
$endgroup$
– Henry
Jan 2 at 22:58
$begingroup$
My best guess would have been to reveal cards until you revealed your first black card and then bet on the very next card. Coming up with the expression for expected value for that and plugging into wolfram alpha suggests that even that would just give an expectation of $50$.
$endgroup$
– JMoravitz
Jan 2 at 23:04
1
$begingroup$
You can argue recursively that "guessing now" is optimal, regardless of what stage you are at. Obviously at stage 52 (with only one card left) guessing now is optimal. But at stage 51 (when we have 2 cards left) we can either guess now (equivalent to betting the top card is black) or reveal (equivalent to betting the bottom is black). Both not-yet-seen cards have the same probability of being black (regardless of history of previous stages), so there is no reason to pick the second over the first, so we might as well guess now. Apply the same reasoning to stage 50, 49, 48, and so on.
$endgroup$
– Michael
Jan 2 at 23:40
add a comment |
1
$begingroup$
It is difficult to see how the answer can be anything other than $$50$
$endgroup$
– Henry
Jan 2 at 22:58
$begingroup$
My best guess would have been to reveal cards until you revealed your first black card and then bet on the very next card. Coming up with the expression for expected value for that and plugging into wolfram alpha suggests that even that would just give an expectation of $50$.
$endgroup$
– JMoravitz
Jan 2 at 23:04
1
$begingroup$
You can argue recursively that "guessing now" is optimal, regardless of what stage you are at. Obviously at stage 52 (with only one card left) guessing now is optimal. But at stage 51 (when we have 2 cards left) we can either guess now (equivalent to betting the top card is black) or reveal (equivalent to betting the bottom is black). Both not-yet-seen cards have the same probability of being black (regardless of history of previous stages), so there is no reason to pick the second over the first, so we might as well guess now. Apply the same reasoning to stage 50, 49, 48, and so on.
$endgroup$
– Michael
Jan 2 at 23:40
1
1
$begingroup$
It is difficult to see how the answer can be anything other than $$50$
$endgroup$
– Henry
Jan 2 at 22:58
$begingroup$
It is difficult to see how the answer can be anything other than $$50$
$endgroup$
– Henry
Jan 2 at 22:58
$begingroup$
My best guess would have been to reveal cards until you revealed your first black card and then bet on the very next card. Coming up with the expression for expected value for that and plugging into wolfram alpha suggests that even that would just give an expectation of $50$.
$endgroup$
– JMoravitz
Jan 2 at 23:04
$begingroup$
My best guess would have been to reveal cards until you revealed your first black card and then bet on the very next card. Coming up with the expression for expected value for that and plugging into wolfram alpha suggests that even that would just give an expectation of $50$.
$endgroup$
– JMoravitz
Jan 2 at 23:04
1
1
$begingroup$
You can argue recursively that "guessing now" is optimal, regardless of what stage you are at. Obviously at stage 52 (with only one card left) guessing now is optimal. But at stage 51 (when we have 2 cards left) we can either guess now (equivalent to betting the top card is black) or reveal (equivalent to betting the bottom is black). Both not-yet-seen cards have the same probability of being black (regardless of history of previous stages), so there is no reason to pick the second over the first, so we might as well guess now. Apply the same reasoning to stage 50, 49, 48, and so on.
$endgroup$
– Michael
Jan 2 at 23:40
$begingroup$
You can argue recursively that "guessing now" is optimal, regardless of what stage you are at. Obviously at stage 52 (with only one card left) guessing now is optimal. But at stage 51 (when we have 2 cards left) we can either guess now (equivalent to betting the top card is black) or reveal (equivalent to betting the bottom is black). Both not-yet-seen cards have the same probability of being black (regardless of history of previous stages), so there is no reason to pick the second over the first, so we might as well guess now. Apply the same reasoning to stage 50, 49, 48, and so on.
$endgroup$
– Michael
Jan 2 at 23:40
add a comment |
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1
$begingroup$
It is difficult to see how the answer can be anything other than $$50$
$endgroup$
– Henry
Jan 2 at 22:58
$begingroup$
My best guess would have been to reveal cards until you revealed your first black card and then bet on the very next card. Coming up with the expression for expected value for that and plugging into wolfram alpha suggests that even that would just give an expectation of $50$.
$endgroup$
– JMoravitz
Jan 2 at 23:04
1
$begingroup$
You can argue recursively that "guessing now" is optimal, regardless of what stage you are at. Obviously at stage 52 (with only one card left) guessing now is optimal. But at stage 51 (when we have 2 cards left) we can either guess now (equivalent to betting the top card is black) or reveal (equivalent to betting the bottom is black). Both not-yet-seen cards have the same probability of being black (regardless of history of previous stages), so there is no reason to pick the second over the first, so we might as well guess now. Apply the same reasoning to stage 50, 49, 48, and so on.
$endgroup$
– Michael
Jan 2 at 23:40