simple dice game played with a single die
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A player pays £1 to play and rolls the die once. Before each subsequent roll the players have to guess if the roll will be higher or lower than the previous roll. If the player guesses correctly for 7 rolls then they will win £5. At any point the player can ignore a roll and pay an extra £1 to re-roll. If a player guesses wrong at any point and choose not to re-roll they are out. If tie occur and player want to carry on playing then they should pay £1 and re-roll.
find the optimal strategy to win the £5 prize eventually, or alternatively find the optimal strategy to make profit
probability dice
asked Nov 11 at 22:30
Nahida Afroz
82
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up vote
-1
down vote
favorite
A player pays £1 to play and rolls the die once. Before each subsequent roll the players have to guess if the roll will be higher or lower than the previous roll. If the player guesses correctly for 7 rolls then they will win £5. At any point the player can ignore a roll and pay an extra £1 to re-roll. If a player guesses wrong at any point and choose not to re-roll they are out. If tie occur and player want to carry on playing then they should pay £1 and re-roll.
find the optimal strategy to win the £5 prize eventually, or alternatively find the optimal strategy to make profit
probability dice
asked Nov 11 at 22:30
Nahida Afroz
82
You have not said what happens with ties. Nor what you have tried yourself
– Henry
Nov 11 at 22:42
If you guess wrong and re-roll, what happens if you are wrong again? Can you pay again to re-roll? Do you compare with the previous roll for higher/lower on a re-roll or with the first roll of the turn. So you roll a $3$ to start. You guess higher but roll $1$. You still guess higher and roll $2$. Is that a win or a loss?
– Ross Millikan
Nov 14 at 23:28
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
A player pays £1 to play and rolls the die once. Before each subsequent roll the players have to guess if the roll will be higher or lower than the previous roll. If the player guesses correctly for 7 rolls then they will win £5. At any point the player can ignore a roll and pay an extra £1 to re-roll. If a player guesses wrong at any point and choose not to re-roll they are out. If tie occur and player want to carry on playing then they should pay £1 and re-roll.
find the optimal strategy to win the £5 prize eventually, or alternatively find the optimal strategy to make profit
probability dice
asked Nov 11 at 22:30
Nahida Afroz
82
A player pays £1 to play and rolls the die once. Before each subsequent roll the players have to guess if the roll will be higher or lower than the previous roll. If the player guesses correctly for 7 rolls then they will win £5. At any point the player can ignore a roll and pay an extra £1 to re-roll. If a player guesses wrong at any point and choose not to re-roll they are out. If tie occur and player want to carry on playing then they should pay £1 and re-roll.
find the optimal strategy to win the £5 prize eventually, or alternatively find the optimal strategy to make profit
probability dice
probability dice
asked Nov 11 at 22:30
Nahida Afroz
82
asked Nov 11 at 22:30
Nahida Afroz
82
edited Nov 14 at 19:52
asked Nov 11 at 22:30
Nahida Afroz
82
asked Nov 11 at 22:30
Nahida Afroz
82
asked Nov 11 at 22:30
Nahida Afroz
82
82
You have not said what happens with ties. Nor what you have tried yourself
– Henry
Nov 11 at 22:42
If you guess wrong and re-roll, what happens if you are wrong again? Can you pay again to re-roll? Do you compare with the previous roll for higher/lower on a re-roll or with the first roll of the turn. So you roll a $3$ to start. You guess higher but roll $1$. You still guess higher and roll $2$. Is that a win or a loss?
– Ross Millikan
Nov 14 at 23:28
add a comment |
You have not said what happens with ties. Nor what you have tried yourself
– Henry
Nov 11 at 22:42
If you guess wrong and re-roll, what happens if you are wrong again? Can you pay again to re-roll? Do you compare with the previous roll for higher/lower on a re-roll or with the first roll of the turn. So you roll a $3$ to start. You guess higher but roll $1$. You still guess higher and roll $2$. Is that a win or a loss?
– Ross Millikan
Nov 14 at 23:28
You have not said what happens with ties. Nor what you have tried yourself
– Henry
Nov 11 at 22:42
You have not said what happens with ties. Nor what you have tried yourself
– Henry
Nov 11 at 22:42
If you guess wrong and re-roll, what happens if you are wrong again? Can you pay again to re-roll? Do you compare with the previous roll for higher/lower on a re-roll or with the first roll of the turn. So you roll a $3$ to start. You guess higher but roll $1$. You still guess higher and roll $2$. Is that a win or a loss?
– Ross Millikan
Nov 14 at 23:28
If you guess wrong and re-roll, what happens if you are wrong again? Can you pay again to re-roll? Do you compare with the previous roll for higher/lower on a re-roll or with the first roll of the turn. So you roll a $3$ to start. You guess higher but roll $1$. You still guess higher and roll $2$. Is that a win or a loss?
– Ross Millikan
Nov 14 at 23:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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The optimal strategy to win eventually is to guess the obvious direction and always pay to reroll. You will win the prize, though you will probably lose money. I am sure the optimal strategy is not to play. Early in the game you should not pay to reroll, but late in the game you should. I will assume you can buy rerolls as many times as you want and always compare with the previous accepted roll.
Whatever you have paid at a given point is sunk cost and should not be considered in the decision of whether to pay for a reroll. First we need the expectation for each next to last roll of $4,5,6$. If you have a $4$ you have $frac 12$ chance of $5$ and $frac 12$ chance of losing $1$ and being where you are. If the expectation here is $E_4$ we have $E_4=frac 12cdot 5 + frac 12 (E_4-1), E_4=4$. If you have a $5$ you have $frac 23$ chance of winning, so $E_5=frac 23cdot 5+frac 13(E_5-1),E_5=frac 92.$ Similarly if you have a $6, E_6=frac 56cdot 5+frac 16(E_6-1), E_6=frac {24}5$
Now if you need two more successes to win and have a $3$, a win has equal chance of being a $4,5,$ or $6$, so a win is worth $frac 13(4+frac 92+frac {24}5)=frac {133}{30}$. You have $frac 12$ chance to win at this point, so your expectation is $E=frac 12cdot frac {133}{30}+frac 12(E-1), E=frac {103}{30},$ which is greater than $1$ so you should pay to reroll if you lose. Having a $1$ or $2$ gives you a better chance to win this roll but more chance to have a $3$ or $4$ to compare with next roll. You can continue the analysis but it gets tedious. I suspect the value will fall below $1$ before you get to $7$ successes needed, which is the source of my claim that you should not play at all.
answered Nov 15 at 0:02
Ross Millikan
287k23195364
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The optimal strategy to win eventually is to guess the obvious direction and always pay to reroll. You will win the prize, though you will probably lose money. I am sure the optimal strategy is not to play. Early in the game you should not pay to reroll, but late in the game you should. I will assume you can buy rerolls as many times as you want and always compare with the previous accepted roll.
Whatever you have paid at a given point is sunk cost and should not be considered in the decision of whether to pay for a reroll. First we need the expectation for each next to last roll of $4,5,6$. If you have a $4$ you have $frac 12$ chance of $5$ and $frac 12$ chance of losing $1$ and being where you are. If the expectation here is $E_4$ we have $E_4=frac 12cdot 5 + frac 12 (E_4-1), E_4=4$. If you have a $5$ you have $frac 23$ chance of winning, so $E_5=frac 23cdot 5+frac 13(E_5-1),E_5=frac 92.$ Similarly if you have a $6, E_6=frac 56cdot 5+frac 16(E_6-1), E_6=frac {24}5$
Now if you need two more successes to win and have a $3$, a win has equal chance of being a $4,5,$ or $6$, so a win is worth $frac 13(4+frac 92+frac {24}5)=frac {133}{30}$. You have $frac 12$ chance to win at this point, so your expectation is $E=frac 12cdot frac {133}{30}+frac 12(E-1), E=frac {103}{30},$ which is greater than $1$ so you should pay to reroll if you lose. Having a $1$ or $2$ gives you a better chance to win this roll but more chance to have a $3$ or $4$ to compare with next roll. You can continue the analysis but it gets tedious. I suspect the value will fall below $1$ before you get to $7$ successes needed, which is the source of my claim that you should not play at all.
answered Nov 15 at 0:02
Ross Millikan
287k23195364
add a comment |
up vote
0
down vote
accepted
The optimal strategy to win eventually is to guess the obvious direction and always pay to reroll. You will win the prize, though you will probably lose money. I am sure the optimal strategy is not to play. Early in the game you should not pay to reroll, but late in the game you should. I will assume you can buy rerolls as many times as you want and always compare with the previous accepted roll.
Whatever you have paid at a given point is sunk cost and should not be considered in the decision of whether to pay for a reroll. First we need the expectation for each next to last roll of $4,5,6$. If you have a $4$ you have $frac 12$ chance of $5$ and $frac 12$ chance of losing $1$ and being where you are. If the expectation here is $E_4$ we have $E_4=frac 12cdot 5 + frac 12 (E_4-1), E_4=4$. If you have a $5$ you have $frac 23$ chance of winning, so $E_5=frac 23cdot 5+frac 13(E_5-1),E_5=frac 92.$ Similarly if you have a $6, E_6=frac 56cdot 5+frac 16(E_6-1), E_6=frac {24}5$
Now if you need two more successes to win and have a $3$, a win has equal chance of being a $4,5,$ or $6$, so a win is worth $frac 13(4+frac 92+frac {24}5)=frac {133}{30}$. You have $frac 12$ chance to win at this point, so your expectation is $E=frac 12cdot frac {133}{30}+frac 12(E-1), E=frac {103}{30},$ which is greater than $1$ so you should pay to reroll if you lose. Having a $1$ or $2$ gives you a better chance to win this roll but more chance to have a $3$ or $4$ to compare with next roll. You can continue the analysis but it gets tedious. I suspect the value will fall below $1$ before you get to $7$ successes needed, which is the source of my claim that you should not play at all.
answered Nov 15 at 0:02
Ross Millikan
287k23195364
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The optimal strategy to win eventually is to guess the obvious direction and always pay to reroll. You will win the prize, though you will probably lose money. I am sure the optimal strategy is not to play. Early in the game you should not pay to reroll, but late in the game you should. I will assume you can buy rerolls as many times as you want and always compare with the previous accepted roll.
Whatever you have paid at a given point is sunk cost and should not be considered in the decision of whether to pay for a reroll. First we need the expectation for each next to last roll of $4,5,6$. If you have a $4$ you have $frac 12$ chance of $5$ and $frac 12$ chance of losing $1$ and being where you are. If the expectation here is $E_4$ we have $E_4=frac 12cdot 5 + frac 12 (E_4-1), E_4=4$. If you have a $5$ you have $frac 23$ chance of winning, so $E_5=frac 23cdot 5+frac 13(E_5-1),E_5=frac 92.$ Similarly if you have a $6, E_6=frac 56cdot 5+frac 16(E_6-1), E_6=frac {24}5$
Now if you need two more successes to win and have a $3$, a win has equal chance of being a $4,5,$ or $6$, so a win is worth $frac 13(4+frac 92+frac {24}5)=frac {133}{30}$. You have $frac 12$ chance to win at this point, so your expectation is $E=frac 12cdot frac {133}{30}+frac 12(E-1), E=frac {103}{30},$ which is greater than $1$ so you should pay to reroll if you lose. Having a $1$ or $2$ gives you a better chance to win this roll but more chance to have a $3$ or $4$ to compare with next roll. You can continue the analysis but it gets tedious. I suspect the value will fall below $1$ before you get to $7$ successes needed, which is the source of my claim that you should not play at all.
answered Nov 15 at 0:02
Ross Millikan
287k23195364
The optimal strategy to win eventually is to guess the obvious direction and always pay to reroll. You will win the prize, though you will probably lose money. I am sure the optimal strategy is not to play. Early in the game you should not pay to reroll, but late in the game you should. I will assume you can buy rerolls as many times as you want and always compare with the previous accepted roll.
Whatever you have paid at a given point is sunk cost and should not be considered in the decision of whether to pay for a reroll. First we need the expectation for each next to last roll of $4,5,6$. If you have a $4$ you have $frac 12$ chance of $5$ and $frac 12$ chance of losing $1$ and being where you are. If the expectation here is $E_4$ we have $E_4=frac 12cdot 5 + frac 12 (E_4-1), E_4=4$. If you have a $5$ you have $frac 23$ chance of winning, so $E_5=frac 23cdot 5+frac 13(E_5-1),E_5=frac 92.$ Similarly if you have a $6, E_6=frac 56cdot 5+frac 16(E_6-1), E_6=frac {24}5$
Now if you need two more successes to win and have a $3$, a win has equal chance of being a $4,5,$ or $6$, so a win is worth $frac 13(4+frac 92+frac {24}5)=frac {133}{30}$. You have $frac 12$ chance to win at this point, so your expectation is $E=frac 12cdot frac {133}{30}+frac 12(E-1), E=frac {103}{30},$ which is greater than $1$ so you should pay to reroll if you lose. Having a $1$ or $2$ gives you a better chance to win this roll but more chance to have a $3$ or $4$ to compare with next roll. You can continue the analysis but it gets tedious. I suspect the value will fall below $1$ before you get to $7$ successes needed, which is the source of my claim that you should not play at all.
answered Nov 15 at 0:02
Ross Millikan
287k23195364
answered Nov 15 at 0:02
Ross Millikan
287k23195364
answered Nov 15 at 0:02
Ross Millikan
287k23195364
answered Nov 15 at 0:02
Ross Millikan
287k23195364
287k23195364
add a comment |
add a comment |
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You have not said what happens with ties. Nor what you have tried yourself
– Henry
Nov 11 at 22:42
If you guess wrong and re-roll, what happens if you are wrong again? Can you pay again to re-roll? Do you compare with the previous roll for higher/lower on a re-roll or with the first roll of the turn. So you roll a $3$ to start. You guess higher but roll $1$. You still guess higher and roll $2$. Is that a win or a loss?
– Ross Millikan
Nov 14 at 23:28