@Jack D'Aurizio suggested throwing the die in a wedge, glass, etc.
If you can throw the die so that it always lands occupying exactly the same square on the table, then you can get more information from a die throw than just the number showing on top: you can also see what number is showing on the face closest to the thrower (or certain designated direction).
Actually, most of the time you can still determine this "closest face" even under regular conditions. Let's assume that the thrower can always detect both which face is up and which face is closest to the thrower, and let's call those faces "top" and "South", respectively.
Then each throw yields not just 6ドル$ but 24ドル$ separate possible outcomes, all equally likely. For example, a 1ドル$ on top may be paired with a 2,ドル 3, 4$, or 5ドル$ on the South face (though not with a 6ドル$ since the 1ドル$ and 6ドル$ are on opposite faces on a standard die). Labeling those 24ドル$ outcomes by the numbers 1ドル$ through 24ドル$, we now effectively have a 24ドル$-sided die.
With our 24ドル$-sided die, we can simulate a 10ドル$-sided die by just taking the result mod 10ドル$ (discard the 10ドル$'s digit and have 0ドル$ mean 10ドル$ if you like), except you have to reroll if the result is 21ドル$ or higher: a 4ドル/24 = 1/6$ chance of rerolling. The expected number of rolls, $E$, using this method is thus calculated by $$ E = (5/6)(1) + (1/6)(1+E), $$ so $E = 1.2$, exactly 1ドル$ roll better than in @Jack D'Aurizio's answer.
For readers interested in group theory, we have just seen that the group of rigid symmetries of a cube must have 24ドル$ elements. Indeed this group is isomorphic to $S_4$ (although you don't need that for this problem). See also: Proof that cube has 24 rotational symmetries Proof that cube has 24 rotational symmetries
@Jack D'Aurizio suggested throwing the die in a wedge, glass, etc.
If you can throw the die so that it always lands occupying exactly the same square on the table, then you can get more information from a die throw than just the number showing on top: you can also see what number is showing on the face closest to the thrower (or certain designated direction).
Actually, most of the time you can still determine this "closest face" even under regular conditions. Let's assume that the thrower can always detect both which face is up and which face is closest to the thrower, and let's call those faces "top" and "South", respectively.
Then each throw yields not just 6ドル$ but 24ドル$ separate possible outcomes, all equally likely. For example, a 1ドル$ on top may be paired with a 2,ドル 3, 4$, or 5ドル$ on the South face (though not with a 6ドル$ since the 1ドル$ and 6ドル$ are on opposite faces on a standard die). Labeling those 24ドル$ outcomes by the numbers 1ドル$ through 24ドル$, we now effectively have a 24ドル$-sided die.
With our 24ドル$-sided die, we can simulate a 10ドル$-sided die by just taking the result mod 10ドル$ (discard the 10ドル$'s digit and have 0ドル$ mean 10ドル$ if you like), except you have to reroll if the result is 21ドル$ or higher: a 4ドル/24 = 1/6$ chance of rerolling. The expected number of rolls, $E$, using this method is thus calculated by $$ E = (5/6)(1) + (1/6)(1+E), $$ so $E = 1.2$, exactly 1ドル$ roll better than in @Jack D'Aurizio's answer.
For readers interested in group theory, we have just seen that the group of rigid symmetries of a cube must have 24ドル$ elements. Indeed this group is isomorphic to $S_4$ (although you don't need that for this problem). See also: Proof that cube has 24 rotational symmetries
@Jack D'Aurizio suggested throwing the die in a wedge, glass, etc.
If you can throw the die so that it always lands occupying exactly the same square on the table, then you can get more information from a die throw than just the number showing on top: you can also see what number is showing on the face closest to the thrower (or certain designated direction).
Actually, most of the time you can still determine this "closest face" even under regular conditions. Let's assume that the thrower can always detect both which face is up and which face is closest to the thrower, and let's call those faces "top" and "South", respectively.
Then each throw yields not just 6ドル$ but 24ドル$ separate possible outcomes, all equally likely. For example, a 1ドル$ on top may be paired with a 2,ドル 3, 4$, or 5ドル$ on the South face (though not with a 6ドル$ since the 1ドル$ and 6ドル$ are on opposite faces on a standard die). Labeling those 24ドル$ outcomes by the numbers 1ドル$ through 24ドル$, we now effectively have a 24ドル$-sided die.
With our 24ドル$-sided die, we can simulate a 10ドル$-sided die by just taking the result mod 10ドル$ (discard the 10ドル$'s digit and have 0ドル$ mean 10ドル$ if you like), except you have to reroll if the result is 21ドル$ or higher: a 4ドル/24 = 1/6$ chance of rerolling. The expected number of rolls, $E$, using this method is thus calculated by $$ E = (5/6)(1) + (1/6)(1+E), $$ so $E = 1.2$, exactly 1ドル$ roll better than in @Jack D'Aurizio's answer.
For readers interested in group theory, we have just seen that the group of rigid symmetries of a cube must have 24ドル$ elements. Indeed this group is isomorphic to $S_4$ (although you don't need that for this problem). See also: Proof that cube has 24 rotational symmetries
@Jack D'Aurizio suggested throwing the die in a wedge, glass, etc.
If you can throw the die so that it always lands occupying exactly the same square on the table, then you can get more information from a die throw than just the number showing on top: you can also see what number is showing on the face closest to the thrower (or certain designated direction).
Actually, most of the time you can still determine this "closest face" even under regular conditions. Let's assume that the thrower can always detect both which face is up and which face is closest to the thrower, and let's call those faces "top" and "South", respectively.
Then each throw yields not just 6ドル$ but 24ドル$ separate possible outcomes, all equally likely. For example, a 1ドル$ on top may be paired with a 2,ドル 3, 4$, or 5ドル$ on the South face (though not with a 6ドル$ since the 1ドル$ and 6ドル$ are on opposite faces on a standard die). Labeling those 24ドル$ outcomes by the numbers 1ドル$ through 24ドル$, we now effectively have a 24ドル$-sided die.
With our 24ドル$-sided die, we can simulate a 10ドル$-sided die by just taking the result mod 10ドル$ (discard the 10ドル$'s digit and have 0ドル$ mean 10ドル$ if you like), except you have to reroll if the result is 21ドル$ or higher: a 4ドル/24 = 1/6$ chance of rerolling. The expected number of rolls, $E$, using this method is thus calculated by $$ E = (5/6)(1) + (1/6)(1+E), $$ so $E = 1.2$, exactly 1ドル$ roll better than in @Jack D'Aurizio's answer.
For readers interested in group theory, we have just seen that the group of rigid symmetries of a cube must have 24ドル$ elements. Indeed this group is isomorphic to $S_4$ (although you don't need that for this problem). See also: Proof that cube has 24 rotational symmetries