I have a riddle for you ****goreo cookies

[LegosInMyEgos]

Mmmkay.

You have a quarter.

You need to create a game of chance with the quarter such that

- there are two players
- one player has a 2/3 chance of winning, and the other has a 1/3 chance of winning

if you solve the riddle, you're smarter than a

[nikolic993]

can you be more specific?

[LegosInMyEgos]

if you flip the coin, there is a 1/2 chance it lands on heads

- I want you to create a game in which there are two players, in which one player has a 2/3 chance of winning, and the other has a 1/3 chance of winning.

[nikolic993]

if you flip the coin, there is a 1/2 chance it lands on heads

- I want you to create a game in which there are two players, in which one player has a 2/3 chance of winning, and the other has a 1/3 chance of winning.

Monty hall problem?Don't know how to make a game out of the coins, but I will post if I think of one.

[LegosInMyEgos]

if the monty hall problem was a coin you would totally have gotten this

[Ronceiro]

Monty hall problem?Don't know how to make a game out of the coins, but I will post if I think of one.

Was thinking this

Have the coin behind one door, and make the other 2 doors empty. You open the door with the coin = you win the game


Or if you're too poverty to have doors, have 3 cups instead of doors, with the coin under a cup and the other 2 cups empty

[LegosInMyEgos]

coin has to be flipped

[nikolic993]

Was thinking this

Have the coin behind one door, and make the other 2 doors empty. You open the door with the coin = you win the game

This is not about finding the coin, it's about heads/tails guessing.

edit: is there only one coin or two?

[LegosInMyEgos]

one coin

[nikolic993]

I'm out, I start to think too hard and always get annoyed when I can't find a solution.

[SumDumGoi]

Best two out of three. There are 3 flips of the coin. One player will have to win two of the flips and the other player can only win a single flip. The only caveat will be that the player who loses two of the flips is actually the winner of the game (2/3 chance of winning) and the player who wins two of the flips is actually the loser of the game (1/3 chance of winning).



If this fits your criteria I have a riddle for you. The winner of this game is smarter than "a" what?




Edit: I'm trying to work out the problem mathematically. During the first flip of the coin you have a 50% chance of guessing heads or tails. However, this guess would only represent 1/3 of the total coin flips. So your chances of winning the first flip would be 16.5% (0.5 x 0.33). They would then have the same chance of winning the subsequent flip which they would have the same 16.5% chance of winning. The sum of the two chances would be 0.33 or 1/3. This would have to assume that only a single person is calling heads or tails during the flip for all three coin tosses.

[awkwrdturtle]

Flip the coin 3 times. There are 8 possible outcomes.

HHH
HHT
HTH
HTT
THH
TTH
THT
TTT

If it is either of the first two--HHH or HHT--no one wins and you go back to the start.

If the outcome is HTH, HTT, THH, or TTH, player 1 wins. If the outcome is THT or TTT player 2 wins.

Player 1 wins 2/3 of the time, and player 2 1/3.

[JasonDB]

How much does each player lift?

[desslok]

Player A flips coin twice. If it lands on heads or tails he wins.
Player B flips coin once. If it lands on heads or tails he wins.

[sawoobley]

Flip the coin. Best 2 out of 3 wins.

[miacanesfan25]

I have a 100% chance of winning if player 2 is dead.

Put that in your riddle pipe and smoke it.

[awkwrdturtle]

Mmmkay.

You have a quarter.

You need to create a game of chance with the quarter such that

- there are two players
- one player has a 2/3 chance of winning, and the other has a 1/3 chance of winning

if you solve the riddle, you're smarter than a

Flip the coin. Best 2 out of 3 wins.

Each player has a 50% chance that way.

[pretend]

Player 1 picks a number between 1 and 3.
Player 2 picks a number between 1 and 3.
One player flips the coin, and when it lands, both reveal their number at the same time.
If they're different (2/3), the flipper wins. If they're the same(1/3), the other player wins.

Players could take turns flipping or you could have the coin flip determine who wins on a match vs no match.

[LegosInMyEgos]

no one has gotten it yet

[JasonDB]

Player one has 3 boxes he puts the coin in one of the boxes without player two getting to see which box he puts it in. Player two has to guess which box the coin is in. If he guesses he wins if not he loses and player one wins. If nothing is allowed in the game other than the players and the single quarter then player one could turn around and put the coin in his mouth or either of his hands turn around and player two has to guess which of the 3 it is in for the same result as with the boxes.

You still didn't answer how much these *******s lift yet! >:-O

[LegosInMyEgos]

Player one has 3 boxes he puts the coin in one of the boxes without player two getting to see which box he puts it in. Player two has to guess which box the coin is in. If he guesses he wins if not he loses and player one wins. If nothing is allowed in the game other than the players and the single quarter then player one could turn around and put the coin in his mouth or either of his hands turn around and player two has to guess which of the 3 it is in for the same result as with the boxes.

You still didn't answer how much these *******s lift yet! >:-O

Each person lifts exactly 350 on the press squat bench and deadlift.

you are only allowed to create the 2/3 and 1/3 possibilities using the probability involving the flipping of a coin. You can use 400 coin flips, infinite coin flips, or only one, but you arent allowed to use other objects to create possibilities.

Hint: think of a game that could have these odds that only uses the coin and the laws of probability.

Double hint: Riemann sums

[awkwrdturtle]

Flip the coin 3 times. There are 8 possible outcomes.

HHH
HHT
HTH
HTT
THH
TTH
THT
TTT

If it is either of the first two--HHH or HHT--no one wins and you go back to the start.

If the outcome is HTH, HTT, THH, or TTH, player 1 wins. If the outcome is THT or TTT player 2 wins.

Player 1 wins 2/3 of the time, and player 2 1/3.

no one has gotten it yet

Explain how this doesn't meet your criteria.

[pretend]

no one has gotten it yet

Just cuz we didn't get *your* answer, doesn't mean we didn't get it. Maybe you didn't phrase it properly to match the answer you have in mind.

Flip the coin 3 times. There are 8 possible outcomes.

HHH
HHT
HTH
HTT
THH
TTH
THT
TTT

If it is either of the first two--HHH or HHT--no one wins and you go back to the start.

If the outcome is HTH, HTT, THH, or TTH, player 1 wins. If the outcome is THT or TTT player 2 wins.

Player 1 wins 2/3 of the time, and player 2 1/3.

That can be reduced to 2 flips per round (just need one to win 2x as often as other):

HH = 1 wins
TH/HT = 2 wins
TT = reflip

[LegosInMyEgos]

Explain how this doesn't meet your criteria.

Just cuz we didn't get *your* answer, doesn't mean we didn't get it. Maybe you didn't phrase it properly to match the answer you have in mind.




That can be reduced to 2 flips per round (just need one to win 2x as often as other):

HH = 1 wins
TH/HT = 2 wins
TT = reflip

because reflipping the coin doesn't create a 2/3 or a 1/3 possibility.

the probability on any given flip is 1/4 or 1/2 for HT/TH and HH.

each flip is an independent event and the probability does not converge to 1/3 or 2/3 as the number of flips approaches infinity.

[awkwrdturtle]

because reflipping the coin doesn't create a 2/3 or a 1/3 possibility.

the probability on any given flip is 1/4 or 1/2 for HT/TH and HH.

each flip is an independent event and the probability does not converge to 1/3 or 2/3 as the number of flips approaches infinity.

Right, but as the number of rounds of 3 flips approaches infinity, the probabilities do converge to 1/3 and 2/3.

Call the first round of 3 flips round i=0. For player 1, the probability that he wins in round i is 1/(2^(2i+1)). Sum the probability that player 1 wins across all rounds from i=0 to infinity and you get 2/3.

[pretend]

because reflipping the coin doesn't create a 2/3 or a 1/3 possibility.

the probability on any given flip is 1/4 or 1/2 for HT/TH and HH.

each flip is an independent event and the probability does not converge to 1/3 or 2/3 as the number of flips approaches infinity.

Or does it...

http://jsfiddle.net/gGNsU/

(Chrome recommended, click Run at top)

[LegosInMyEgos]

Or does it...

http://jsfiddle.net/gGNsU/

(Chrome recommended, click Run at top)

No it doesn't.

The function you plugged into the program solved the riddle by accident on your part.

The variables are not treated as independent in your program, but in actuality each flip is an independent event and the probability NEVER actually gets to 2/3 or 1/3

[sawoobley]

Two people take turns flipping a coin and the first one to get heads wins.

[LegosInMyEgos]

Two people take turns flipping a coin and the first one to get heads wins.

Winner

smartest nut miscer of the year award goes to sawoobley.

explain it to the other dumb dumb miscers if you would be so kind.

[desslok]

Just cuz we didn't get *your* answer, doesn't mean we didn't get it. Maybe you didn't phrase it properly to match the answer you have in mind.p

This^ my answer covered it according to what was written. But I'm sure the OP had undisclosed stipulations that only he had in mind from when he learned it in class this week.