Probability game help

plopz

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Joined
Dec 17, 2014
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Cost of the game: $50
2 players
A player can either bet 10 or 20 dollars
The winner will gain $50 and the additional amount they bet and vice versa for losers (lose $50 and the amount they bet)
The player that rolls the higher dice wins (each player rolls only 1 dice)
If both players roll the same dice, they both lose the amount they bet
The player with the most money at the end will win.
The winner at the end receive the money they earned and an additional $50.
For example:
If a player places a $20 bet, they have a chance of winning or losing $20.
If a player places a $10 bet, they have a chance of winning or losing $10.

Made this game but can someone point me in the right direction?: Is this a binomial/discrete distribution and how to calculate theoretical/experimental probability distributions, and expected value. Please and Thanks, I am unsure. Can someone link me to a similar game to this?
 
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Cost of the game: $50
2 players
A player can either bet 10 or 20 dollars
The winner will gain $50 and the additional amount they bet and vice versa for losers (lose $50 and the amount they bet)
The player that rolls the higher dice wins (each player rolls only 1 dice)
If both players roll the same dice, they both lose the amount they bet
The player with the most money at the end will win.
The winner at the end receive the money they earned and an additional $50.
For example:
If a player places a $20 bet, they have a chance of winning or losing $20.
No, they have chances of winning $70 or of losing either $20 or $70.

If a player places a $10 bet, they have a chance of winning or losing $10.
No, they have chances of winning $60 or of losing either $10 or $60.

Made this game but can someone point me in the right direction?: Is this a binomial/discrete distribution and how to calculate theoretical/experimental probability distributions, and expected value. Please and Thanks, I am unsure. Can someone link me to a similar game to this?
You say "The player with the most money at the end will win.
The winner at the end receive the money they earned and an additional $50."
so how much money a person wins depends on when the game "ends". How is that decided. No, this is not a "binomial/discrete distribution" because it is not a "zero sum game"- at the end of the game, no matter how much money has changed hands during the game, the two people playing it will have, in total, $50 more than they started with. Where does that money come from?
 
No, they have chances of winning $70 or of losing either $20 or $70.


No, they have chances of winning $60 or of losing either $10 or $60.


You say "The player with the most money at the end will win.
The winner at the end receive the money they earned and an additional $50."
so how much money a person wins depends on when the game "ends". How is that decided. No, this is not a "binomial/discrete distribution" because it is not a "zero sum game"- at the end of the game, no matter how much money has changed hands during the game, the two people playing it will have, in total, $50 more than they started with. Where does that money come from?
thanks for your reply, if i changed it to only winning or losing 10/20 rather than 60/70, then would it be binomial/discrete distribution? And also I forgot to include that each game is 10 rounds.
 
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