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Topic: Simple Dice Game (Read 654 times) |
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Given the following rules for a simple dice game :
Player1 puts X dollars into the pot. Player2 puts twice that amount (2X dollars) into the pot. Player1 then rolls one 6-sided die. This number becomes the "point." Player2 then guesses if their next roll will be higher or lower than the "point". After announcing their guess, Player2 rolls the 6-sided die. If they guessed correctly, they get the pot, if they guessed incorrectly, or if the numbers are the same then player1 gets the pot.
Question 1: Assuming Player2 calls higher/lower with the best chance of winning, then which player has the best odds, or is this a fair game?
Question 2: If the die rolled is weighted so that it lands on the number "3" twice as often as any other number. Then is it a fair game? If not then which player is favored to win more money?
Question 3: If the die rolled is weighted so that it lands on the number "6" twice as often as any other number. Then is it a fair game? If not then which player is favored to win more money?
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Lightboxes
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Re: Simple Dice Game
« Reply #1 on: Dec 5th, 2003, 10:52am » |
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Well well.
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1) The second player has to win 66.66% of the time OR more to NOT lose money because for every game the player loses, the player will lose 2 dollars. For every game the player wins, the player will win 1 dollar. To even out, the player must win twice as much as the player loses.
Doing some calculations with tree diagrams:
The second player will win 66.66% of the time when the game is played a lot. So, I would say the game is fair.
2) The second player will win only 65.3061224% of the time because the #3 on the die now has a P(2/7) whereas all the others 1,2,4,5,6 have P(1/7). This gives the new tree, 32/49 chance of winning. Player one is in favor of winning.
3) (I want to try guessing to work on my skills) without doing the calcuations...I say player 2 will win, around, 87% of the time.
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| « Last Edit: Dec 5th, 2003, 10:52am by Lightboxes » |
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TenaliRaman
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Re: Simple Dice Game
« Reply #2 on: Dec 5th, 2003, 11:37am » |
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on Dec 5th, 2003, 9:19am, guest wrote:
.... If they guessed correctly, they get the pot....
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how much does each of them get in this case?
If i were player 2 i would never play this game in the first place.
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ThudnBlunder
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Re: Simple Dice Game
« Reply #3 on: Dec 6th, 2003, 10:11pm » |
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1) For the possible throws 1, 2, 3, 4, 5, and 6 chances of Player 2 calling correctly
are 5/6, 4/6, 3/6, 3/6, 4/6, and 5/6, respectively.
So average probability = 24/36 = 2/3
Hence the game is fair.
2) For the possible throws 1, 2, 3, 3, 4, 5, and 6 chances of Player 2 calling correctly
are 6/7, 5/7, 3/7, 3/7, 4/7, 5/7, and 6/7, respectively.
So average probability = 32/49 < 2/3
Hence the game favours Player 1.
3) For the possible throws 1, 2, 3, 4, 5, 6, and 6 chances of Player 2 calling correctly
are 6/7, 5/7, 4/7, 3/7, 4/7, 5/7, and 5/7, respectively.
So average probability = 32/49 < 2/3
Hence the game still favours Player 1.
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| « Last Edit: Dec 6th, 2003, 10:11pm by ThudnBlunder » |
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guest
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Very good - Now here is an interesting twist. (and a little more difficult question)
Suppose you are player2, and that you can make a loaded die that is weighted in any manner you choose. How would you make a die that gives you the best advantage in the game. And what would your best odds be.
Give a ratio (or percentage) for each of the numbers to be rolled on average.
For example, a fair die would be:
rolling a 1: 1/6 (~16.67%)
rolling a 2: 1/6 (~16.67%)
rolling a 3: 1/6 (~16.67%)
rolling a 4: 1/6 (~16.67%)
rolling a 5: 1/6 (~16.67%)
rolling a 6: 1/6 (~16.67%)
P.S.
TenaliRaman,
I guess I wasn't to clear about it. There would be 3X dollars in the pot (1X from player1 and 2X from player2)
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ThudnBlunder
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Re: Simple Dice Game
« Reply #5 on: Dec 7th, 2003, 4:43am » |
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Quote:| that is weighted in any manner you choose. |
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Rolling a 1: 1 (100%)
Rolling a 2: 0 (0%)
Rolling a 3: 0 (0%)
Rolling a 4: 0 (0%)
Rolling a 5: 0 (0%)
Rolling a 6: 0 (0%)
Player 2's chances of winning are now 5/6.
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| « Last Edit: Dec 7th, 2003, 9:11am by ThudnBlunder » |
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towr
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Re: Simple Dice Game
« Reply #6 on: Dec 7th, 2003, 7:40am » |
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How do you figure his chance of winning is 5/6th?
If he throws with your die he allways throws 1, so it's never higher, nor ever lower, so he can never correctly call higher/lower, and win.
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ThudnBlunder
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Re: Simple Dice Game
« Reply #7 on: Dec 7th, 2003, 8:22am » |
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on Dec 7th, 2003, 7:40am, towr wrote:How do you figure his chance of winning is 5/6th?
If he throws with your die he allways throws 1, so it's never higher, nor ever lower, so he can never correctly call higher/lower, and win. |
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Quite right, towr.  At least it's now a fair game. 
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Eigenray
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Re: Simple Dice Game
« Reply #8 on: Dec 8th, 2003, 2:13am » |
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Well, if player 2 calls higher on 1,2,3, and lower on 4,5,6, then the probability of winning is:
(1 - [sum]xi2)/2 + S(1-S),
where S = x1 + x2 + x3.
Setting partial derivatives to 0 shows that this is maximized when x1=...=x6=1/6. This is essentially because if [sum]xi=1 is constant, [sum]xi2 is minimized when the xi are equal, and S(1-S) is simultaneously maximized when S=1/2.
This doesn't take care of the cases when, for example, x1+x2 > x4+x5+x6, in which player 2 should call lower on 3.
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James Fingas
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Re: Simple Dice Game
« Reply #9 on: Dec 10th, 2003, 7:48am » |
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on Dec 8th, 2003, 2:13am, Eigenray wrote:| This doesn't take care of the cases when, for example, x1+x2 > x4+x5+x6, in which player 2 should call lower on 3. |
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I think this is easily generalized from your answer. If the distribution is such that we call higher on 1,2 and lower on 3,4,5,6, then we define S = x1 + x2. Nothing else about the equation need change. Note that again S(1-S) is maximized for S = 1/2, but the expectation of rolling the same thing twice will be higher than it was before.
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Eigenray
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Re: Simple Dice Game
« Reply #10 on: Dec 10th, 2003, 11:40pm » |
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I guess those cases aren't as annoying as I'd thought they'd be.
If you call lower on 3+, and higher on 2-, then looking at the equation it's clear that the probability of winning won't decrease if we make x1=x2, x3=...=x6, while keeping S=x1+x2 fixed. Then there's really only one variable to optimize over, giving x1=x2=5/22, x3=...=x6=3/22, for a probability of 29/44 < 2/3.
If we call lower on 2+, and higher on 1, we maximize the probability at x1=3/8, x2=...=x6=1/8, to get a probability 5/8 < 2/3.
Note that the optimal die weighting for this strategy has x1 < x3+...+x6, so a different strategy would actually be better, but this doesn't matter. It's sufficient to show that the strategy of lower on 2+ never wins more than 5/8 of the time for any die weighting, including those where another strategy would make more sense.
It's easily demonstrated that the optimal strategy for any die is one of the above three (or a symmetrically equivalent one). It follows that the die whose optimal strategy gives the best chance of winning is the fair die.
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