New game with dice. Never seen before.
July 13th, 2014 at 2:25:31 AM
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There is a guy in a local bar who offers a game with 2 6 sided dice. I want to know the odds of the game and if there is a way to win in the long term.
Rules.
Before each roll you have to say 2 numbers between 1 and 6 but different numbers (eg: 1 and 2 but not 1 and 1 etc).
You then roll the 2 dice at the same time. If one of the 2 numbers you selected comes out you lose. (eg: You say 3 and 6 and 2 and 6 come out you lose as 6 appeared.. If 1 and 5 comes out none of your numbers appear so you beat the first roll).
After each roll he will offer you a payout to walk away or you can gamble and try and beat the next roll to increase your payout. If at any time you lose a roll, you lose your initial bet and obviously he will not pay you any money. You can walk at any time after each roll and take the current payout.
Payouts.
Beat 1 roll: Even money
Beat 2 rolls: 3 to 1
Beat 3 rolls: 8 to 1
Beat 4 rolls: 20 to 1
Beat 5 rolls: 40 to 1
Beat 6 rolls: 100 to 1
Example.
I bet $10 and beat the first roll, he will offer me $10 to walk away now. So I make $10 and keep my $10 initial bet. If I walk, game over and done. If I continue I roll again and win, he will offer me $30 dollars plus my bet to walk away now and so on. If at any time I lose a roll, I instantly lose my initial $10 stake and obviously will not be paid any money. It's a nervey game, you bet $10 and have beaten 4 rolls and the offer is 20 to 1 so $200 to walk now. But one more roll and you double to $400. Gamble and double my potential winnings or take?
Anyways, what are the stats to this game. Is it good for him or the player?
Rules.
Before each roll you have to say 2 numbers between 1 and 6 but different numbers (eg: 1 and 2 but not 1 and 1 etc).
You then roll the 2 dice at the same time. If one of the 2 numbers you selected comes out you lose. (eg: You say 3 and 6 and 2 and 6 come out you lose as 6 appeared.. If 1 and 5 comes out none of your numbers appear so you beat the first roll).
After each roll he will offer you a payout to walk away or you can gamble and try and beat the next roll to increase your payout. If at any time you lose a roll, you lose your initial bet and obviously he will not pay you any money. You can walk at any time after each roll and take the current payout.
Payouts.
Beat 1 roll: Even money
Beat 2 rolls: 3 to 1
Beat 3 rolls: 8 to 1
Beat 4 rolls: 20 to 1
Beat 5 rolls: 40 to 1
Beat 6 rolls: 100 to 1
Example.
I bet $10 and beat the first roll, he will offer me $10 to walk away now. So I make $10 and keep my $10 initial bet. If I walk, game over and done. If I continue I roll again and win, he will offer me $30 dollars plus my bet to walk away now and so on. If at any time I lose a roll, I instantly lose my initial $10 stake and obviously will not be paid any money. It's a nervey game, you bet $10 and have beaten 4 rolls and the offer is 20 to 1 so $200 to walk now. But one more roll and you double to $400. Gamble and double my potential winnings or take?
Anyways, what are the stats to this game. Is it good for him or the player?
July 13th, 2014 at 3:49:24 AM
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Probability of beating 1 roll: (4/6)^-2 = 1 in 2.25
Probability of beating 2 rolls: (4/6)^-4 = 1 in 5.06
Probability of beating 3 rolls: (4/6)^-6 = 1 in 11.39
Probability of beating 4 rolls: (4/6)^-8 = 1 in 25.63
Probability of beating 5 rolls: (4/6)^-10 = 1 in 57.67
Probability of beating 6 rolls: (4/6)^-12 = 1 in 129.75
The payouts get worse if you keep rolling and only a sucker would play.
Probability of beating 2 rolls: (4/6)^-4 = 1 in 5.06
Probability of beating 3 rolls: (4/6)^-6 = 1 in 11.39
Probability of beating 4 rolls: (4/6)^-8 = 1 in 25.63
Probability of beating 5 rolls: (4/6)^-10 = 1 in 57.67
Probability of beating 6 rolls: (4/6)^-12 = 1 in 129.75
The payouts get worse if you keep rolling and only a sucker would play.
July 13th, 2014 at 3:58:27 AM
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i hope someone else can prove whether what i am saying is right or not, but i still think this is correct
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2/3 of the time you will not get hit on dice (1) and the % of the time that you are not hit on dice (1) you will also not get hit on dice (2) 2/3 of the time,
2/3 x 2/3 = 44.44%
therefore you will not get hit 44.44..% of the time, so the first roll will be a losing roll for the player 55.55..% of the time
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on the first roll there is a -11.111..% player edge but the bet would win 44.44..% of the time, "so you could continue" to roll two if you wanted to
on the second roll there is a -20.98..% player edge and the bet would win 19.75...% of the time, "" to roll three if you wanted to
on the third roll there is a -20.98..% player edge and the bet would win 8.77..% of the time, "" to roll four if you wanted to
i don't think i need to prove it any further, as you can see it will not be possible to reach a positive player edge for roll's 1 to 6
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extra info below:
for the game to have a break-even point
you would have to be offered 5 to 4 on first roll to consider playing that first roll,
you would have to be offered 4.0625 to 1 on the second roll to consider playing to the second roll,
you would have to be offered at least 10.39... to 1 on the third roll to consider playing to the third roll,
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general 'break-even' formula below:
you would have to be offered at least $2.25 to the power of N, if you wanted to play to the 'Nth' roll
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again i am not sure if this response is correct, so wait until someone else proves it(or disproves it), before thinking i gave the correct answer or not
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2/3 of the time you will not get hit on dice (1) and the % of the time that you are not hit on dice (1) you will also not get hit on dice (2) 2/3 of the time,
2/3 x 2/3 = 44.44%
therefore you will not get hit 44.44..% of the time, so the first roll will be a losing roll for the player 55.55..% of the time
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on the first roll there is a -11.111..% player edge but the bet would win 44.44..% of the time, "so you could continue" to roll two if you wanted to
on the second roll there is a -20.98..% player edge and the bet would win 19.75...% of the time, "" to roll three if you wanted to
on the third roll there is a -20.98..% player edge and the bet would win 8.77..% of the time, "" to roll four if you wanted to
i don't think i need to prove it any further, as you can see it will not be possible to reach a positive player edge for roll's 1 to 6
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extra info below:
for the game to have a break-even point
you would have to be offered 5 to 4 on first roll to consider playing that first roll,
you would have to be offered 4.0625 to 1 on the second roll to consider playing to the second roll,
you would have to be offered at least 10.39... to 1 on the third roll to consider playing to the third roll,
------
general 'break-even' formula below:
you would have to be offered at least $2.25 to the power of N, if you wanted to play to the 'Nth' roll
------
again i am not sure if this response is correct, so wait until someone else proves it(or disproves it), before thinking i gave the correct answer or not
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July 13th, 2014 at 4:08:29 AM
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Quote: drjohnnyProbability of beating 1 roll: (4/6)^-2 = 1 in 2.25
Probability of beating 2 rolls: (4/6)^-4 = 1 in 5.06
Probability of beating 3 rolls: (4/6)^-6 = 1 in 11.39
Probability of beating 4 rolls: (4/6)^-8 = 1 in 25.63
Probability of beating 5 rolls: (4/6)^-10 = 1 in 57.67
Probability of beating 6 rolls: (4/6)^-12 = 1 in 129.75
The payouts get worse if you keep rolling and only a sucker would play.
looks like my answer's were right, according to drjohnny, we must have been answering at about the same time, because when i originally clicked on your post and started writing there was no response from anyone showing.
Edit @ 4:40 am
for the bit where you said "If one of the 2 numbers you selected comes out you lose.", i assumed you meant to say "if 1 or both of the 2 numbers you selected comes out you lose", for my original answer ?!?
because, for two numbers to come out there is an 11.11..% chance of that happening,
if what you said was EXACTLY what you meant, then the game would be a player edge game, with an EV of +11.11..% on the first roll
also to cover all bases, if both of the 2 numbers you chose come out is a PUSH, then you would need to go to at least three rolls, before you had a player edge at the odds you stated
July 13th, 2014 at 5:00:26 AM
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Hello ksdjdj and drjohnny. Thank you for replying, very appreciated.
If you were to select 1 and 2 for example and rolled a 1 and 5, you would lose as 1 appeared. If you rolled a 2 and 3 you lose as 2 has appeared. if you rolled a 11 you lose as one of your numbers has appeared. If you rolled a 1 and 2 you still lose as one of your numbers has appeared. To win, the 2 numbers you have selected MUST NOT appear. So if ONE or MORE of your numbers were to appear, it's a lost roll.
If you were to select 1 and 2 for example and rolled a 1 and 5, you would lose as 1 appeared. If you rolled a 2 and 3 you lose as 2 has appeared. if you rolled a 11 you lose as one of your numbers has appeared. If you rolled a 1 and 2 you still lose as one of your numbers has appeared. To win, the 2 numbers you have selected MUST NOT appear. So if ONE or MORE of your numbers were to appear, it's a lost roll.
