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High or low?


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11 replies to this topic

#1 phillip1882

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Posted 11 January 2012 - 10:14 PM

looking for a good gambling game, you come upon a merchant who offers what seems to be a fair game called high or low.
the rules are this. he'll roll 3 dice. then he'll bet that you either get a total higher than him, or lower than him after also rolling 3 dice. if you bet the same way, and are right it's +50% of your money. if wrong, then you lose 50%. if you bet the opposite way, then its a double or nothing bet. if the result is a draw, then nothing is won or lost by either player.
as an example. he rolls 1,2,4 and bets that you get higher than him. you stake 2 dollars that you also get higher.
if right, you would now have 3 dollars. how fair is this game?
(what's your expected earnings/losses ?)
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#2 CaptainEd

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Posted 11 January 2012 - 10:47 PM

I think I'm clear on the rules, but let me play back a couple of details:
Merchant rolls, chooses a direction (high or low).
If you bet in that direction, and roll a number further in that direction, you get +50%; if you roll same total, it's a draw, you roll total in opposite direction, you get -50%.
If you bet in opposite direction, and roll a number in your direction, you get +100%, if you roll same total, it's a draw, you roll total in same direction, you get 0%.
Is this true?
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#3 phillip1882

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Posted 11 January 2012 - 10:55 PM

exactly. also i should clarify something. you need to place a stake before the merchant bets. i.e. you place a stake of 2 dollars, and the merchant rolls and bets high or low, then you bet high or low. that is you must risk something in order to have the merchant roll.
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#4 Molly Mae

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Posted 11 January 2012 - 11:38 PM

Spoiler for I didn't find the catch

Edited by Molly Mae, 11 January 2012 - 11:41 PM.

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A recipe for honey-pickled apples


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Spoiler for Molly's Rules to Live By

#5 akajaria

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Posted 12 January 2012 - 11:20 AM

Spoiler for My guess at this

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#6 phillip1882

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Posted 12 January 2012 - 01:29 PM

Spoiler for Pretty sure you're right

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#7 ak4su

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Posted 12 January 2012 - 08:33 PM

The merchant is always at the disadvantage.

Actually it's always a double or nothing bet, just the amount of stake is halved if you bet the same way.

Ex.- you stake a 2$ bet. Win the game you get 3$, loose and you get back 1$.
It's equivalent to staking 1$ for a double or nothing bet.

Since you are able to know the sum of dices of the merchant, your probability of winning are always > more than .5
So your expected earnings are aways +ve

Calculating the expected earnings is probably a lot of calculation.

Edited by ak4su, 12 January 2012 - 08:38 PM.

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#8 Molly Mae

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Posted 12 January 2012 - 09:03 PM

Spoiler for I donno...

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A recipe for honey-pickled apples


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#9 ak4su

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Posted 12 January 2012 - 09:45 PM

Your chances of loosing - (2*(1*0 + 2*1 +3*3 + 4*6 + 5*10) + 6*15)/ 36^2 = .200 when you know the sum of previous roll.

Draw - (2(1*1 + 2*2 + 3*3 + 4*4 + 5*5) + 6*6)/36^2 = .113

Edited by ak4su, 12 January 2012 - 09:55 PM.

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#10 ak4su

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Posted 12 January 2012 - 10:11 PM

Winning (2*(1*35 + 2*34 + 3*30 + 4*26 + 5*21) + 6*15)/ 36^2 = .689

Placin a bet of 2$ would mean that on winning he would get 1$ on winning and 1$ on loosing

expected earning =
1*.685 - 1*.113

It has been assumed that if the sum on the first roll is above 7 you choose low and vice versa.

Edited by ak4su, 12 January 2012 - 10:17 PM.

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