High or low?

[Guest]

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 ?)

[CaptainEd]

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?

[Guest]

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.

[Molly Mae]

If he rolls a 4 and picks higher, you should naturally pick higher. Per the OP, though, if you roll three 1's, you still lose 50% since you were wrong. (For clarification: Would the above scenario lose you money or be considered a draw since nobody picked correctly?)

Rethink Edit: Mm...maybe that's not so unfair, though, since you win if you both choose the same and you're right.

Edited January 11, 2012 by Molly Mae

[Guest]

I feel I have the advantage - as one can know what odds are there for beating the dealer. If the dealer picks the side with the higher probability, i still have the option to go with him and win 50%. If he picks the wrong side, then I have a double advantage. Assuming probabilities play out.

[Guest]

yeah i ran several simulations, and it seems if the player bets high if M gets 10 or less, and low if M gets 11 or more, no matter what M bets on, player wins. (has an advantage)

for some odd reason, i was thinking that this game was to the merchant's advantage as going from 2 dollars to 3 dollars is not the same as going from 2 dollars to 1 dollar. that is if the player won and lost an equal number of times, the player should lose money.

[Guest]

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 January 12, 2012 by ak4su

[Molly Mae]

We can assume that there is no double or nothing bet and that both the player and the merchant will choose the side with best odds.

You are certainly at an advantage to win any single game. Is it enough to overcome the overwhelming losses, though? Starting with $10, if you play 3 games and win two, you'll end up positive (but not by much). Because the number isn't a random number from 1-18, though, the focus of the rolls will be between 9 and 12. That will drastically reduce your odds of winning by making the margins closer to equal.

If there is an advantage for the player, it isn't by much.

[Guest]

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 January 12, 2012 by ak4su

[Guest]

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 January 12, 2012 by ak4su

[Guest]

This is not so hard, I assume you can see what he rolled.

If he plays correct he will on a 3,4,5,6,7,8,9,10 give you the higher and on a 11,12,13,14,15,16,17,18 give you the lower

(so you end up playing for 50% of your stake).

You will always follow him and have a pretty +EV on those bets:

3: 1/216 * (215/216*1+1/216*0)

4: 3/216 * (212/216*1+3/216*0+1/216*-1)

5: 6/216 * (206/216*1+6/216*0+4/216*-1)

6: 10/216 * (196/216*1+10/216*0+10/216*-1)

7: 15/216 * (181/216*1+15/216*0+20/216*-1)

8: 21/216 * (160/216*1+21/216*0+35/216*-1)

9: 25/216 * (135/216*1+25/216*0+56/216*-1)

10: 27/216 * (108/216*1+27/216*0+81/216*-1)

It makes a total of 0.25 which you have to multiply by 2 (adding 11 to 18).

This gives you an EV = +0.5

So if you stake $2, you play for $1 and your EV=$0.5

It would be a little tougher when you don't know what he rolled ... Altough it won't be difficult to find a nash equilibrium.

Edited January 12, 2012 by Fish11

[plainglazed]

I get the same as Fish11 assuming his expected value is for a $2 bet.