[Guest]

as before, alex has a board with 6 squares,

+---+---+---+

| 1 | 2 | 3 |

+---+---+---+

| 4 | 5 | 6 |

+---+---+---+

he'll roll 3 dice. if you get 1 of the number you bet on, you get double back,

if you get 2 of the number, 3 times back, and 3 of the number 4 times back.

you can only bet on a maximum of three numbers.

as an additional rule however, there are 20 squares around the edge of the board.

every time you get paid, you move around the board, equal to the money multiplier.

if you lose any amount of money you must either start over, or stop playing all together.

if you get all the way around the board, however, alex will pay out 20 times the amount bet for the whole round.

1) now is it worth while to play the game?

2) if not, how much should the pay out be?

3)say you have 1,000 dollars, and the pay out for making it around the board is what it should be to make it a fair game,

how much should you bet on each roll to maximize your winnings?

[bushindo]

as before, alex has a board with 6 squares,

+---+---+---+

| 1 | 2 | 3 |

+---+---+---+

| 4 | 5 | 6 |

+---+---+---+

he'll roll 3 dice. if you get 1 of the number you bet on, you get double back,

if you get 2 of the number, 3 times back, and 3 of the number 4 times back.

you can only bet on a maximum of three numbers.

as an additional rule however, there are 20 squares around the edge of the board.

every time you get paid, you move around the board, equal to the money multiplier.

if you lose any amount of money you must either start over, or stop playing all together.

if you get all the way around the board, however, alex will pay out 20 times the amount bet for the whole round.

1) now is it worth while to play the game?

2) if not, how much should the pay out be?

3)say you have 1,000 dollars, and the pay out for making it around the board is what it should be to make it a fair game,

how much should you bet on each roll to maximize your winnings?

I would like some clarification

1) You can only bet on a maximum of 3 numbers, but are you required to bet on all 3 number?

2) When you get to the end of the board, are you required to wait until you get the correct multipler before you finish the round? For instance, you're at square 18, you'll need to wait for a 2x multiplier until you get around the board.

3) What do you mean by '20 times the amount bet for the whole round', does that mean Alex will pay 20 times the cummulative amount that you bet so far in the game? Or does that mean that Alex will pay 20 times the bet on the winning round that allows you to complete the 20 squares?

[Guest]

Another clarification needed:

Suppose you bet on 1,2 &3 and all 3 dice show 1. How is this repaid? 2x,3x or 4x?

[Guest]

I would like some clarification

1) You can only bet on a maximum of 3 numbers, but are you required to bet on all 3 number?

no you can bet on a single number if you wish.

2) When you get to the end of the board, are you required to wait until you get the correct multipler before you finish the round? For instance, you're at square 18, you'll need to wait for a 2x multiplier until you get around the board.

no, if your two squares away and get 3x mltiplier, that also puts you around the board.

3) What do you mean by '20 times the amount bet for the whole round', does that mean Alex will pay 20 times the cummulative amount that you bet so far in the game? Or does that mean that Alex will pay 20 times the bet on the winning round that allows you to complete the 20 squares?

20* the cumualuative amount for the whole round.

that is lets say you are extreamly lucky. you bet 20 dollars 5 times and each time get the 4 times multiplier. you would get an additional 20*100 = 2000 dollars back for your fifth and final roll. but if you lose any money at any point, you must start over, both in total bet and in board position.

4)

Another clarification needed:

Suppose you bet on 1,2 &3 and all 3 dice show 1. How is this repaid? 2x,3x or 4x?

you get 4x whatever you bet on the three 1's but lose whatever you bet on 2 and 3.

your move number would be 2 squares, assuming all three had an equal number of bet.

if you got 1,2,3 as the roll, you would double your money on all 3 squares, and the move number would be 6.

Edited August 2, 2009 by phillip1882

[Guest]

1st off, don’t try to sell this game since the rules are way to confusing. If I understand right, the # squares you move is based on the SUM of the multipliers where a losing number counts as a -1? This is my guess based on your clarification #4 if the roll is 1,1,1 and you bet (equally) on 1,2,3 then you get 2 squares. So that would be 4 back on 1 and a -1 for 2 and a -1 for 3 for a net of 2? Also you say if 1,2,3 are bet and come up then you move 6 (i.e. 2+2+2). So when do you go back to start? I assume that would be when your net $ for the roll is a loss (i.e. bet on 1,2,3 and a 1,5,6 come up so you get 2 but lost 3), right?

As for is 200x fair, is that based on the optimum betting or equal betting?

Here are the probabilities for the payouts on 1,2, or 3 bets: Bet on #6 then probablity you get 1 #6 is 1/6*5/6*5/6 + 5/6 * 1/6 * 5/6 + 5/6 * 5/6 * 1/6 = 75 / 216 And that you get 2 #6s is (1/6*1/6*5/6) * 3 = 15/216 And that you get 3 #6s is 1/6*1/6*1/6 = 1/216 And 0 #6s is 5/6*5/6*5/6 = 125/216 Now ROI is 2*75 + 3*15 + 4*1 = 199/216 which is <1 so on every $216 you bet you only get back $199 But if you bet on 2#s (i.e. #5 & #6) then: Get only 1 of the 2#s is (2/6*4/6*4/6 )*3 = 96/216 Get 2 diff #s is (2/6*1/6*4/6) * 3 = 24/216 Get 2 same #s is same as above = 24/216 Get 2 same & 1 the other (is prob that you get 1 of the 2 for each die – prob that you get all 3 the same) = 2/6*2/6*2/6 – 2/216 = 6/216 Get 3 same for either is 2/6*1/6*1/6 = 2/216 Get neither of the #s is 4/6*4/6*4/6 = 64/216 Now ROI assuming = bets on each is 2*96+(2+2)*24+3*24+(2+3)*6+4*2 = 398/432 Now betting on 3# Get only 1 is 3/6*3/6*3/6 * 3 = 81/216 Get 2 same #s is 3/6 * 1/6 * 3/6 * 3 = 27/216 Get 3 same #s is 3/6 * 1/6 * 1/6 = 3/216 Get 2 diff #s is = (3/6*2/6*3/6)*3 = 54/216 Get 2 same and 1 diff = 3/6*1/6*2/6+3/6*2/6*2/6 = 18/216 Get 3 different = 3/6*2/6*1/6 = 6/216 Get none of the 3 is 3/6*3/6*3/6 = 27/216 Now ROI is 2*81+3*27+4*3+(2+2)*54+(3+2)*18+(2+2+2)*6 = 597/648 NOTICE how the sum of the individual parts add up to 216 in each case and how betting on 1, 2, or 3 doesn’t change your ROI payout. BUT now if I am right that you go back to start if you have a net loss for the roll, then you should: BET on 2 since all you need is 1 to come up and you break even (i.e. don’t lose) and move 1 square (I think). The probabilities of ‘staying alive’ for 1 bet = 91/216, for 2 is 152/216, and 3 is 108/216. This is because if you bet on 3 then getting only 1 is not enough (you bet 3 and get back 2 and have a net loss of 1). Now the probability of getting around the board using 2 bets and based on my assumed movement using the addition of each multiplier and a -1 for not getting a number is: Average step size: (96*1step+24*4steps+24*2steps+6*5steps+2*3steps)/(96+24+24+6+2) = 276/152 = 1.8 So it will take 20/1.8 number of rolls = 3040/276 = 11 rolls Now the probability that you can make 11 rolls without busting is (152/216)^11 = 0.02 Now the best betting is the bare minimum all the way around until you are possibly going to cross the finish and then it would depend on how many squares from the finish you end up the turn before. Then you have to budget your remaining money to maximize your return if you cross but leave enough if you don’t. A gutsy play and easy play is to bet all but all but 4 times the # squares left as soon as you are within 5 away so that you can have enough to play those last rolls but guarantee all the max money has been played.

[HoustonHokie]

Are you sure this is the same Alex? Offering 20 times the cumulative bet is, I think, a big mistake for him. Here's why:

The ROI for any turn, as Doctor Moshe pointed out, is always the same, no matter how many numbers you bet on (assuming you place equal bets on all numbers). That ROI is 199/216. Stated another way, you expect to lose about 8 cents for every dollar you bet (1 - 199/216 = 0.08). So, if I bet a dollar each time, whether its all on one number, or fifty cents on 2, or 33 cents on 3, I expect to lose 8 cents with each roll of the dice.

If I bet on 2 numbers with each roll, my probability of getting all the way around the board in one "streak" is also calculated by Doctor Moshe. It's 0.02, which means that I should get around the board after 1/0.02 = 50 streaks. Each streak is, on average, less than 11 rolls, but let's say that they are 11 just for simplicity.

In those 50 streaks, if I keep placing one dollar bets, I expect to lose 49 * 11 * 0.08 = 43 dollars. But then I get all the way around the board, and it takes me about 11 turns to do it. I've placed 11 one-dollar bets in that time, and Alex is going to give me 20 * 11 = 220 dollars when I get around the board. So I've lost 43 dollars (or less), but I'll make 220!!

I simulated 1000 games in a spreadsheet, and the expected winnings pan out. If I start with $1000 and keep placing $0.50 on 1 and $0.50 on 2, I end up with about $1185 on average. The lowest holdings I had in any of those games was $865. So I figured I could increase my bets. I tried $5 dollars ($2.50 each on 1 & 2), and ended up with an average of $1920 at the end, with a low of $375. Increasing again to $10, I ended up with $2835 on average, but in one of the 1000 games, I lost everything before I could make it around the board. So I think a $10 bet each time, split evenly on 2 numbers, would be a pretty safe way to increase your holdings about three fold each game that Alex lets you play. You could probably do even better if you increased your bet amount as you got further around the board.

Alex should stick with the original version of the game.