[Guest]

You are given two dice. One is "loaded" so that '6' appears 50% of the time, and the other numbers (1,2,3,4,5) appear 50% of the time with equal probability. I will pay $1 for every point you roll (e.g. if you roll a 2 and 3, I will pay you $5). What would you pay for the right to roll the set of dice? Explain your answer.

[Guest]

I would pay no more than $2...which is the minimum amount I could win for playing.

Of course you are probably looking for the highest amount that sometone would pay based on the probability of a "safe" risk. To that I would say, uh, I don't know.

[Guest]

I'd pay 5 bucks, because I'd feel good about making money... with 2 or 3 rolls of course

[Guest]

$7.99 is a winning wager (on average).

Average payout $8.00

prob/n die 1+ prob/n die 2

45/10 = $4.50 + 21/6 = $3.50

[Guest]

$7.99, assuming risk neutrality.

expected value on loaded die should be $4.50, right?

.5*6 + .1*5 + .1*4 + .1*3 + .1*2 + .1*1 = 4.5

expected value on the second die should be:

(1/6)*6 + (1/6)*5 + (1/6)*4 + (1/6)*3 + (1/6)*2 + (1/6)*1 = 3.5.

total expected value then should be $8.00, right?

Edited August 26, 2008 by HotMarmalade

[Guest]

$7.99, assuming risk neutrality.

expected value on loaded die should be $4.50, right?

.5*6 + .1*5 + .1*4 + .1*3 + .1*2 + .1*1 = 4.5

expected value on the second die should be:

(1/6)*6 + (1/6)*5 + (1/6)*4 + (1/6)*3 + (1/6)*2 + (1/6)*1 = 3.5.

total expected value then should be $8.00, right?

Thanks for your responses. How did you get the $7.99 risk neutral?

[Guest]

You are given two dice. One is "loaded" so that '6' appears 50% of the time, and the other numbers (1,2,3,4,5) appear 50% of the time with equal probability. I will pay $1 for every point you roll (e.g. if you roll a 2 and 3, I will pay you $5). What would you pay for the right to roll the set of dice? Explain your answer.

I would not trust any one who said they gave me a loaded dice - intresting pzzle though presume it's higest bid

equalize for 0/0 return, and see if I have luck on my side

that's either $7 or $8 not sure, actually I am - $8 is the wiining bet according to others - I win the roll - yeah me

Edited August 26, 2008 by Lost in space

[Guest]

$8

unless

The OP doesn't specify that the other die is fair. If that one is loaded as well, all bets are off.

[soop]

Hah, I'd pay $1 max. That way I ensure I get more profit than you guys.

[Guest]

Hah, I'd pay $1 max. That way I ensure I get more profit than you guys.

That's not at all the case. If the price is set at $1.50, your profit will be $0 because you will not play. Anyone who plays would profit a minimum of $.50, with an expected (average) profit of $6.50.

[Guest]

i won't pay anything. Gambling just isn't my style

[unreality]

The way the OP is stated, it sounds like he will pay you out no matter what, so what would I pay? $0, of course, if he'll do it no matter what! If the OP is asking for 'what would you pay in advance to make it an even game', it would be $8, right?

[Guest]

$7.99 is a winning wager (on average).

Average payout $8.00

prob/n die 1+ prob/n die 2

45/10 = $4.50 + 21/6 = $3.50

agree with maths, $8.00 is average return. "How much would you pay?" is a separate question relating to how much return per roll do you consider sufficient to make the bet. For me, $0.01 is not enough. Maybe $0.50?

[Guest]

i won't pay anything. Gambling just isn't my style

I too am against gambeling. But a "gamble" is a risk, while this could be taken advantage of has a trustworthy moneymaker. Odds are impossible to deny in the long run.