what is the fair payout for this game?

Question

theres this game. you pick a three digit number whose digits must be from 1-6 and then you roll 3 dice and if your digits are on the dice [any order cuz you roll them all at once] then you win some money. it costs a dollar to play this game and the question is to determine how much you win. fair payout so that you're expected to lose as much as you gain.

Let it be so that the game is only 1 game and there is only one prize amt. of money for winning. So you can't say that there are differen't payoffs for different cases.

Answer 1

There would have to be different payouts depending on the combination of numbers picked.

FOR 3 NUMBERS THE SAME:

For any particular combination of all 3 numbers the same, there is only one way to get that combination.
111, 222, 333, 444, 555, 666.

Thus you're expected to win 1/216 of the time. So for that the payout should be $216.

FOR TWO NUMBERS THE SAME:

For one pair of numbers (e.g. 11) and one other digit, there are only 3 ways to arrange the numbers:
112, 121, 211.

3/216 = 1/72. So in this case you should be paid $72.

FOR ALL 3 NUMBERS DIFFERENT

In this case, there are 6 ways to win:
123, 132, 213, 231, 312, 321.

Since you are expected to win 6/216 (or 1/36), the payout should be $36.

(By the way, in terms of payout, casinos will take your $1 bet regardless of whether you win or lose. So if you bet $1, the payout is $216, but really $1 of that was your original bet. As you can see it will even out in the end. 215 losses = $215 lost. 1 win = $216 payout on a $1 bet = $215 won.)

If you want to express it as odds, case #1 is 215 to 1, case #2 is 71 to 1 and case #3 is 35 to 1...

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Going with your new stipulation of only one payout... there have to be some assumptions on how the player can get their numbers. If they can pick it themselves, they would be smart to pick different numbers since that happens twice as often as two numbers the same and six times as often as all the same. Anyone that bets all three numbers the same is unfamiliar with statistics.

Let's assume instead the number is picked randomly for you.
6 of the numbers will yield you a 1/216 chance.
90 of the numbers will yield you a 3/216 chance.
120 of the numbers will yield you a 6/216 chance.

So the weighted payout should be:
6/216 * 216 = $6
90/216 * 72 = $30
120/216 * 36 = $20

This would be a total of $56 paid out on a $1 bet. (Or $55 winnings, if you want to express it that way).

If this is the answer desired by your teacher, you now have several confirmations on the amount. If you want the easiest way to get there, don't choose my roundabout way however. Jim's method is much clearer.

Again, I disagree with this concept of a single payout, unless the user is forced to accept the numbers randomly. A smart player will pick different numbers and win more than expected each time. You won't see that game in Vegas any time soon...

Answer 2

I guess there would be:

* 6 ways to choose all 3 digits the same
* 6 x 5 = 30 ways to choose 2 digits the same and one digit different (6 choices for the doubled digit and 5 for the different one), and
* 6C3 = (6x5x4)/(3x2x1) = 20 ways to choose 3 different numbers where order makes no difference.

That's a total of 6 + 30 + 20 = 56 possible outcomes, so your choices of winning are 1 in 56.

So I guess a win should net you $55.

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Puzzling's answer is the best to make it a fair game. Both I and math_ninja would be assuming that each outcome would be equally likely, and as Puzzling points out, that's not true...You would be more likely than 1/56 to win if you went with 3 different numbers than 3 numbers that were the same.

Although I think he should have subtracted $1 off each payout.

OK, he's right again. :P never mind.

Answer 3

Your payout will correspond to the number of UNIQUE dice rolls possible. To figure that out, write the three dice possibilities in acsending order.
111,112,113....116
122,123,124...126
133....
...
222....
233....
...
666

You get 21+15+10+6+3+1 = 56.
So you have a 1 in 56 chance of guessing the dice combination.
Solve for x being the payout, to be a fair game wins and losses equal zero, so

1/56 x - 55/56 = 0. Payout = 55.

Answer 4

Well, no matter what you pick, the digits have to match the dice exactly, and the chance of that happening is (1/6)^3 or 1/216... So a fair payout would be $216.