# Probability of rolling 6's on 3 dice - adjusted gambling odds?

I'm struggling to work out odds on a game that were working on. It's probably best if I write an example as I'm really not a mathematician!

I'm working on a dice game where the player bets 1 coin and rolls 3 dice. If any of the dice are a 6 we payout based on the following.

If 1 dice is a six we pay x If 2 are a six we pay y and if all of them are a 6 we pay z

This seemed simple at first as the odds of rolling 1 six are 1/2, 2 sixes would be 1/14 and 3 sixes would be 1/216 so they would be our odds.

The issue is that in theory would have to roll the dice 216 times to get the jackpot and receive the £216 payout (at £1 per bet). However, if I did do that I would also win 108 times at 2/1, 15 times at 14.4/1 and once at 216/1 (ignoring the luck factor etc).

Therefor for my £216 stake i would win a total of £648.

So what should my payouts be for 1x6, 2x6, and 3x6 (assuming 0% house edge and all that) to be fair to the player?

Thanks for you time.

Mark

## 2 Answers

The probabilities you have are wrong. Denote by $X$ the number of sixes rolled. Then the correct probabilities are:
$P(X = 0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$

$P(X = 1) = P$ (first dice six) $+ P$(second dice six) $+ P$(third dice six) $= \frac{1}{6} \frac{5}{6} \frac{5}{6} + \frac{5}{6} \frac{1}{6} \frac{5}{6} + \frac{5}{6} \frac{5}{6} \frac{1}{6} = 3 \cdot \frac{25}{216} = \frac{75}{216} = \frac{25}{72}$.

$P(X = 2) = P$ (first dice not six) $+ P$(second dice not six) $+ P$(third dice not six) $= \frac{5}{6} \frac{1}{6} \frac{1}{6} + \frac{1}{6} \frac{5}{6} \frac{1}{6} + \frac{1}{6} \frac{1}{6} \frac{5}{6} = 3 \cdot \frac{5}{216} = \frac{15}{216} = \frac{5}{72}$.

$P(X = 3) = \left(\frac{1}{6}\right)^3 = \frac{1}{216}$

To answer your question, if you want a fair game (expectation zero profit/loss) you can pick $x,y,z$ however you want, as long as they satisfy $75x + 15y + z = 216$. Namely, on average in 216 rolls, you will get one six $75$ times, two sixes $15$ times and three sixes once. Reasonable values could be $x = 2, y = 3.4, z = 15$.

• Thanks Arthur! Running through that lot now. Will post back with any queries/questions. – Mark Oct 14 '13 at 10:22
• It seems so simple now and makes perfect sense. Wish i tried harder in school! – Mark Oct 14 '13 at 10:33

I cant comment so I will answer.

Are you sure of your percentages for one six? I got this result:

There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2x3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.

• There are $15$ combinations, not $18$. Three possibilities for which dice has no six and five possibilities for what number is on that dice. – Arthur Oct 14 '13 at 10:13
• @arthur Ok I got it thx :) – EpicGuy Oct 14 '13 at 10:16
• @EpicGuy 5/72 = 1/14.4. – Mark Oct 14 '13 at 10:18
• @Mark Yeah I realised that deleted that part. – EpicGuy Oct 14 '13 at 10:21