Lucky 6 is the game where person picks 6 numbers $[1,48]$, then 35 numbers are drawn, and if a player guesses all 6, he earns money.

I was trying to calculate how much money would be expected to lose when playing the game, but somehow I got the positive number, so there must be something wrong with my calculation, but I can't figure out what.

So first of all I think that probability of earning anything is ${35 \choose 6}$ of possible winning combinations divided by ${48 \choose 6}$ total combinations gives me $1623160/12271512\approx0.13$ chance.

But in reality most of the times if person wins, he will win more than they bet, and the game looks like this: enter image description here

First 5 numbers go into the middle, and then if person guesses the correct combination within the first 6 drawn numbers he will get $10000\times money spent$, if in 7th draw he will get $7500\times money spent$, and so on.

So to calculate the game value I was multiplying the money earned with the probability of that event, and this is where it went wrong. This was my logic:

The probability of scoring in the exact n-th drawn ball = $({n \choose 6} -{n-1 \choose 6})/ {48 \choose 6}$

for example this would give me that guessing the correct number within first 6 draws is equal to ${1 \over 12271512} $. To get game value I multiplied this number with amount of money won for each case (in the first case 10000).

  • 6th draw: ${1 \over 12271512} \times 10000$
  • 7th draw: ${6 \over 12271512} \times 7500$ (based on the image above). And so on..

In the end I took the sum of all this multiplications and got number greater than 1. It was $\approx1.019$

I know that this is not correct because this would mean that house is designed to lose.


Let $P_n$ denote the probability of scoring on the nth draw.

$P_n = {n \choose 6}/ {48 \choose 6} -\sum_{i=5}^{n-1} P_{i} $ for $n >5$. With $P_0=P_1= \cdots=P_5=0$.

I'm not familiar with the game, but I believe in order to "score" you must get all 6 digits correct.

Define $N_n$ as the number of ways to win on the nth draw it is:

$N_6 =1$

$N_7 =7-1=6$

$N_8 = {8 \choose 6}-6-1= 28-6-1=21$

Someone has already written the odds for this here: https://wizardofodds.com/games/lucky-6-35-48/

Except the winnings listed are different.

  • $\begingroup$ the formula I presented also works for $N_8$, it goes by the fact that the combinations for $n-2$ are also contained in the $n-1$, yours is also correct. For $N_8$ you get 21 (you forgot to subtract), my formula is also the same: $({n \choose 6} -{n-1 \choose 6})$, so for $n=8$ its $28-7 =21$. $\endgroup$ – user334473 Oct 18 '19 at 22:03
  • $\begingroup$ Yup, you're right. the formulas are the same. $\endgroup$ – Mark Oct 18 '19 at 22:10
  • $\begingroup$ Thank you for taking your time to look into this, I'm glad that I didn't have a problem in this logic, but in the code I used to implement it. Your link was useful to figure the probabilities out. $\endgroup$ – user334473 Oct 18 '19 at 22:35
  • $\begingroup$ Actually I was right, the game on your link has the same result as mine when I scrolled down. Now I wonder does this mean that house is on the loss, but that's the question for another forum. :) $\endgroup$ – user334473 Oct 18 '19 at 22:53

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