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Can we do better than the official cop-out that the “Odds of Winning” are “Variable*”? As some variables (like number of players) are unknown, we can’t calculate a number. But can we deduce a formula?

*Your odds of winning the Gold Ball Draw are one out of the total number of Gold Ball Draw Numbers issued for that draw. The odds of the winning Gold Ball Draw Number taking home the Gold Ball Jackpot are one in the total number of balls in the draw. In the first draw, the odds of the Gold Ball being drawn are 1 in 30. If the jackpot continues to carry over and all the white balls are drawn, the odds will be 1 in 1.

CLASSIC JACKPOT

  • The Classic Jackpot is fixed at \$5 million.

  • You get a set of six Classic Draw numbers from 1 – 49 for each \$3 LOTTO 6/49 play.

  • Six numbers are drawn from 1 – 49.

  • Match all six main Classic Draw Numbers on one line to win the jackpot.

GOLD BALL DRAW

  • Each \$3 LOTTO 6/49 play will include a unique 10-digit Gold Ball Draw Number in addition to your six Classic Draw Numbers.

  • One Gold Ball Draw winning number will be drawn from all selections for each draw.

  • Once a Gold Ball Draw winning number is drawn, a separate Gold Ball Jackpot Draw will determine the prize - the Gold Ball Jackpot OR the guaranteed \$1 million prize.

GOLD BALL JACKPOT DRAW

  • The Gold Ball Jackpot starts at \$10 million and can grow to \$68 Million!

It starts with 30 balls – 29 white balls each representing the guaranteed \$1 million prize and 1 Gold Ball representing the growing jackpot.

  • If a white ball is drawn, the guaranteed \$1 million prize is won. That ball is removed leaving 28 white balls and 1 gold ball for the next draw, and the jackpot grows by \$2 million.

  • Draws continue in this manner until the Gold Ball is drawn and the jackpot is won.

  • Once the Gold Ball Jackpot is won, the draw resets to 30 balls and the jackpot starts back at \$10 million.

  • If the Gold Ball Jackpot keeps growing until no white balls remain and only the Gold Ball is left – the jackpot can reach \$68 million!

Here's my attempt.

Expected Value of each play of Lotto 6/49 with Gold Ball Draw = Expected Value from Classic Draw + $\color{red}{\text{Expected Value from Gold Ball Draw}}$ – Cost of Ticket.

$\color{red}{\text{Expected Value(Gold Ball Draw)}}$ = Probability of winning Gold Ball Jackpot + Probability of winning $1M.

Thus Expected Value each play $= \dfrac{5 \; million}{13,983,816} + {\color{red}{\dfrac{\dfrac{(\text{1 gold ball × Gold Ball Jackpot) + [number of white balls × \$1 million}] }{\text{number of white balls + 1 gold ball}}}{\text{Number of players}}}} - 3$

$= {\color{red}{\dfrac{\dfrac{\text{Gold Ball Jackpot + number of white balls} }{\text{number of white balls + 1 gold ball}}}{\text{Number of players}}}} - 2.64244$

1 Answers1

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There are $\binom {49}6 = \frac{49\cdot48\cdot47\cdot46\cdot45\cdot44}{1\cdot2\cdot3\cdot4\cdot5\cdot6} = 13983816$ draws possible for the regular jackpot. Each are equally likely, so your odds of winning it are $\frac 1{13983816}$, with an expected payout of $\frac{5000000}{13983816}\approx\$0.36$. This is independent of the rest of the lottery.

If for a particular drawing,

  • there are $T$ tickets sold (and presuming each ticket has a unique number), and
  • there are $B$ balls left in Gold Ball Jackpot draw.

Then the probability of a particular ticket being selected for the draw is $\frac 1T$. This comes with an automatic $\$1$ million. The gold jackpot draw is for $\ge\$9$ million (the other $\$1$ million is the automatic prize). Since $30-B$ balls have already been drawn, this jackpot is $\$(69 - 2B)$ million.

So the expectation for this part of the lottery is $$\frac1T\left(1 + \frac 1B(69-2B)\right)\cdot \$1000000 = \$1000000\cdot\dfrac{69 - B}{BT}$$

Thus the overall expected gain (payout - cost) from a single ticket is $$ \$1000000\cdot\dfrac{69 - B}{BT} + \$0.36 - \$3 = \$1000000\cdot\dfrac{69 - B}{BT} - $2.64$$

To go any further, you would need to take into account how many people play each lottery, which is likely to increase as the gold ball jackpot grows. But that is not predictable by mathematics alone.

Paul Sinclair
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  • Many thanks! I have 3 follow up questions. $\color{limegreen}{(1)}$ Did you define B balls for white balls? But you didn't write "white"? $\color{limegreen}{(2)}$ Why did you write "$$(69−2B)$ million"? Is your 69 a typo for 68? The rules published that "The Gold Ball Jackpot starts at $10 million and can grow to $68 Million!" The rules never refer to "69"? –  Apr 25 '23 at 05:53
  • $\color{limegreen}{(3)}$ Why did you write $9 million in "The gold jackpot draw is for $≥$9$ million (the other $1 million is the automatic prize)"? Again, the rules published that "The Gold Ball Jackpot starts at $10 million". But why did you deduct the automatic $1 million from this published $10 million? If you draw the Gold Ball when the Gold Ball Jackpot is $10E6, then you win $10E6 (NOT $9E6). –  Apr 25 '23 at 05:54
  • (1) No, $B$ is the total number of balls, including the gold one. (2) No, If you follow the math, it simplifies down to $69 - 2B$. (3) $9+1 = 10$. I even specifically explained that right in the sentence you quoted. The other million was already accounted for in the earlier contribution. At its lowest payout, the gold ball only adds $9$ million to that separately included $1$ million (in the expectation formula, it provides the "$1+$" at the start. – Paul Sinclair Apr 25 '23 at 11:35
  • I'll try to be a little clearer. As soon as the ticket is chosen for the gold ball draw, it wins $$1000000$, regardless of the outcome of that draw. What the draw determines is whether they win some amount more. That additional amount is what gets multiplied by the probability of drawing the gold ball, not the total. That additional amount is $9$ million plus $60 - 2B$ million, where $B$ is the total number of balls in the draw (which will be at least $1$). And the probability of drawing the gold ball is $1$ out of the total number of balls $B$, not white balls. – Paul Sinclair Apr 25 '23 at 15:17
  • Many thanks again! $\color{limegreen}{(4)}$ Where does the 60 come from, in "$60 - 2B$ million"? –  Apr 25 '23 at 23:42
  • You know, you should really try it out yourself. Then you would understand it a lot better. The $1 + 9$ gives you $10$ million, $60 - 2B$ is how much more you get because of how many balls are left. On the first draw, $B = 30$, so $60-2B = 0$, and $10$ million is all you get. The next draw $B =29$, so $60-2B = 2$, for $12$ million. The next draw $B = 28, 60-2B = 4$ for $14$ million, etc. If you get all way to the end, $B = 1$, so $60-2B = 58$, so you get $10 + 58 = 68$ million. So where do you think the $60$ comes from? – Paul Sinclair Apr 26 '23 at 00:56
  • $\color{limegreen}{(5)}$ I still can't deduce the origin of your $60$. When $B = 1$, you win the most in the amount of $$68$ million $= [29$ white balls removed) $\times 2] + $9$ million + guaranteed $$1$ million. But $29$ white balls removed $\times 2 = 58$, not 60. $\color{limegreen}{(6)}$ To check my understanding, the odds of winning the Classic Draw OR the Gold Ball Draw $= \dfrac1{13,983,316} + \dfrac1T + \dfrac1{BT}$. Correct? –  Apr 29 '23 at 05:34
  • First, "odds" and "probability" do not mean exactly the same thing. "Odds" is the ratio $\text{Lose : Win}$ while (uniform) "probability" is the ratio $\dfrac{\text{Win}}{\text{Win}+\text{Lose}}$. Probability is the more useful mathematically and is what I've been talking about. Second, you can only add probabilities when they are exclusive, which is not true here. You have to break it into distinct cases: (Win classic & lose gold) + (win classic & win gold & lose jackpot) + (win classic & win gold & win jackpot) + the same cases for gold and jackpot, but you lose the classic. – Paul Sinclair Apr 29 '23 at 12:11
  • The easiest way to calculate the probability of winning at least one thing is that it will be $1$ minus the pobability of losing everything, which is $1 - \dfrac{13,983,315}{13,983,316}\dfrac{T-1}T$. (Note that if you win the gold draw, you will win something, whether or not you win the jackpot. But if you lose the gold draw, you don't get to play the jackpot. So $B$ does not come into this calculation.) As for the other question, the formula I gave you for calculating the total for a $B$-ball win was $69 - 2B = 1 + 9 + (60 -2B)$ million. Not just $60-2B$. One piece is not the whole thing! – Paul Sinclair Apr 29 '23 at 12:24
  • Sorry - $69-2B$ is not the total winnings, but the additional amount you get from winning the jackpot, and is $= 9+ (60-2B)$, not $1 + 9 + (60-2B)$. Maybe it would be easier to see where the $60$ comes from if I pull out the $2$? $60 - 2B = 2(30 - B)$. Now in the first jackpot draw, when the jackpot is at its lowest, how many balls are in the draw? – Paul Sinclair Apr 29 '23 at 12:52
  • Many thanks. I meant "probability", NOT "odds". I agree that $\Pr($winning at least $1E6) $= 1 - \Pr($losing Classic & Gold Ball Draws). $\color{sienna}{(7)}$ Are you contending that $\Pr($losing Classic & Gold Ball Draws) $= \dfrac{13,983,315}{13,983,316}\dfrac{\color{limegreen}{T}\color{red}{-1}}T$? I grok that $\color{limegreen}{\dfrac{13,983,315}{13,983,316}\dfrac{T}T= \Pr(}$losing Classic Draw). But what's $\color{red}{\dfrac{13,983,315}{13,983,316}\dfrac{-1}{T}}$? What does this mean? –  Apr 29 '23 at 22:06
  • $\color{peru}{(8)}$ If you win the first Gold Draw, then the jackpot is at its lowest and $B = 30$. Hence you win $1 + 9 + 2(30 - 30) = $10E6$. If you win the last Gold Draw, then the jackpot is at its lowest and $B = 1$. Hence you win $1 + 9 + 2(30 - 1) = $68E6$. For these initial and end boundary values, I grok that your formula works. But we are working backwards from two possible outcomes! I am still dumbfound! How can we work forwards? How did you deduce this 60? –  Apr 29 '23 at 22:14
  • $B$ is the number of balls. Let $J$ be the number of millions of dollars in the jackpot. With each unsuccessful draw, $J$ goes up by $2$ and $B$ goes down by $1$, and therefore $2B$ goes down by $2$. Thus there sum $J + 2B$ doesn't change at all. Its value is the same at every draw. At the start, $J = 10$ (including the one million that will be won no matter the outcome), and $B = 30$, so $J + 2B = 70$. So $J = 70+2B$. Removing the $1$ million that is obtained just from having the ticket selected, which accounted in a different term, this leaves $69 - 2B$ obtained by winning the draw. – Paul Sinclair Apr 29 '23 at 23:05
  • Concerning the prior comment: $\dfrac{13,983,315}{13,983,316}$ is the probability you lose the classic, and $\dfrac{T-1}T$ is the probability you lose the ticket selection (yours was one of the $T-1$ losing tickets). These two events are independent of each other (one occurring does not influence how likely the other is to occur), the probability of your losing both is the product of the two probabilities. So your overall chance of losing everything is $\left(\dfrac{13,983,315}{13,983,316}\right)\left(\dfrac{T-1}T\right)$, Thus your change of winning something is $1$ minus that product. – Paul Sinclair Apr 29 '23 at 23:12