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# Mega Millions - is a \$1, \$2, or \$3 ticket the best value (EV)?Prev TopicNext Topic

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• Sugar Land
United States
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August 28, 2019
86 Posts
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In my state there are three types of ticket options for Mega Millions:

(1) A regular ticket at \$2,

(2) A Megaplier ticket for \$3 with enhanced non-Jackpot prizes, or

(3) "Just The Jackpot", \$1 - no non-Jackpot prizes.

Which is the best "value"? To make this a tangible question, we need to define "value" in a statistical probability sense. I'm going to define this as the highest Expected Value (EV) per unit bet option. I've explained EV in previous post.

The "best" option in terms of EV per unit bet is going to be dependent on the value of the Jackpot. Just As the Jackpot goes up, the less the value of the regular and Megaplier tickets become relative to Just The Jackpot. This is just with respect to the EV (per unit bet), as the Standard Deviation of Just the Jackpot is going to be higher for the same EV.

Let's take a regular \$2 ticket. Since the non-Jackpot prizes have fixed return (at least in my state, not so in California for the 5-0 2nd prize outcome, which is pari-mutuel there), we can calculate the precise EV of the non-Jackpot prizes. I've worked out that is \$0.2470. For the Megaplier option, it's 3x that value.

What are the cross-over points of equal EV?

Let's define the EV of the non-Jackpot prizes as the variable "A" for the regular \$2 ticket. Again, for Mega Millions, that is \$2470. Furthermore, let's define the factor "k" as the reduction fraction [0,1] of the Jackpot that accounts for the pari-mutuel aspect of the Jackpot, meaning if we win, there's a chance of having to share the Jackpot with other winners. "k" will be a function of the number of tickets sold. The more tickets sold, the lower the value of "k". I've also talked about this in a previous post. Finally, let's define "p" as the probability of a single ticket winning the jackpot. For Mega Millions that is 1/302575350. In other words, given that we are a Jackpot winner, the expected value of our (pre-tax) winnings is kJ, not J. (I'm defining J here as the nominal, advertised cash value of the Jackpot).

Let's compare Just The Jackpot vs. Regular Ticket first. The value of J (our share of the cash Jackpot) where they have equal EV is as follows:

(p)(kJ) = (1/2)(p)(kJ) + A/2

The "1/2" factor on the right-hand side is to account for a \$2 ticket instead of \$1 for Just The Jackpot. Therefore, we can easily see that J = A / ( kp). Below that the regular ticket has the highest EV, and above Just The Jackpot.

Remember, A and k are constants given a fixed N tickets sold. "k" assumes all tickets are Quick Picks in my calculations (to follow), so we'll increase it by, say, 20% to account for people buying multiple unique combinations at a time.

What about Just The Jackpot vs. Megaplier ticket? Here, the Megaplier ticket is \$3, and the non-Jackpot EV is 3A. The equation for equal EV is:

(p)(kJ) = (1/3)(p)(kJ) + A.

We can therefore conclude J = (3/2) [A / (kp)].

Next, let's compare the regular ticket with the Megaplier ticket. The equation is:

(1/2)(p)(kJ) + A/2 = (1/3)(p)(kJ) + A

Therefore, J = 3A / (kp)

What are the theoretical values of "k" - given all tickets sold are Quick Picks. Here are the numbers I've worked out for Mega Millions. "N" is the number of tickets.

 k N 0.7401 200M 0.6446 300M 0.5660 400M 0.5008 500M 0.4465 600M 0.4007 700M 0.3619 800M 0.3286 900M 0.2998 1000M (1B)

As mentioned, to account for non-Quick Pick tickets, let's reduce these numbers further by 20%, or a factor of 0.8.

Take an example of 500M tickets sold. Adjusted "k" is therefore 0.5008 x 0.8 = 0.4006. Compare Just The Jackpot to regular \$2 ticket. Therefore

J = \$0.2470 x 302575350 / 0.4006 = \$186560437. Since our current Jackpot cash value is higher than this, we have a higher EV / unit bet with Just The Jackpot. Again, we'll naturally have a higher Standard Deviation with Just The Jackpot, so it's fair to point that out as a detracting factor.

Do taxes change this calculation? No, but only if we assume all prize tiers are calculated at the same (presumed ~maximum) income tax rate. This is because the EV for both sides of the equation is reduced by the same (1 - tax rate) factor. Most people probably won't report a \$2 net win, for example, but in theory (in the U.S) we are obligated to report it legally.

• Sugar Land
United States
Member #200,559
August 28, 2019
86 Posts
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Sorry, I messed up the "Just the Jackpot" (JTJ) pricing. It's \$3 for 2 tickets, not \$1 for 1 ticket.

Comparing JTJ to regular ticket, then, we have the following equation for the equal EV / unit bet:

(2/3)(kJ)p = (1/2)(kJ)p + A/2

Therefore, J = 3A/(kp).

Comparing JTJ to Megaplier (\$3) ticket, we have the following:

(2/3)(kJ)p = (1/3)(kJ)p + (3A)/3

Therefore, again, J = 3A/(kp)

To re-iterate Regular Ticket vs Megaplier:

(1/2)(kJ)p + A/2 = (1/3)(kJ)p + (3A)/3

Once again, J = 3A/(kp)

A is \$0.2470 in our case and p = 1/302575350.

According to https://<snip>/mmsales.htm there were 166280263 tickets sold in the last July 26 drawing. This corresponds to a "k" factor of 0.7694. I suggested we should reduce this by another 20% or so to account for non-QP sales. That would roughly put the adjusted "k" at 0.6.

Therefore the value of J (cash Jackpot) equal to 3A/(kp) is \$373680557. So with our current cash Jackpot above \$500M (\$1.2B annuity), JTJ would appear to have the best EV / unit bet.

This post has been automatically changed by the Lottery Post computer system to remove inappropriate content and/or spam.

• United States
Member #161,360
November 27, 2014
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Quote: Originally posted by Orange71 on Jul 27, 2022

Sorry, I messed up the "Just the Jackpot" (JTJ) pricing. It's \$3 for 2 tickets, not \$1 for 1 ticket.

Comparing JTJ to regular ticket, then, we have the following equation for the equal EV / unit bet:

(2/3)(kJ)p = (1/2)(kJ)p + A/2

Therefore, J = 3A/(kp).

Comparing JTJ to Megaplier (\$3) ticket, we have the following:

(2/3)(kJ)p = (1/3)(kJ)p + (3A)/3

Therefore, again, J = 3A/(kp)

To re-iterate Regular Ticket vs Megaplier:

(1/2)(kJ)p + A/2 = (1/3)(kJ)p + (3A)/3

Once again, J = 3A/(kp)

A is \$0.2470 in our case and p = 1/302575350.

According to https://<snip>/mmsales.htm there were 166280263 tickets sold in the last July 26 drawing. This corresponds to a "k" factor of 0.7694. I suggested we should reduce this by another 20% or so to account for non-QP sales. That would roughly put the adjusted "k" at 0.6.

Therefore the value of J (cash Jackpot) equal to 3A/(kp) is \$373680557. So with our current cash Jackpot above \$500M (\$1.2B annuity), JTJ would appear to have the best EV / unit bet.

This post has been automatically changed by the Lottery Post computer system to remove inappropriate content and/or spam.

WTF?

If my brain was more "Seagullish"

Maybe I could understand.

• United States
Member #202,920
December 14, 2019
689 Posts
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Quote: Originally posted by Orange71 on Jul 27, 2022

Sorry, I messed up the "Just the Jackpot" (JTJ) pricing. It's \$3 for 2 tickets, not \$1 for 1 ticket.

Comparing JTJ to regular ticket, then, we have the following equation for the equal EV / unit bet:

(2/3)(kJ)p = (1/2)(kJ)p + A/2

Therefore, J = 3A/(kp).

Comparing JTJ to Megaplier (\$3) ticket, we have the following:

(2/3)(kJ)p = (1/3)(kJ)p + (3A)/3

Therefore, again, J = 3A/(kp)

To re-iterate Regular Ticket vs Megaplier:

(1/2)(kJ)p + A/2 = (1/3)(kJ)p + (3A)/3

Once again, J = 3A/(kp)

A is \$0.2470 in our case and p = 1/302575350.

According to https://<snip>/mmsales.htm there were 166280263 tickets sold in the last July 26 drawing. This corresponds to a "k" factor of 0.7694. I suggested we should reduce this by another 20% or so to account for non-QP sales. That would roughly put the adjusted "k" at 0.6.

Therefore the value of J (cash Jackpot) equal to 3A/(kp) is \$373680557. So with our current cash Jackpot above \$500M (\$1.2B annuity), JTJ would appear to have the best EV / unit bet.

This post has been automatically changed by the Lottery Post computer system to remove inappropriate content and/or spam.

in practical terms, jtj is the worst play because you are more likely to get bit by a shark while attending bigfoot's wedding than win the jackpot, and you have 0 chance of winning a smaller prize that makes up for what you spent on tickets. With regular plays and megaplier you will sometimes get back your "investment".

• Sugar Land
United States
Member #200,559
August 28, 2019
86 Posts
Offline

That's not how Expected Value works. Your statement is valid, however, from the standpoint of Standard Deviation. No question JTJ has the highest SD, by far.

Let me give an expository example from Video Poker. Let's say this is Jacks or Better Draw. Suppose we have the following starting hand dealt to us before the draw:

The payout chart is per unit bet is:

Royal Flush - 800

Straight Flush - 50

4 of a Kind - 25

Full House - 9

Flush - 6

Straight - 4

3 of a Kind - 3

2 Pair - 2

Jacks or Better Pair - 1

Obviously if you keep the pair of Jacks on the draw, you have a sure winner. You will at least have a high pair, possibly 2 Pair, 3 of a Kind, Full House or 4 of a Kind. However, this would be dumb from an Expected Value perspective, as any statistically-based Video Poker tutorial will tell you. (Assuming you have sufficient bankroll to play in the first place, you always make the highest EV move.) Discarding the Jack-Spades is the much better move from an EV perspective, or, in other words, drawing for the Straight Flush. The EV of that move is 3.57 units (including possible lower tier payouts). Keeping the pair of Jacks has an EV of 1.54 per unit bet. The draw for the Straight Flush is going to be a higher standard deviation, however, mainly because only 2 out of the remaining 47 cards will complete the Straight Flush. You could also get a High Pair (by drawing another Jack) or a Flush (any other Club drawn except 7 or Queen).

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