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Chances of Winning Mega Millions

Lottery odds from first principles - combinations, prize tiers and expected value
Odds of matching 5 white balls from 70 and 1 gold ball from 25
Odds of matching 3 white balls plus the gold ball
Probability of winning any prize in a 5/70 + 1/25 game
Expected value of a $2 ticket with a $400 million jackpot at 1 in 302,575,350

Counting the Tickets

Lottery odds are a counting problem. A ticket draws 5 white balls from a pool of WW and 1 gold ball from a pool of GG. Order does not matter, so the number of distinct tickets is a combination times a choice:

T=(W5)×G,(W5)=W!5! (W−5)!T = \binom{W}{5} \times G, \qquad \binom{W}{5} = \frac{W!}{5!\,(W-5)!}

Mega Millions has used a 5 from 70 white plus 1 from 25 gold format, giving (705)×25\binom{70}{5} \times 25 tickets. Ball-pool sizes are game rules and have been changed before, so confirm WW and GG against the current official rules and substitute them - the formula does not change.

Exactly one ticket wins the jackpot, so the probability is 1/T1/T and the odds are quoted as 1 in TT.

Lower Tiers and Expected Value

To match exactly ww of the 5 white balls, choose which ww you hit, then fill the rest from the balls you missed:

N(w,gold)=(5w)(W−55−w)×{1gold matchedG−1gold missedN(w, \text{gold}) = \binom{5}{w}\binom{W-5}{5-w} \times \begin{cases} 1 & \text{gold matched} \\ G-1 & \text{gold missed} \end{cases}

Divide by TT for the probability of that tier. Summing every paying tier gives the chance of winning anything, which is far higher than the jackpot odds and dominated by the smallest prizes.

Expected value per ticket is

E=∑ipi⋅prizei−costE = \sum_i p_i \cdot \text{prize}_i - \text{cost}

Prize amounts are set by the operator and vary by draw, by tier and sometimes by state, so this page does not publish them - enter the figures from the official prize schedule for the draw you care about and the arithmetic above turns them into an expected value. Note also that a jackpot advertised as an annuity is not the cash you receive, and any withholding applies on top; both belong in prizei\text{prize}_i before you compare it with the ticket cost.

Common Mistakes to Avoid

  • Using permutations: order is irrelevant, so it is (705)\binom{70}{5}, not 70×69×68×67×6670 \times 69 \times 68 \times 67 \times 66 - a factor of 5!=1205! = 120 too large.
  • Forgetting the gold-ball multiplier: white-ball combinations alone undercount the tickets by a factor of GG.
  • Counting "at least ww" as "exactly ww": the tiers are exclusive; add them if you want a cumulative chance.
  • Treating two tickets as doubling a meaningful chance: 2/302,575,3502/302{,}575{,}350 is still about 1 in 151 million.
  • Believing past draws matter: each draw is independent, so "due" numbers do not exist.
  • Comparing an annuity jackpot with a ticket price: discount it first, or use the cash figure.

Examples

Step 1: (705)=70×69×68×67×665!=1,452,361,680120\binom{70}{5} = \dfrac{70 \times 69 \times 68 \times 67 \times 66}{5!} = \dfrac{1{,}452{,}361{,}680}{120}
Step 2: =12,103,014= 12{,}103{,}014 white-ball combinations
Step 3: Each pairs with any of the 25 gold balls: 12,103,014×25=302,575,35012{,}103{,}014 \times 25 = 302{,}575{,}350
Step 4: One ticket wins, so p=1/302,575,350≈3.3049×10−9p = 1/302{,}575{,}350 \approx 3.3049 \times 10^{-9}
Answer: 1 in 302,575,350302{,}575{,}350

Step 1: Choose 3 of the 5 winning whites: (53)=10\binom{5}{3} = 10
Step 2: Fill the other 2 from the 65 non-winning whites: (652)=65×642=2080\binom{65}{2} = \dfrac{65 \times 64}{2} = 2080
Step 3: Gold matched, so 1 way: 10×2080×1=20,80010 \times 2080 \times 1 = 20{,}800 tickets
Step 4: p=20,800/302,575,350p = 20{,}800 / 302{,}575{,}350, and 302,575,350/20,800≈14,546.9302{,}575{,}350 / 20{,}800 \approx 14{,}546.9
Answer: About 1 in 14,54714{,}547

Step 1: With the gold ball: any of the 12,103,01412{,}103{,}014 white combinations qualifies
Step 2: Without it, 3+ whites: (53)(652)+(54)(651)+(55)=20,800+325+1=21,126\binom{5}{3}\binom{65}{2} + \binom{5}{4}\binom{65}{1} + \binom{5}{5} = 20{,}800 + 325 + 1 = 21{,}126
Step 3: Each of those pairs with 24 losing gold balls: 21,126×24=507,02421{,}126 \times 24 = 507{,}024
Step 4: Total winning tickets: 12,103,014+507,024=12,610,03812{,}103{,}014 + 507{,}024 = 12{,}610{,}038
Step 5: 302,575,350/12,610,038≈23.99302{,}575{,}350 / 12{,}610{,}038 \approx 23.99
Answer: About 1 in 24 - though the overwhelming majority of those wins are the smallest prize tier

Frequently Asked Questions

For the 5-from-70 plus 1-from-25 format the count is C(70,5) × 25 = 12,103,014 × 25 = 302,575,350 possible tickets, so one ticket has a 1 in 302,575,350 chance. Ball-pool sizes are game rules that have changed before, so check the current rules and re-run the count if they differ.

Count the tickets that match exactly w white balls: C(5,w) × C(65,5−w), multiplied by 1 if the gold ball matches or 24 if it does not. Divide that count by 302,575,350. Matching 3 whites plus the gold ball, for instance, gives 20,800 tickets, or about 1 in 14,547.

The probability scales linearly, so k distinct tickets give k/302,575,350. Ten tickets move the jackpot chance from 1 in 302.6 million to 1 in 30.3 million — a tenfold improvement on a number that stays vanishingly small.

Multiply each tier's probability by that tier's prize, add them up and subtract the ticket price: E = ÎĢ p_i · prize_i − cost. Prizes vary by draw and jurisdiction, so take them from the official prize schedule. Use the cash value of a jackpot, not the annuity headline, or the comparison is not like-for-like.

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