Mega Millions, how many combinations

Page 2 - Seeking answers? Join the AnandTech community: where nearly half-a-million members share solutions and discuss the latest tech.

GasX

Lifer
Feb 8, 2001
29,033
6
81
Originally posted by: profmath
Originally posted by: ElFenix
Originally posted by: profmath


The lottery is NEVER and even bet. Odds are always 1 in 135 million

do you know what an even bet is?

Yes an even bet is one that you have a 50% chance of winning.

A bad bet is one that you have less than a 50% chance of winning.....like the lottery or you being right on this.
not quite right. If you got paid a guaranteed $135million, then the lottery is an even bet. Until your guaranteed payoff approaches the odds of winning, it remains a sucker bet.
 

profmath

Junior Member
Mar 10, 2004
20
0
0
Originally posted by: ElFenix
Originally posted by: profmath


Yes an even bet is one that you have a 50% chance of winning.

A bad bet is one that you have less than a 50% chance of winning.....like the lottery or you being right on this.

no, an even bet is when the cost is equal to the expected payout. if the expected payout of mega millions is $135,145,920 and the cost is $1 then your expected value is $1 assuming only 1 person wins. since there is a chance of multiple winners the even bet point needs to be adjusted upward by that chance, but it is still there. and it's actually lower than that because there are minor prizes that have a better chance of being won (picking 5 balls and whatnot).

Trust me. I am a math professor. An "even bet" is one that you have a 50% chance winning. You are getting this confused with a bet that has an expected value of 0 (fair game). I know what you meant but thats not what you said. The expected value only comes into play for a long period of time or many trials. For this to be relavent to the lottery you would have to buy 135 million or so tickets.
 

GasX

Lifer
Feb 8, 2001
29,033
6
81
Originally posted by: profmath
Trust me. I am a math professor. An "even bet" is one that you have a 50% chance winning. You are getting this confused with a bet that has an expected value of 0 (fair game). I know what you meant but thats not what you said. The expected value only comes into play for a long period of time or many trials. For this to be relavent to the lottery you would have to buy 135 million or so tickets.
fair enough. I know enough math to know that the lottery is a sucker bet and that I should max odds whenever possible at the craps table... :)
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
81
Originally posted by: profmath
Originally posted by: ElFenix
Originally posted by: profmath


Yes an even bet is one that you have a 50% chance of winning.

A bad bet is one that you have less than a 50% chance of winning.....like the lottery or you being right on this.

no, an even bet is when the cost is equal to the expected payout. if the expected payout of mega millions is $135,145,920 and the cost is $1 then your expected value is $1 assuming only 1 person wins. since there is a chance of multiple winners the even bet point needs to be adjusted upward by that chance, but it is still there. and it's actually lower than that because there are minor prizes that have a better chance of being won (picking 5 balls and whatnot).

Trust me. I am a math professor. An "even bet" is one that you have a 50% chance winning. You are getting this confused with a bet that has an expected value of 0 (fair game). I know what you meant but thats not what you said. The expected value only comes into play for a long period of time or many trials. For this to be relavent to the lottery you would have to buy 135 million or so tickets.

Trust me, I am a mathematician that designs gambling devices for a living, and ElFenix is correct. When referring to any type of gaming an "even bet" is something where, on average, you will win the same amout of money that you expend participating in it. Give me a second to type it up and I'll provide a complete formula on how to decide when the lottery is an "even bet"
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
81
This problem has three variables to consider.
tickets(T) - The number of tickets sold for the given drawing
winnings(W)- The prize availible to win
probability(P) - The probability your numbers being the winnning numbers

The P is constant for a given game, it onlt depends on how they draw the numbers. So for this game:
P = 1/(52*CHOOSE(52,5)) = 1/135155920 = 7.399x10^-9

From P and T you can calculate the average number of winners (AvgN) for a given drawing
AvgN = SUM(i=0,T, i * CHOOSE(T,i) * P^i * (1-P)^(T-i) )

So if you have 135155920 Tickets sold the average number of winners is exactly 1. This doesn't mean that if you sell 135155920 tickets that you are guaranteed a winner however.
p(0) = 36.79%
p(1) = 36.79%
p(2) = 18.39%
p(3) = 6.13%
p(4) = 1.53%
p(5) = 0.31%
p(6) = 0.05%
There is still a 36.79% chance nobody will win and a 0.00832% chance that >6 people will win.


For this game your expected return for participating (EV) is this
EV = W*P/AvgW

By this equation if W=135155920 and T=135155920 then the lottery is an even game.

Of course none of this takes volitility in to account. Since the lottery is a highly volitile game you would be required play a massive amount of games before you could expect to converge to the EV. In other words you would need to play about 1,000,000 games before you could expect to be +-0.05 from the EV. I don't know about you, but I'm sure not ready to sink that kind of money in to the lottery.
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
81
I though of a better way to illustrate whether a game is an even game or not. Pretend I was running a casino on the street with these 3 games. It costs you $1.00 to play any of them.

1) You flip a coin. If it is heads I'll give you $1, otherwise you get nothing.
2) You roll a dice. If it is 1 I'll give you $6, otherwise you get nothing.
3) I pick a card from a deck of 52 and you try to guess. If you get it I'll give you $60, otherwise you get nothing.

According to the 50/50 definition 1) is an even bet. I can tell you flat out it is not and nobody would ever play that game. 2) is an even bet, even though your chances of winning are only 1/6. If you play it for a long time you can expect not to lose any money. 3) is a great bet. If you play this game enough you can expect to make money steadily.
 

StageLeft

No Lifer
Sep 29, 2000
70,150
5
0
Originally posted by: Kyteland
Originally posted by: profmath
Originally posted by: ElFenix
Originally posted by: profmath


Yes an even bet is one that you have a 50% chance of winning.

A bad bet is one that you have less than a 50% chance of winning.....like the lottery or you being right on this.

no, an even bet is when the cost is equal to the expected payout. if the expected payout of mega millions is $135,145,920 and the cost is $1 then your expected value is $1 assuming only 1 person wins. since there is a chance of multiple winners the even bet point needs to be adjusted upward by that chance, but it is still there. and it's actually lower than that because there are minor prizes that have a better chance of being won (picking 5 balls and whatnot).

Trust me. I am a math professor. An "even bet" is one that you have a 50% chance winning. You are getting this confused with a bet that has an expected value of 0 (fair game). I know what you meant but thats not what you said. The expected value only comes into play for a long period of time or many trials. For this to be relavent to the lottery you would have to buy 135 million or so tickets.

Trust me, I am a mathematician that designs gambling devices for a living, and ElFenix is correct. When referring to any type of gaming an "even bet" is something where, on average, you will win the same amout of money that you expend participating in it. Give me a second to type it up and I'll provide a complete formula on how to decide when the lottery is an "even bet"
Trust me, I am Skoorb :p
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
167
111
www.slatebrookfarm.com
I'm much too late, but to explain why it's 52C5 * 52 and not 52*51*50*49*48 * 52

If you had to get them in the exact same order, the chance of getting the first one right would be 1 in 52, followed by 1 in 51 (there's only 51 left) followed by...

But, since you can have them in any order, the first 5 numbers you chose would occur in 5*4*3*2*1 of the total number of ordered possibilities. Thus, your odds are1 in 52*51*50*49*48 *52 / (5*4*3*2*1) (which most people would simply calculate using the nCr button on a calculator as 52C5 * 52)