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.
Kyteland
Diamond Member
- Dec 30, 2002
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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
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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.
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
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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.
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.
DrPizza
Administrator Elite Member Goat Whisperer
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)
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)
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