If you pay $1 to buy one lottery ticket, and 7,500 other people have also bought a lottery ticket, for which there is only one winning ticket worth $2,500...what is your expected value?
2006-06-05 10:56:50 · 14 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Ok, I spelled statistics wrong...haha. But my math for this problem was:
2,500(1/7,500) - 1(7,499/7,500) = -0.6665
Is this way off? Did I do it right?
2006-06-05 11:04:23 · update #1
your math is 100% correct!
2006-06-19 08:17:53 · answer #1 · answered by navymilitarybrat76 5 · 0⤊ 1⤋
"Expected Value" is useless when applied to the Lottery
Here's a lesson about how sometimes you have to put the math into perspective.
Some gambling authors are so caught up in the numbers that they won't buy a lottery ticket unless the jackpot gets high enough for their bet to have a high enough "expected value". That is, if the odds are 20 million to 1 and the jackpot is at $10 million, they won't buy a ticket, because the expected value of that dollar bet is low. (10/20 = $0.50. And by the way, we're excluding all the smaller prizes to keep things simple). But once the jackpot grows to $19.5 million, with the same odds, they'll buy. Why? Because the "expected value" af that $1 ticket is now almost a dollar -- 19.5/20 = 0.95, or $0.95.
But this is crazy! When you buy a $1 lottery ticket, you can't get just $0.95 back -- you'll likely lose the whole dollar OR you'll win a whole boatload of money, not 95 cents. So the expected value in this case is irrelevant. It's irrelevant because (1) you're playing in the short term, with only one or a small handful of plays, and (2) playing such a small number of times, you're most likely to win nothing. That 95% expected return would be meaningful only if you bought 20 million lottery tickets!
There's another thing to consider about our example in which the Expected-Value fan only buys a ticket when the jackpot grows from $10 million to $20 million: The odds of winning the jackpot didn't change. Sure, you'd win less in the first case, but so freaking what?! Wouldn't you be nearly just as happy with $10 million instead of $20 million? What exactly could you do with $20 million that you could not do with ten?
The only thing that really matters when buying a lottery ticket is this: Is the chance of winning $X jackpot worth a dollar to you, if the odds are Y to 1 against your winning? If yes, then buy one. If not, then don't. But don't even consider the expected value, because it's self-defeating.
Looking at expected value has a lot of benefit in games where you're playing hundreds of hands and the amount you can win on one bet is small (like blackjack or slot machines), but it shouldn't become a religion that's applied uselessly to games with few plays where you can win a bundle off one bet, like the lottery.
But there's yet one more twist about experts who say that it's mathematically proper to buy lottery tickets only when the jackpot reaches a certain amount: their math is wrong. What they're failing to consider is that when the jackpot grows, more people buy lottery tickets. Not because those buyers are calculating the expected value (most people have no idea what that is), but because they're more excited about winning a bigger pile of money. The more people play, the more likely it will be that more than one person picks the winning numbers, so the prize will be split among all the winners. If two people win a $20 million jackpot, they each get $10 million -- the same as if one person won a $10 million jackpot. Therefore, you can't assume that the expected value goes up in relation to the amount that the jackpot goes up, because if you do win, you're less likely to have won the whole thing yourself.
Don't mistake all this to mean that the lottery is a good deal. It's not. The odds are bad no matter how you cut it. The point is just this: when the jackpot gets big, you aren't any more likely to win. And for that reason there's no point in waiting for a bigger jackpot. If you want to play the lottery, play it whenever you like, it won't affect your odds. Those odds will be terrible no matter when you play.
2006-06-05 11:00:58 · answer #2 · answered by philk_ca 5 · 0⤊ 0⤋
Well... exactly 7.500 persons or almost? So... let it be P the persons, except you, who bought a lottery ticket. Then the persons who bought a lottery ticket are P+1. Because only one person will win the $2.500, the winning price of the man will be $2.499, because he boughgt a lottery ticket too. Also all the others will lose $1 because of the lottery ticket they bought. With these clues, the expected value will be:
(1/P+1)*$2.499-(P/P+1)*$1=$[(2.499-P)/P+1].
Place the exact persons and you will find the true expected value.
2006-06-19 01:41:41 · answer #3 · answered by Leonidas Nikolakis 3 · 0⤊ 0⤋
You are close.
EV = 2,499(1/7,501) - 1(7,499/7,501)
You have to subtract the $1 that you spent on the ticket.
You win $2500, but you spent $1, so your overall winnings are $2499.
Also, if 7500 other people also bought a ticket, then there would be a total of 7501 tickets because you would have to include the ticket you bought.
2006-06-05 11:11:36 · answer #4 · answered by MsMath 7 · 0⤊ 0⤋
-1$
2006-06-05 11:00:21 · answer #5 · answered by Anonymous · 0⤊ 0⤋
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2016-12-06 09:50:39 · answer #6 · answered by mance 3 · 0⤊ 0⤋
1 out of how many people bought a ticket since you only bought one so
1 out of 7500
2006-06-18 13:30:08 · answer #7 · answered by ramsarabia94 2 · 0⤊ 0⤋
1 out of 7501
2006-06-18 20:54:52 · answer #8 · answered by IT 4 · 0⤊ 0⤋
(income) * (probablity of income) + (expence) * (probability of expence) =
2500 * (1/7501) + (-1) * (1) = - 2/3 (nearly)
2006-06-17 09:10:22 · answer #9 · answered by djfox_2001 3 · 0⤊ 0⤋
1*7500=7500
2500/7500=0.333%
2006-06-18 18:14:15 · answer #10 · answered by dodo 1 · 0⤊ 0⤋
It is going over my head.
2006-06-18 23:18:01 · answer #11 · answered by nayanmange 4 · 0⤊ 0⤋