I got really excited on Friday when I got mail from Pepsi remembering that the winners of a car would be notified by mail. Turns out it was a check for $20! Still, was happy. My dad and I got to discussing the odds of winning. Was it better to save your caps and enter them all on one day or to enter them one per day? I argued one way my dad argued another. We called my uncle Mike, a mathematician, for the answer and proof. He emailed it to me. I'll post it when done. Here is the contest: The first 3 correct answers with valid reasons win a $25 gift card to Solera. What are the better odds, excluding your entries? A) Make one entry in each of 10 successive days. B) Make 10 entries on a single day. For the sake of this contest, we'll say that the exact same number of total entries are made each day, 100,000. People can pm me if you want to keep answers private. The contest lasts for 2 days. That way more people can enter. I'll anounce the winners and post the PM'd answers later tonight. Anyone else win? Note: Uncle Mike, my dad or any other Thoma are excluded from entering this contest. Ps. Thanks Uncle Mike! Josh
wouldnt it be exactly equal chance no matter what of those 2? if u put all 10 on 1 day and lets say 100 was submitted total.. thats 10 % chance then the following 9 days non were so still 10%. and if 1 % was submitted a day for the next 10 days it would also be 10% chance... wrong?(im probably far off)... but id do the 10 entries on a single day...
Option A) I can't give specific math as I don't know the amount of entrants per day, but I would guess it is alot like the lottery in that your odds aren't much better if you play once or 10 times. As there is only one correct winning code chosen at random per day.
Also, you'd have ten different opportunities to win, as opposed to one. It's like being in 10 different drawings instead of one.
well basically you have 10 chances to win @ 1 out of 100,000 or a 1 time chance to win at 10 of 100,000. so there is 10 x (1/99999) or 1 x (10/99999) they both come out the same at .01% chance of winning. although i am sure i did that wrong cause i know i forgot a step somewhere.... stupid calculus..... "the more calculus you know the less real math you know"
This is correct. There is a law of probability where you can add two or more events' probabilities together that are independent, which these are. So the probability is the same.
Yep, the odds are the same. Math > me. Here is what my uncle sent: The Pepsi challenge. Josh can enter a raffle one of two ways: case A) make one entry in each of 10 successive days. case B) make 10 entries on a single day. Which is better? Terminology V The value of the prize. An identical prize V, is awarded each day. One prize per day no matter how many entrants. E1, E2, E3, .... E10: The total number of entrants on day n is En. P1, P2, P3,,, P10: (applies only to case B) The probablity of Josh making all 10 entries on day n is Pn. The sum of all Pn must equal 1. Pn = 1/10 for each n seems reasonable because Josh has no way of knowing on which day he should enter. We are interested in the expected value of each of case A and case B. Expected value is the average over many trials (if you could repeat the process many times.) Expected value is not the only way to compare two cases. But is widely used when monetary amounts are involved. Expected value of Josh's winnings for case A (one entry on each of 10 separate days.) Expected_A = sum of { V * (1/En } = V * (1/E1) + V * (1/E2) + V * (1/E3) + .... this is because the chance of winning is 1/En Expected value of Josh's winnings for case B (all 10 entries on day n with probability Pn of choosing day n.) Expected_B = sum of { Pn * V * (10/En) } = P1 * V * (10/E1) + P2 * V * (10/E2) + P3 * V * (10/E3) + .... this is because the chance of entering on day n if Pn and the chance of winning if you enter on day n is now 10/En. (10 entries) Remember (P1 + P2 + ... + P10 = 1.0) Observations: 1) If the number of entries each day is the same, En, then, as I mentioned on the phone, Expected_A = Expected_B = V * 10 / En. And Pn does not affect the answer. Josh can enter on any day with equal exopected outcome. 2) If the number of entries excluding Josh is the same, then we need to add Josh's entries to the other entries. Let Rn be the rest of the entries, excluding Josh. En = Rn + no. of Josh entries. Assuming Rn is the same each day Expected_A = V * 10 / En = V * 10 / (Rn+1) because Josh enters 1 time each day. Expected_B = V * 10 /(Rn + 10) In this case, Expected_A is vet slightly bigger. To confirm this, imagine Rn is very small, say 10 other people enter each day. Expected_A = V * 10/11 (Josh has 1/11 chance of winning on 10 separate days) Expected_B = V * 10/20 ((Josh has 10/20 chance of winning , all on the same day.) More realistically, if Rn = 100,000 entrants then Expected_A = V * 10/100001 = V * .000099999. (Josh has 1/100001 chance of winning on 10 separate days) Expected_B = V * 10/100010 = V * .000099990. (Josh has 10/100010 chance of winning , all on the same day.) 3) If the number of entrants varies from day to day, but Josh has no way of knowing how many entrants there will be on any given day then, if Josh picks a random day to enter ten times (Probability 1/10 of selecting each day) Expected_A = sum of { V * (1/En } = V * (1/E1) + V * (1/E2) + V * (1/E3) + .... Expected_B = sum of { (1/10) * V * (10/En) } As in case 2, if the number of entries each day in case A and case B is unaffected by the fact that Josh enters on a given day, Expected_A = Expected_B. Otherwise (it seems most likely that other people are independent of Josh: Expected_A is slightly bigger.) 4) If Josh knows that some days are more heavily bet than others, he should make his bet on the day with the smallest no. of bets by other people (small Rn.) No Surprise here. Yep, Uncle Mike > Math. Thanks for playing. I'll pm the winners. JT
You are correct that math > you since you totally misread your Uncle's proof. The odds are only the same when the total number of entries is exactly the same for each day. However, the probability that there would be exactly 9 less entrants on the given day that you would enter all 10 of your caps is highly unlikely. Thus, as in both cases 2 & 3 it's pretty easy to see that entering once a day is the way to go.
Oh snap, I totally missed this in the first post, lol. That thows a sticky wicket into things, hehehe. So let me rephrase the correct answer: The correct answer is that it depends based on how you interpret the above quote. Does it mean that there are exactly 100,000 entries made by other people or does that include the ones you made on a given day? :biggrin: reading > me lol.
The Winners are: And Pete for going with the flow So no one else won $20? PM me your address and I'll send out the gift cards or you can pick them up at Solera. Thanks! Josh
My uncle covered this with me on the phone. The odds are almost identical if you don't include me. In the contest I meant that with me, there were 100,000 per day. The almost part would be covered in this: 2) If the number of entries excluding Josh is the same, then we need to add Josh's entries to the other entries. Let Rn be the rest of the entries, excluding Josh. En = Rn + no. of Josh entries. Assuming Rn is the same each day Expected_A = V * 10 / En = V * 10 / (Rn+1) because Josh enters 1 time each day. Expected_B = V * 10 /(Rn + 10) In this case, Expected_A is vet slightly bigger. To confirm this, imagine Rn is very small, say 10 other people enter each day. Expected_A = V * 10/11 (Josh has 1/11 chance of winning on 10 separate days) Expected_B = V * 10/20 ((Josh has 10/20 chance of winning , all on the same day.) More realistically, if Rn = 100,000 entrants then Expected_A = V * 10/100001 = V * .000099999. (Josh has 1/100001 chance of winning on 10 separate days) Expected_B = V * 10/100010 = V * .000099990. (Josh has 10/100010 chance of winning , all on the same day.)
