# Which offer would you take - and show your math! **URL:** \ **Category:** In My Humble Opinion\ **Created:** [April 12, 2013, 3:50am UTC](https://boards.straightdope.com/t/which-offer-would-you-take-and-show-your-math/655473 "2013-04-12T03:50:49Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [April 12, 2013, 4:23pm UTC](https://boards.straightdope.com/t/which-offer-would-you-take-and-show-your-math/655473/21 "2013-04-12T16:23:15Z")
> [@septimus](#): > > I hope OP comes back and tells us whether option 2 is repeatable or not, but even if he says Yes that option figures to cost you more. By buying up to three tickets in option 1, you win with probability 98.44% and save an average of $656 compared with option 2. In the limit, if you’re willing to buy an unlimited number of tickets in option 1, your savings will average $650 compared with option 2. > > To solve the problem exactly, one needs the details. (You should know that by now!) Specifically we want to know your total capital and what its utility is to you. > > (_Utility = logarithm Total\_Capital_ is a frequent, but by no means generally accepted, assumption.) I agree that being able to repeat the purchase is an important factor I missed. I assumed both purchases were repeatable but the OP doesn’t explicitly state this. I also agree that, on average, you’d spend less with the first option. But I felt the advantage of the second plan in having a fixed upper limit outweighed the first plan’s advantage of having a lower average cost but no fixed upper limit. In theory, you could run out of money before winning a prize under the first option but that couldn’t happen in the second. ---
**Author:** ![Ravenman](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ravenman/32/2929_2.png) [@Ravenman](https://boards.straightdope.com/u/Ravenman)\ **Post date:** [April 12, 2013, 4:46pm UTC](https://boards.straightdope.com/t/which-offer-would-you-take-and-show-your-math/655473/22 "2013-04-12T16:46:29Z")
> [@RTFirefly](#): > > If you can’t explain to your friend that there’s only a 1/64 chance of spending more in option A than in option B, and less than 1 chance in 1000 that option A would be at least $750 worse than Option B, while you’ve got a 75% chance that option A will be $750 better than Option B, I’m not sure I can help you. Ooh, that’s a really good summary of the two options. I like that. ---
**Author:** ![Wolverine](https://avatars.discourse-cdn.com/v4/letter/w/ba9def/32.png) [@Wolverine](https://boards.straightdope.com/u/Wolverine)\ **Post date:** [April 12, 2013, 5:23pm UTC](https://boards.straightdope.com/t/which-offer-would-you-take-and-show-your-math/655473/23 "2013-04-12T17:23:24Z")
> [@Ravenman](#): > > However, the compadre with whom I’m discussing this scenario fundamentally does not agree that the odds are quite high that one would probably (but not certainly) win after a few attempts at option 1. As in, simply doesn’t believe that there’s less than a 10% chance of “losing” a couple times in a row, because obviously there’s a 25% of losing on every ticket, and I’m committing a gambler’s fallacy. I’m not sure how to explain my disagreement with that conclusion. While it is true that there is a 25% chance of losing with each ticket, no one buys a 2nd or 3rd ticket if they’ve already won. This means that 75% of all losers will win on the next ticket. To find the probability of winning only once, let’s take 1 - P(all losing tickets) 1 ticket @$1000 = 0.75 2 tickets @$1300 = 0.9375 3 tickets @$1600 = 0.984375 98% of people would save money on option A compared to option B. The 6 tickets only give an extra 0.015381 (6 tickets = 0.999756). That’s not worth an extra $150. For fun, I assumed that all 7 billion people in the world bought tickets. After 17 tickets, there were no losers remaining. Everyone in the world should have won. (7000000000\*(1-0.25^x tickets) \< 1) 17 tickets costs $5800 (700+300\*x tickets) which is our global maximum. Your pool should be signifacantly lower so I would expect the maximum number of tickets would be lower. If 5000 people buy tickets until they win, only 7 tickets are needed for a maximun $2800. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [April 12, 2013, 5:57pm UTC](https://boards.straightdope.com/t/which-offer-would-you-take-and-show-your-math/655473/24 "2013-04-12T17:57:26Z")
I just don’t see saving money as the primary goal. In either case, you’re going to end up with more money when you win the undefined big prize, so quibbling over the initial investment seems trivial. To me, the primary goal is making sure you win the prize, which is why I favored the second option (assuming it’s repeatable). ---
**Author:** ![Doctor\_Jackson](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/doctor_jackson/32/32_2.png) [@Doctor\_Jackson](https://boards.straightdope.com/u/Doctor_Jackson)\ **Post date:** [April 12, 2013, 6:59pm UTC](https://boards.straightdope.com/t/which-offer-would-you-take-and-show-your-math/655473/25 "2013-04-12T18:59:01Z")
> [@Ravenman](#): > > However, the compadre with whom I’m discussing this scenario fundamentally does not agree that the odds are quite high that one would probably (but not certainly) win after a few attempts at option 1. As in, simply doesn’t believe that there’s less than a 10% chance of “losing” a couple times in a row, because obviously there’s a 25% of losing on every ticket, and I’m committing a gambler’s fallacy. I’m not sure how to explain my disagreement with that conclusion. The disagreement could have less to do with mathematical proof and more to do with communication. It sounds like your compadre is viewing this game as individual trials like roulette, where the odds to not change based on number of attempts, while you see it as a zero sum game, like lottery scratch offs, where odds change in your favor based on the number of tickets bought. Maybe? [Previous page](https://boards.straightdope.com/t/which-offer-would-you-take-and-show-your-math/655473.md?page=1)