Timeline for Can a casino system prove my (divine) luck as cheating?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2017 at 2:26 | comment | added | Deepak | @charlie_pl Irresistible force vs immovable object lol. In any case, "divine luck" will probably act in such a way that only the winning wager gets accepted, e.g. the "wrong" bet is never entered into the system because of a computer glitch or human error. | |
| Sep 1, 2017 at 14:42 | comment | added | SZCZERZO KŁY | @LarsH That's why my answer was that the luck would act in a way to nullify the bet. | |
| Sep 1, 2017 at 14:39 | comment | added | LarsH | I think the point of @charlie_pl's question was, how would the (magic?) luck function (not, would this be a good way to win money even without special luck), if you bet on two opposing outcomes? It would be impossible for you to win both. The OP says "Of course, if it's impossible, it won't happen." | |
| Aug 31, 2017 at 13:32 | comment | added | OrangeDog | @SZCZERZOKŁY the point is if you play red/black evens the house will still win (i.e. on average, make a profit at your expense) because there are 0s. | |
| Aug 30, 2017 at 7:16 | comment | added | SZCZERZO KŁY | @DeveloperInDevelopment I assume the luck works the way that you are handled the winning card. | |
| Aug 30, 2017 at 3:52 | comment | added | DeveloperInDevelopment | Presumably, the scratch cards would have been printed before the lucky day, so wrt "but only if it's possible to happen in a day", the best you could do in scratch cards is happen to have enough other people buy a ticket that day that you end up with the best card on the roll. This is very unlikely to be the big jackpot (on average it will be significantly less than the cost of a ticket times the number of tickets on a roll). The sports bet would work especially well with an accumulator - multiple individual outcomes that must all happen, leading to very high pay out. | |
| Aug 29, 2017 at 12:35 | comment | added | SZCZERZO KŁY | @Nolonar A consecutive winner in poker is also not good for house because he attract people watching and not bidding while at the same time people lower their bidding on other tables because they assume that if someone else wins the casino will cheat on the to get even. | |
| Aug 29, 2017 at 11:52 | comment | added | Nolonar | If you really want to win without the casino chasing you out or marking you as "not welcome", play Poker. Whether you win or you lose, the house still wins, so they don't care if you're cheating or not (at least until the other players start suspecting you). | |
| Aug 28, 2017 at 20:14 | comment | added | Michael | @Draco18s for reference, the numbers they actually pick are 24 red, 212 green, and 87 plaid; the first two come up but the chase scene starts before the last one is actually spun. | |
| Aug 28, 2017 at 19:54 | comment | added | Draco18s no longer trusts SE | @Hurkyl Here's that segment. | |
| Aug 28, 2017 at 17:08 | comment | added | user2781 | @charlie_pl: Reminds me of the episode of The Monkees where they desperately needed to lose a roulette bet, so they bet on cyan 100. And won. | |
| Aug 28, 2017 at 14:33 | comment | added | SZCZERZO KŁY | @ViktorMellgren yes but it's not payable in the same way as red and black. So if you put 1 on red, black and 0 zero is always the more profitable option. | |
| Aug 28, 2017 at 14:12 | comment | added | Viktor Mellgren | @SZCZERZOKŁY, there is 0 and 00 which are green. | |
| Aug 28, 2017 at 13:17 | comment | added | SZCZERZO KŁY | @charlie_pl The problem only exist if you play in game with only two outcomes with the same payout. Which for example exist in roulette. You can gamble on red and black. But the luck for you may mean you won't loose money. So the table may broke physically during your bet. | |
| Aug 28, 2017 at 12:56 | comment | added | ratchet freak | @charlie_pl a casino/bookie is never going to give 2:1 payout to 2:1 odds. Often by allowing a significant chance for a third possibility | |
| Aug 28, 2017 at 12:32 | comment | added | Rdster | @Vylix Actually, if you bet an equal amount on all the options of a single game, you'd still lose...not win. Don't forget the house edge. Don't believe me, go bet 1 dollar on all 38 numbers on a roulette wheel and tell me how much you won when they give you $35. | |
| Aug 28, 2017 at 12:29 | comment | added | Vylix | @charlie_pl I imagine the host will not host (pun intended) a gamble that gives net result more than you have bet. For instance, in a 2 options gambling (win-lose), the odds will never be totaled more than 2. In 3 options, 3. And so forth. If it does, then you can just bet like you said: $1 on every option and always gain, albeit small, winning. | |
| Aug 28, 2017 at 12:28 | comment | added | Secespitus | @charlie_pl Fate would choose the one where you would get more money. And if they both happen to be exactly equal - maybe the butterfly effect kicks in and you will be lucky to meet someone with the same interests by yelling loud enough that you won/lost? | |
| Aug 28, 2017 at 12:20 | comment | added | charlie_pl | What would happen if you have placed an equal bet on two opposite outcomes I wonder? | |
| Aug 28, 2017 at 12:15 | history | edited | Secespitus | CC BY-SA 3.0 |
Fixed typos and tried to improve wording
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| Aug 28, 2017 at 12:06 | history | answered | SZCZERZO KŁY | CC BY-SA 3.0 |