How long would it take to get a free room at Aria, Venetian or even a quasi ****hole like PH or Flamingo by just playing a cost mitigation roulette strategy. $5 on black, $5 on red. or $5 on odd, $5 on even. You get the picture.

Asking for a friend.

But no really, could you really rack up enough to trick em into comping you a room? I mean I literally cannot win playing like this, but I won't lose much either. Dare it to go green baby. I'll just drink their free drinks in the meantime. Outside of poker, I do not play casino games because there is NO chance in the long run I will win. So I utilize a cost mitigation strategy instead. I've been able to sit for over an hour before and not lose a penny playing like this. Having tons of fun. People telling me I'm an idiot, yet I always get to "win" some chips almost every spin rofl. I always forget to give them my card, so I've never paid attention to how much points I might rack up.

It makes more sense for your friend to pay for a room.

They don't even want to give out free parking anymore. And if you're only betting $5 then I can't see them caring about you at all.

This cant be a serious post......

Well since were on the subject, how many 1$ chips do I have to tip the waitress if I want to be comped a free room?

wait a minute, you have actually done this WITHOUT giving them your card?

Eff it I will play along. Let's say you have $5 on all 4, red/black/odd/even. That gives you a $20 average bet, which equals roughly $33 per hour in theoretical loss. Casinos comp a third of your theo (roughly) so you would earn around $11 per hour in comps. Now you factor in casinos consider comped rooms as higher per night than rack rate, you would need around 40 hours of this strategy to get a room at Aria, per night.

Now, what are the actual odds of seeing either 0 or 00 twice in one hour putting you upside down? It is a 5.26% chance of it happening on any given spin, so let's just say it happens every 20 spins. According to Wizard of Odds, a full table will usually get you around 35 spins per hour. You need to plan on losing $20 at least once to try and "win" your $11 in comps.

Im not doing any more googling on this, this is the dumbest ****ing idea ever.

You do realize the only person you're "tricking" is yourself. Betting red/black or odd/even is posting a guaranteed loss. Zero shot of a win, can only lose or break even. In other words, you're letting the casino freeroll you. I'd imagine the casino would be more than happy to comp you out of proportion to your bet size, bc you're basically the biggest idiot in the casino. Even more so if you don't use your card, bc now they can't even track you.

The analogy for this is running your heater full blast in the dead of winter, then opening up all the doors and windows. But let us know how it turns out!

You're not mitigating cost. You're reducing variance. Your expected loss is exactly the same each spin. However, the casino will view you as a tool and probably not comp you anything if you do this.

priceline.com >>>>> Your Plan

You'd be better off betting $5 on black and doubling your bet every time you lose. When you win, you win all your money back + $5. Sure, red/green can come up a few times in a row, but black always comes eventually. They'll rate you at a much higher average bet too.

go to craps one person bet 25$ on pass the other friend bet 25$ on dont pass. neither of you take ur odds and just split the losses btwn to two of you when a 12 comes on the comeout. there will prob be 35-45 come out rolls in an hour so you expected loss for each person shouldnt exceed 15$ each hour played where as your roulette idea would be a much higher expected loss

He could do it alone, doesn't need a partner.

How do you lose 15 there? Don't pass wins on 12 so wtf? Anyways it could work but they're on it I even saw it on an episode of Las Vegas long time ago. The roulette idea is soooo dumb

no

if two players are doing it and splitting the 25$ loss at 40-45 rolls an hour you can expect a loss a lil over 12.50$

Betting $5 on red and $5 on black will not get you a $10 avg bet. The same is true for the pass / don't pass. Also, any competent box / floor will be able to figure out if two people are betting together and you will be rated accordingly.

I love every special person who comes up with ideas like this and want to smile and pat them on the head.

Cheapest way to earn a "free" room is to pay for it.


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While some managers think this is the way to handle this if the casino has a clue they welcome play like this rate as $10 bet and comp accordingly. They should encourage this sort of play at the low levels. At high levels it creates suspicious activity as it could be money laundering and nobody wants that paperwork hassle.

The play is obviously welcome because it's a complete freeroll. However, I don't see a reason to rate at $10 then add messages in their account to correct, which can be overlooked, and lead to overcomping, the player receiving an inflated tier score, and inflated player worth leading to additional offers he/she shouldn't be eligible for. Additionally, if enough players are not rated correctly over a long enough period it will skew gaming data for analytics. There is also the money laundering issue as you mentioned.

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It's not inflated. No need to correct anything. The players theoretical loss is losing 5.26% of $10 per spin. Exactly the same as if he bet $10 on a single bet.

psandman is correct. Comps are based on theo, and that doesn't change with either method. My local casino appropriately rates both bets.

12 everyone loses huh? F that

This. Sometimes I really wonder where supposed "shrewd" gamblers come up with these ideas. The good news is that the 2+2 community came back with some very solid math behind why it's a terrible idea.

You are both 100% correct in your mathematical assessment. If player A bets $10 on red and never changes his bet and Player B bets $5 on black and $5 on red and never changes his bet the house expects the same win from both players in the long run if they continue to bet this way forever. However, due to human nature Player A is significantly more valuable to the house than Player B since there is a much greater potential upside (referring to increasing bets, chasing losses, etc, NOT variance.) I have a work acquaintance who was asked a related question in an interview for a casino manager position. He was asked something along the lines of who is more valuable: 5 players betting $5,000 or 1 player betting $25,000? Most people answer that they're equal or the 5 players betting $5,000; however, the answer they were looking for was 1 player betting $25,000 since the upside is significantly greater, the bigger player is more likely to go off for much bigger losses. So that being said, a guy betting black and red for $5 each trying to grind out free rooms or improve his tier level is not very valuable to the house and some managers feel they should not be rated the same.