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Old 11-28-2005, 01:07 AM
yoshi_yoshi yoshi_yoshi is offline
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Location: Cambridge, MA
Posts: 54
Default Re: 10% refund question

[ QUOTE ]
Yoshi,

When I play craps or blackjack in a Vegas Casino I get back 40% of my projected losses in comps. Yet that doesn't make the casino table games +EV. So to answer your question as to where the percentage lays I know it must be greater than 40%. Right? Keep in mind I get that whether I win or lose.

[/ QUOTE ]

Hi Jimbo, sure, if you play enough then 40% rebate will still be -EV. I'm just saying that how much rebate you need to make it a breakeven game depends on how many hands you plan to play there during each rebate period. If you play infinite hands per rebate period, the rebate you need approaches 100%. If you play one hand per rebate period, the % rebate you need should be pretty low, definitely lower than 10%, as some people have calculated above.

So for your situation, (I am guessing that) the casino will make sure you play too many hands for 40% rebate to make you +EV. A player that plays a couple hands a month is surely not going to get the same rebate a regular player like yourself would.

So to summarize, my only real point is that if you play below a certain number of hands per rebate period at 10% rebate, you will be a +EV player. I hope we can all agree on that.
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  #42  
Old 11-28-2005, 01:19 AM
AcmeSalesRep AcmeSalesRep is offline
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Join Date: Aug 2004
Posts: 25
Default Re: 10% refund question

[ QUOTE ]
So to summarize, my only real point is that if you play below a certain number of hands per rebate period at 10% rebate, you will be a +EV player. I hope we can all agree on that.

[/ QUOTE ]

Prett much everyone BUT Jimbo can agree on this...

Acme
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  #43  
Old 11-28-2005, 03:04 AM
pzhon pzhon is offline
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Join Date: Mar 2004
Posts: 66
Default Re: 10% refund question

This works a lot like a sticky bonus. It can be +EV, but you have to be careful.

You should set a stop-loss and a stop-win, and make large bets when between these values.

Suppose you win 1 unit with probability .495, and lose with probability .505. Suppose you set a stop-loss of L and a stop-win of W. The probability of winning W units is (c^L - 1)/(c^(L+W)-1), where c=.505/.495~1.0202. The probability of losing L*9/10 units is (c^(L+W)-c^(L))/(c^(L+W)-1). (Derivation.) The optimum occurs at L=3, W=3, where the EV is +0.0645 units. The EV is positive for any L,W adding up to 10 or less, and negative for L+W>10.
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