Pharaoh's Casino bonus worth going for?

[FredMSloniker]

I'm looking at playing blackjack at Pharaoh's Casino because of their advantageous single-deck format. I'm also, however, looking at their signup bonus of $30-- which I can't collect by playing single-deck. In order to get it, I'd have to bet a minimum of $1500 (25 x $60) at the regular blackjack tables, where I'm at the mercy of the house. The question is, how likely am I to actually get my hands on that bonus? In other words, what is the probability that I can successfully wager $1500 before my $30 deposit runs out?

It's a tricky thing to compute, but I'm taking a stab at it. Let's define a function P(n,t,x) to be the probability of being able to play a certain number of hands of blackjack, t, with a certain initial bankroll, n (the number of dollars in the initial bankroll divided by the minimum bet), and a certain player chance of winning, x. (Am I correct in believing that a house advantage of, say, .1% translates to x being 49.9%?) In this specific case, n is 30 and t is 1500; I'm not sure what x is. (Note that I'm assuming the best way to go about this is to always make minimum bets. If I'm wrong, please let me know. I'm also ignoring the possibility of a push or blackjack; if a push is considered 'not winning' and a blackjack 'winning' when it comes to determining x, the actual probability of going the distance is slightly higher than P(n,t,x). This shouldn't be significant.)

P(n,0,x) is obviously 1 (anyone can not play blackjack). When t>0, P(0,t,x) is obviously 0 (you can't play blackjack without betting). To make the problem interesting, of course, we need to be able to solve for arbitrary n, t, and x. I note that if n>=t, P(n,t,x)=1. (Even if you lose every hand, you can play as many hands as you have money to play.) Therefore the first interesting problem is P(1,2,x).

After a little noodling, I figured out that, when n<t, t>0, and n>0, P(n,t,x)=x*P(n+1,t-1,x)+(1-x)*P(n-1,t-1,x). In other words, if we win (probability x), we now have one more betting unit; if we lose (probability 1-x), we have one less. Either way, we have one less turn to complete; our odds of success can be figured from that point. (I'm hoping anyone serious about playing blackjack for money knows at least some probability math and followed that, but just in case...)

Now we can compute P(n,t,x) for any n, t, and x, at least in theory. The tricky part is actually doing the computation. I need two things to finish this problem: the exact value of x, and appropriate software to handle the whole P-defined-in-terms-of-P... thing. I'm working on part two (hoping I can get access to Mathematica on the college campus, and, if that fails, coding something up myself), but while I do, I figured I'd let people know what I'm doing. Certainly someone can provide a value for x in this case; maybe someone even has the software lying around, or even a solution gathering dust. If you can help out, post to the thread!

[raymondc]

I tried last week and failed. I deposited $30 and got extra $30 to play with.

However, perhaps it's my poor skill. I found $60 too little to conduct any effective money management plans.

Try Cowboy Casino or Player's Club Casino, they offer $75-$100 bonus. So you could have $150-$200 to begin with.

I thought as you do that the single deck blackjack could be easy but after a week's intensive playing with it I don't really feel that it's easier than the multi-deck game and my game records showed that I usually lose more money on the single deck than on the multi deck. The hands dealt in the multi-deck game seemed more even. While the single deck could be very biased; there are more continuous losing or winning streaks. When you lose on the single deck game it can be worse than the multi-deck.

Besides not counted towards the wagering requirement, playing the single-deck game does not get comp points reward. So I have already given up the single deck.