foxfan20
foxfan20
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February 23rd, 2011 at 6:10:03 PM permalink
A local casino is running a slot promo: If you lose up to $100 at slots, they essentially reimburse you twice your losses in slot dollars to a maximum of 200 (for losing $100). What's the correct strategy for playing my $100 at video roulette (double zero)?

My thoughts are: The expectation on 200 slot dollars at video roulette is $189 and change. I should "go-for-broke" with an $89 spin on black, and if I hit it, I'm done. If I lose it, I start betting whatever I've got until I'm over half of $+89, etc. Essentially, each spin is for either the amount I have left or the amount I need to win to reach $189, whichever is less. Ideally you win the first spin and walk; if you lose, you get $189 EV worth of slot dollars.

Is there a better strategy?
foxfan20
foxfan20
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February 24th, 2011 at 10:51:17 AM permalink
OK so I've been thinking about this way too much.

First: A better strategy is a modification of the one above, where if an even-money bet doesn't hit your target, you make a 2:1 bet. If that doesn't get you there, make a 5:1 bet, etc. If you can't get there at all, put all your money on the 35:1 and pray. Remember: If you lose it all, you're getting $190 EV worth of slot dollars back.

The "right" answer to this problem is very complex. Essentially what you need to construct is a list of probabilities. Each item in the list is your current amount of money and the probability that you can reach your target from that amount of money. Each amount >= $190 will have a probability of 100%, since you'll stop playing when you reach your target. Then the "correct" bet at each point is the one that maximizes the expected probability of the resulting amount.

For example (the numbers in this example are COMPLETELY made up!), let's say you're at $100, your original amount. Let's take the case where you get $90 on black. You have a 18/38 chance of reaching your target right away, and a 20/38 chance of getting to $10. Let's say the value of the entry for $10 is ... oh, I don't know, 5%. (COMPLETELY made up percentage.) That would make the percentage for $10 be (18/38)*(100%) + (20/38)*(5%) = 50%. Each bet you place should be with the goal of maximizing the expected resulting percentage.

If it weren't for the cyclic nature of such a list, I'd know how to generate it. Who wants to take a stab?
ChesterDog
ChesterDog
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February 24th, 2011 at 12:42:47 PM permalink
Quote: foxfan20

...Then the "correct" bet at each point is the one that maximizes the expected probability of the resulting amount....



The "correct" bet might instead be the bet with the highest expectation. Compare two different strategies: 1) Bet $100 on black. Walk away if you win, or receive the $200 coupon (worth $189.47) if you lose. 2) Bet $100 on an individual number. Walk away if you win, or receive the $200 coupon (worth $189.47) if you lose.

The expectation for strategy 1 is (18/38)($100)+(20/38)($189.47-$100) = $94.46.

The expectation for strategy 2 is (1/38)($3500)+(37/38)($189.47-$100) = $179.22.

So, it looks like strategy 2 might be the "correct" bet.
foxfan20
foxfan20
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February 24th, 2011 at 2:41:59 PM permalink
Not sure how I didn't see that, but I like that strategy. It has a nice side effect of being resolved in one spin too :-P

However it has some interesting consequences. Specifically, if you had less than $3,505.19, you'd "want" to bet it all on one number. Consider $3,500. You expect to make $0.27 by betting it all on one number.

Thankfully, after winning the first spin you already have $3,600, so betting it all is negative EV (specifically you expect to lose $94.81).
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