My local casino sends me a free bet voucher most months. You can use it for any table game, but I tend to go for roulette as it’s quick and easy. If your free bet wins, you get the usual payout but you don’t get your stake back.
In the past I have usually gone for an even money bet, on the basis that it’s the greatest chance of turning something free into actual cash. But discussing this with an acquaintance recently, they said you should always go for a bet with longer odds. Their reasoning was that with the even money bet, if it wins you will receive X (where X is the value of your free bet). This is only half your usual return for making a bet of X (because you don’t get your stake back). Whereas if you bet on a number, you will receive 35X if it wins, which is nearly as much as the 36X you normally get back. Therefore this way, you are getting much more ‘value’ out of your free bet. Of course, with this method it’s much less likely you will actually win, which was basically my counter-argument. But he was so sure of this, he actually offered to buy my free bet ticket for X/2 if all I was going to do was put it on an even money shot. Obviously, from a pure EV point of view I should have accepted this offer, but instead I decided to follow their advice and place it on a single number (it didn’t come in). I’m not upset by this outcome, but I do find myself wondering if their argument is sound. It seems to me that the fact the initial stake is paid for doesn’t change the EV, so as with a normal bet it just depends on whether you fancy a relatively high chance of a relatively small return, or a relatively low chance of a relatively high return. You pays your money (or not, in this case) and takes your choice. What say you, Dopers?
After thinking a bit, I believe the friend is right. Because the stake isn’t returned the EV of the even money bet is much less than the long-shot.
Ignoring the 0 and 00 to make things simpler,
Betting on red wins half the time and returns one bet on a win.
Betting on #36 wins 1/36 of the time and returns 35 units on a win.
EV (red) = 1 unit * 0.5 = 1/2 units
EV (#36) = 35 units * 1/36 = 35/36 units.
If you need an explanation, it’s that losing the initial stake hurts EV a lot more on the even-odds bet, because you lose that stake more often (half the time), whereas on the longshot, you only lose the stake once in a while.
Your friend has it wrong. If you place a $5 bet on EVEN, if you win, you get an additional $5 chip next to yours, for a total of $10. Plus, if it comes up 0 or 0-0, you get half your bet back ($2.50).
Let’s analyze what happens when you play the game 38 times and get one of each possible outcome:
Red: You win 18 times and lose 20 times, walking away with $18.
#36: You win one time and lose 37 times, walking away with $35.
The fact that your original stake (the coupon) is not returned on a win makes a big difference. In casinos this is referred to as “Pay and take”, as opposed to “Pay and lay”, where you don’t lose the promotional item on a win.
I think you misunderstand. The OP says “If your free bet wins, you get the usual payout but you don’t get your stake back.”
If you win a normal $5bet, you get an additional $5, plus keep your stake. You double up your money.
If you win a free $5 bet, you get $5, but they take away your free chip. You change free chips into real money.
But you lose the ability to place another $0 bet that wins you real money. That’s the “stake” you lose. If you got to keep & reuse it on a win, the EV calculations would be different, because you could keep playing for longer if you won more games, even if each win netted you less money.
Your “stake” is actually zero. What you have is a voucher that can only be used in one casino to place a bet at one of their gaming tables. Unless the voucher has some other monetary value. Can you use it to tip a waiter? Trade it for a cup of coffee? Down payment on a bus ride back home?
You have 380 vouchers good for a single dollar play that expire tomorrow, so you’re going to play all of them. Payout is 1:1 for a color bet, 35:1 for a number bet.
Case 1 - you always lose the voucher after each game.
Bet Red - you win 180 of 380 games. Net gain, $180.
Bet #11 - you win 10 of 380 games. Net gain, $350.
Case 2 - you keep the voucher on a win. It still expires tomorrow, so you’re going to keep playing until you’re out of vouchers.
Bet Red - you win 180 of 380 games. Gain $180, 180 vouchers left.
You win 85 of the next 180 games. Gain $85, 85 vouchers left.
Keep going - $40, $19, $9, $4, $2, $1, for a grand total of $340.
Bet #11 - you win 10 of 380 games. Net gain, $350, 10 vouchers left, You bet those 10, lose them all.
So interestingly, it’s still better to bet the highest odds bet available, but the EV winds up much closer.
I’m not seeing how the EV would be different between a free bet and one that was $5 for both the even/odd bet and the single number bet. How does the initial bet change the EV?
It doesn’t for a single game. But see my post above yours - if you just keep playing the game until you run out of vouchers, then the game that returns the voucher on a win gets you a higher EV for the game with smaller, more frequent wins, since you get to keep playing longer.
Assuming the casino would allow their own staff to bet the vouchers, then a waiter would probably be happy to accept it as a tip. No to coffee or bus ride, though.
I agree, but the EV for the odd/even bet compared to the single number bet remains the same, regardless of which way the vouchers are used. In other words, if you have a voucher, there is no difference in EV if you bet on EVEN, or if you bet on 17.
I agree that in both games, there is a difference in EV between losing the voucher each time, or keeping it when you win. In fact, keeping the voucher when you win is no different then just a regular bet with your own money.
“Welcome coupons” may differ from casino to casino (IIUC most do NOT allow high-odds bets like a roulette single-number) but the coupon itself is lost even when you win. If your win is unlikely (i.e. betting the longshot #18(*)) then your EV suffers less from that lost coupon.
An often-overlooked fact is that, for appropriate objective functions, the single-number bet is better than the even-money bet at American roulette even in the normal case — without “welcome coupons.” How can this be when the House vigorish is the same in each case? Turn the arithmetic around, and compare your vigorishes (albeit negative) in the two cases.
Disclaimer: Past performance is no guarantee of future results but #18 was the winning number in a scene from the excellent Nick Nolte movie Good Thief. It deserves more than 6.6 IMDB stars. A little meme in the film was that Nutsa Kukhianidze’s character’s age, 17, was a prime number and Nick placed a stack of chips on #17. “I’m not a prime anymore. Today is my birthday,” she said as she moved the stack to #18. Ah, SPOILER ALERT — the movie does have a happy ending. Not only does #18 come up, but the nubile Nutsa is now licit!
Ignoring the house edge, your expected returns from a $1 voucher are:
1/2 bet: 1/2 * $1 = $0.50
1/36 bet: 1/36 * $35 = $0.97
More generally, if we assume that the EV of all bets is the same (i.e. the house edge is the same for all bets) we can look at the situation in two steps after you are given a $1 voucher:
(1) You can make any one of a selection of normal $1 bets, in which you DO get your stake back; normal bets (with stake returned) have different probabilities of winning, but EV is the same for all bets.
(2) If you win the bet in (1), you must pay $1. If you lose the bet in (1), you pay nothing.
Whatever bet you choose in (1), the EV of step (1) considered alone is always the same.
But since the probability of paying $1 in step (2) depends on the probability of winning in step (1), you should choose the bet in step (1) with the lowest probability of winning.