# Best betting strategy for simple high/low bet of variable amount, but if you lose it all you're out?

Say there was a game like where you rolled 2 6-sided dice and made a custom dollar bet based on if you thought the next roll would be higher or lower. The game had the following conditions:

• Each round you bet whether you think the next die roll will be higher or lower than the last one
• In case of a tie your bet carries over to the next round
• You can bet any amount in your account, including 0
• If you lose all your money, the game is over
• The game is over after 10 rolls. Whatever is in your account at that time you get to keep.

What would be the optimum betting strategy so you ended up with the most money at the end? The obvious bets are when the current dice are showing 1 or 12, then bet it all on higher or lower. But what about when the current dice are showing 2-11? If you just bet straight odds and lose a couple of rounds, you could end up losing a big chunk of your money. Some of the scenarios I can think of:

No bet: Always bet \$0 and go home with 1000 Obvious bet: Only bet on the 1 or 12. Most likely go home with ??? (minimum \$1000)
Odds bet: Based on the high low probability, bet P and most likely go home with ???

What would the optimum betting strategy be in a game like this?

[Moderating]
Moving to GQ.

[Not moderating]
What you’re looking for is called the Kelly Criterion.

Should we assume that your TOTAL bankroll is much greater than \$1000, even though you’re not allowed to bet it? If so, we can make the assumption that your money utility function is linear.

On the very last shake you may as well bet everything: your chance will be at least even-money. But your optimal bet size in earlier rounds will vary depending not only on what the shake-to-beat is, but how far the game has progressed. Trying to beat 11 (or 3) on the 9th round, I think you’re best off betting everything, but you’d be much more conservative on the 1st round.

I think you mean “2 or 12.” Snake-eyes (1+1 = 2) is the lowest shake.

There’s no “best” answer, it all depends on your risk aversion. Strictly by EV with a high tolerance for risk the best strategy is bet it all on anything that isn’t 7.

On 2nd thought, assuming there’s no limit on your bet-size and utility is linear, my risk-averse suggestion is incorrect.

Yes, I think Calavera is correct. To make the problem interesting, we need some constraint – e.g. on maximum bet or maximum winning – or non-linear utility.

Yep, a more interesting question might be, you’ve got \$1000 and owe the mob \$10000, payable in the next 10 minutes. What’s the best strategy to avoid concrete overshoes?

I think it would be impossible to beat the positive EV you get if you bet zero except when you are betting that the next roll will be over 2 or under 12. You would bet the whole lot then. So you either end up with your \$1,000 or double it sometime after any 2 or 12 is rolled.

Not if there’s a tie, which presumably would be exceedingly common in cases of a 2 or a 12. And when your \$1,000 bet rolls over to the next round and it’s a 7, what do you do then?

There’s only 10 rolls and a 1/18 chance of getting 2 or 12. If you wait for it to happen chances are you won’t even get one go. Strictly on an Expected Value basis, the high risk bet on anything except 7 strategy will have the highest mean result over a large number of trials. Betting on 7 won’t affect the EV, just increase the variance.

At least, if you had a sufficiently-large bankroll. If the first roll were a 6, you wouldn’t want to bet it all on it being higher, because even though that’s a positive EV bet, you’d run a significant risk of losing everything (including the opportunity to make further bets).

The framework for this question would be a typical TV gameshow where you start with a certain amount of money and that’s all you get. You can’t add to your bankroll from the outside. You play 10 rounds and you keep whatever you end up with.

The question came to me as I was flipping around and came across a show called “Card Sharks”. The final round is a game where contestants start with a certain amount of money and bet a variable amount if they think the next card will be higher or lower. Card Sharks has a bunch of special rules which complicates the statistical question, so I’m not really wondering about that game specifically. I was more wondering about the statistical answer in a general sense. I suggested dice so the each round would be random rather than having to account for card counting.

It seems like this would only work if I could keep adding money from the outside. In the game show scenario, the first wrong guess would mean I would lose everything, the game would be over, and I’d go home with \$0.

A game might go like this:

• Bet \$200 on higher, roll comes up 3, bankroll now \$800
• Bet \$400 on higher, roll comes up 10, bankroll now \$1200
• Bet \$1200 on lower, roll comes up 12, bankroll now \$0. Game over.

It seems easy enough to understand the odds if I can keep getting money from outside, but I’m not seeing what the strategy would be when I start with a fixed bankroll and need to go a certain number of rounds before I can get the winnings out.

It depends on what your objective is. If you absolutely don’t want to go home with less than \$X and/or you have a particular \$Y target you want to reach then you need a hybrid strategy, but you really need to define your objectives.

If your goal is strictly to get the highest mean then big risk big reward is the way. I would guess if you chose “gamble every round”, including 7, you’d have around a 5%* of winning a cool \$1,024,000, (EV~\$50000). No other strategy will increase the average payout.

*This figure is a complete guess, could be miles out.

I guess I’m wondering what the perfect strategy would be in this kind of game in the same way that there is a perfect strategy for a game like blackjack. So rather than what a particular individual may want to achieve, what is the statistically best strategy to use in this kind of game?

In the case of a game show, the show needs to budget an appropriate amount for the winnings. Sure, some players will get lucky and win every round and some players will bet it all and lose everything, but over the long run the show should expect that the average player will take home a certain amount in winnings. The show would need to know the optimum betting strategy so they could calculate what the typical winnings will be.

But if it won, those further bets would be worth more. As I say it depends on what your objectives and risk aversion is, but in order to increase your average payout the bet it all every round strategy cannot be beaten.

The follow-on puzzle is to find the most elegant problem redefinition that avoids this trivial solution. (Assuming the \$1000 is Player’s entire wealth and applying Kelly Criterion is one way.)
I did run simulations. The program took only 35 LOC so I’ll give myself a 50-50 chance the code is correct despite my obvious infirmity. (Once again direct calculation would be faster and better, but more tedious to code.)

``````

Do Not Bet on Seven
0.29879%  \$1024k
1.85831%  \$512k
4.78142%  \$256k
6.60880%  \$128k
5.35585%  \$64k
2.58202%  \$32k
0.75017%  \$16k
0.12085%  \$8k
0.01078%  \$4k
0.00038%  \$2k
77.63263%  0
Mean Result = \$37,658

Bet on Seven
1.63896%  \$1024k
2.73548%  \$512k
2.09309%  \$256k
0.97883%  \$128k
0.31120%  \$64k
0.07225%  \$32k
0.00216%  \$16k
0.00131%  \$8k
0.00010%  \$4k
0.00003%  \$2k
92.15659%  0
Mean Result = \$37,623

``````

(In another thread we debate whether rounding or truncation is the norm, so I finesse that issue by showing the EXACT percentages that arose from the 10 million trials. )

Thanks for the visualization. That makes it easier to understand what the perfect strategy would be. It would occasionally have high payouts, but most of the time the players would go away with nothing. In the real world, I would doubt that the players would use those strategies on the game show since players want to go home with some winnings. So along those lines, what strategy would give the best median value for the winnings? Even though the mean winnings is about \$38k in your calculations, the median is \$0. If instead we wanted a strategy so that the median value was maximized, what strategy would that be? I suspect that’s the strategy that players would be more likely to use since it would be the one that would mean the average player would typically go home with the most amount.

This is a MUCH harder problem. Note, just for starters, that a player might want to stop betting once he gets above the target median! And strategy in the early rounds would differ from strategy in the later rounds.

One overly simple policy would be to always bet a percentage, B[sub]2[/sub], of bankroll when playing against a prior roll of 2, B[sub]3[/sub] against 3 and so on. Clearly this cannot be optimal for any {B[sub]x[/sub]} since it doesn’t take into consideration size of bank or number of remaining rounds, but it is a starting point.

The mean-maximizing strategies given above, when expressed in this form, are simply:
< B[sub]2[/sub], B[sub]3[/sub], B[sub]4[/sub], B[sub]5[/sub], B[sub]6[/sub], B[sub]7[/sub]> = <100%, 100%, 100%, 100%, 100%, 0>
and
< B[sub]2[/sub], B[sub]3[/sub], B[sub]4[/sub], B[sub]5[/sub], B[sub]6[/sub], B[sub]7[/sub]> = <100%, 100%, 100%, 100%, 100%, 100%>

For median-maximization the optimum among strategies with this trivial form may be close to
< B[sub]2[/sub], B[sub]3[/sub], B[sub]4[/sub], B[sub]5[/sub], B[sub]6[/sub], B[sub]7[/sub]> = <100%, 99%, 93%, 66%, 33%, 0>

In this case, simulation produces a \$10463 median along with
1 percentile = \$34
5 percentile = \$287
10 percentile = \$832
20 percentile = \$2574
30 percentile = \$4808
40 percentile = \$7402
50 percentile = \$10463
60 percentile = \$14579
70 percentile = \$19955
80 percentile = \$28559
90 percentile = \$45229
95 percentile = \$65298
99 percentile = \$124913
Arithmetic Mean = \$18855

I think when you bring money into the situation, it throws off the math. People start introducing psychological issues.

I feel the probability question is clearer if you keep it abstract. Just say that you start with a thousand points and your goal is to achieve the highest total of points. But the points are abstract; you will not keep anything at the end of the game. It’s just a mathematical problem to solve.

Thanks! This is the sort of the strategy that most players use on Card Sharks. The amount they bet is typically linked to the odds the next card will be higher or lower.

Here’s a clip from Card Sharks where the player got a great run of cards and bet the whole thing almost every time.

His cards were 2, 3, 7 (switched to K), 4, A, 8 (switched to 2), A, 7. Most players aren't this lucky and end up doing something like that median optimization strategy.

With the stated strategy as a “baseline” but making some adjustments (e.g. setting the round 9 and 10 bets with the target median in mind) I could get the median winnings up to \$16,000. There’s room for further improvement but I’ll stop here.