What is your strategy?

[kut2k2]

As it turns out, there is a well-known solution to this problem. See the "many horses" section in this article. In our particular case, the Red-16, Red-14, and Red can be thought of as 3 different horses, with the well-defined odds and edges. So, the Kelly criterion can be applied to this "multiple horses" situation as a combination bet just as well as it can be applied to just one horse as a single bet.

OK but we don't have the many-horses situation. Red is not mutually exclusive with either R-14 or R-16. I had to solve the Kelly equation the hard way, so I made the assumption that the bet on Red was exactly half of the betting fraction. This reduced the 16 unfavorable red outcomes to pushes, greatly simplifying the final equation.

But let's not forget my point: Kelly trumps standard deviation!

 

[nonlinear5]

OK but we don't have the many-horses situation. Red is not mutually exclusive with either R-14 or R-16.

Right, I will take my suggestion back.

 

[nonlinear5]

But let's not forget my point: Kelly trumps standard deviation!

I stll think there is a certain equivalence here, because the continuous Kelly is a generalization of the descrete Kelly, and the continuous Kelly is simply return divided by squared standard deviation. I have an idea of how about how to test for this equivalence, which I will do tomorrow.

 

[Visaria]

As it turns out, there is a well-known solution to this problem. See the "many horses" section in this article. In our particular case, the Red-16, Red-14, and Red can be thought of as 3 different horses, with the well-defined odds and edges. So, the Kelly criterion can be applied to this "multiple horses" situation as a combination bet just as well as it can be applied to just one horse as a single bet.

But the bets in our scenario aren't mutually exclusive...If the ball goes on either red14 or red16, RED is also a winner.

 

[SIUYA]

But the bets in our scenario aren't mutually exclusive...If the ball goes on either red14 or red16, RED is also a winner.

As a math probability question - does that just make the odds of R-16, R14 .....37-1 v 36-1 or does this seriously change the whole situation and you really should have to calculate for the two scenarios separately? Red/Black, or Red + R-16 or R-14

 

[Visaria]

As a math probability question - does that just make the odds of R-16, R14 .....37-1 v 36-1 or does this seriously change the whole situation and you really should have to calculate for the two scenarios separately? Red/Black, or Red + R-16 or R-14

you have 4 chances out of 37 of r16 coming up and 3 out of 37 for r14, 23 out of 37 for red to come up

 

[SIUYA]

you have 4 chances out of 37 of r16 coming up and 3 out of 37 for r14, 23 out of 37 for red to come up

thanks I understand that, I was just wondering about the maths in terms of separating out each number, and simply assigning the payout per slot, + the payout per colour or combining them as per number, and colour .......not the odds of them actually coming up on a spin....I guess in this case its pretty much the same.
Running through a couple of examples in my head I was thinking if the odds of a number coming up was 1-35, but its payout was different eg; 38-1 for a number, or 1.5-1 for a colour - would the calculations be radically different in how it would be best to calculate - as a combination or as separate.....probability maths is always fun.

But dont mind me - too much cake for lunch, and a bit bored . Cheers

 

[SplawnDarts]

It's impossible to know what the interviewers were looking for. The OP of the original thread never responded to any of the replies to his thread.

There are several "good" answers and many bad answers so maybe any of the good answers would suffice. It's difficult to believe the interviewers would expect The Absolute Best Answer in a 20-minute time limit.

Can we even determine The Absolute Best Answer? Your thoughts.

There's two pieces of KEY missing information:
Are we guaranteed the game is not rigged? Because there's no other reason for the house to offer it that I can see.
Do we have a "life bankroll" behind the $1000? I assume yes, but its size is a key parameter.

If those questions are both answered in reasonable ways, then there is in fact a "best" answer to the problem, although it can't be solved exactly on paper. It can however be reasonably approximated.

 

[nonlinear5]

Okay, I think I am closing on the solution.

First, I found a bug in my Monte-Carlo simulation, so the previously posted results should be discarded.

Second, I fixed the bug, and verified the code by running it against the various single bets on R16 (I have not tried the combo bets yet).

The results are below. Notice that although my simulation doesn't know anything about Kelly, it correctly identified the 7% or 8% bet on R16 as the best strategy. The score (i.e. the utility function) is the median profit after the 10 spins (thanks Kut2k2 for the suggestion).

	R16	R14	Red	Score
	7	0	0	795.41
	8	0	0	794.21
	6	0	0	776.28
	9	0	0	775.90
	10	0	0	743.39
	5	0	0	733.18
	11	0	0	699.22
	4	0	0	662.08
	12	0	0	645.68
	13	0	0	584.77
	3	0	0	558.47
	14	0	0	518.23
	15	0	0	447.60
	2	0	0	417.37
	16	0	0	374.22
	17	0	0	299.23
	1	0	0	233.24
	18	0	0	223.62
	19	0	0	148.22
	20	0	0	73.74

Here is how these results look in a graph:

And notice how the shape of the graph above corresponds to the theoretical shape implied by Kelly (below):

 

[Visaria]

thanks I understand that, I was just wondering about the maths in terms of separating out each number, and simply assigning the payout per slot, + the payout per colour or combining them as per number, and colour .......not the odds of them actually coming up on a spin....I guess in this case its pretty much the same.
Running through a couple of examples in my head I was thinking if the odds of a number coming up was 1-35, but its payout was different eg; 38-1 for a number, or 1.5-1 for a colour - would the calculations be radically different in how it would be best to calculate - as a combination or as separate.....probability maths is always fun.

But dont mind me - too much cake for lunch, and a bit bored . Cheers

Actually, i see where u are coming from. So let's say we bet $10 on both r16 and on red. We work out the payout if it lands on r16 as $350+$10=$450 with a probability of 4/37. We can work out the payout if it lands on red but not on r16, $10-$10=0, probability of 19/37, anything else is -$10 with prob of 14/37.