It’s soccer: I’m trying to fuse together two different types of bets in order to find the optimal division of stake. Here we go …
Bet A: has a 50% chance at decimal odds of 2.1. If it wins I lose the stake on Bet B
Bet B: has a 30% chance at decimal odds of 6.0. If it wins I lose the stake on Bet A
I know the numbers are crazy but we can all dream
It may or may not confuse the issue that Bet A is called ‘Draw No Bet’ which means the stake is returned if the result is a draw. If it matters, the usual all-market spread of results is Home Win: 36%, Draw 28%, Away Win 36%
Bet B is a correct score prediction, so a conventional all-market, win-lose market
As above, how would I go about optimising the stake as between the two aspects of the bet?
It’s certainly tricky but I’m not sure i understand what you’re asking that isn’t there.
Trying to put it slightly differently; the two bets are different in nature: in A the stake is returned if there is a draw. In B, the bet is only won if the draw is of a particular kind (1-1).
I think what I’m trying to do is create two probability (minus the stake of the other bet - which will be a different amount), and then somehow apportion the stake between the two probabilities for maximum potential.
One major fly in the calculus ointment for me is the returned stake in A
I think you are saying that p_1 = 0.5 and p_2 = 0.3, but I don’t know what p_3 or p_4 are. If you can supply those numbers you can find the maximum of E(X) for X taken from [0,1]. Since E(X) is a line segment, it will probably be one of the endpoints (or every point in rare pathological cases), most likely X = 0 since that is a ridiculous bet.
When a bet of X is placed at decimal odds D, winning that bet results in a a net gain of (D-1) * X. Losing the bet results in a loss of X (which is the same as a gain of -X).
So for bet A in the original post you listed decimal odds of 2.1. We have p_1 the probability of both winning bet A where we bet X and losing bet B where we bet 1 - X. So we gain 1.1 * X and lose 1 - X (gaining -(1 - X)) with probability p_1. So the first term of our expectation function is p_1 * (1.1 * X - (1 - X)).
For the second outcome (1-1 draw), you gave decimal odds of 6 for bet B. So if this occurs (probability p_2), we push bet A (gaining nothing) and gain 5 * (1 - X) from bet B. This gives the second term of our expectation function p_2 * (1 - X).
The other terms are derived similarly.
Then there’s this.
This doesn’t make sense to me.
We have four mutually exclusive outcomes that comprise all possible outcomes. p_1 + p_2 + p_3 + p_4 must equal 1 exactly. The probabilities listed above add to 1.33. One or more must be incorrect.
Furthermore the odds listed don’t match the original post and we only need to know the odds for bets which we wish to make.
Perhaps if you link me to a page where the odds are posted and tell me which bets you wish to make I can give you a better answer.
IF you put a dollar down on Bet A, you win the dollar.
If you put a dollar down on Bet B, you expect to win (A little more than) two dollars.
So if you had lots of games where you can do this, forget bet A , go for bet B. But thats too simple the case. This question is about a single game of soccer, and A is team A winning and B is team B winning.
And so “optimum” means to win as much as possible, but also avoid making any loss if it all possible.
The reason that the odds of the draw wasn’t stated was that it must be the remaining 20%… Anyway, in the draw, the money is refunded from bet A, but there is no way to avoid the loss from bet B. So there’s no avoiding the loss from a draw if bet B is used. If Bet B is not used, there’s no avoiding making a loss if team B wins.
So the optimal bet pattern is to hope to make a profit from bet B, but put the same stake on team A, so that if team A wins, no loss occurs .
There’s no optimising the loss from a draw in that case, but if you can bet on ten different games, you would likely make a profit overall by hedging.
Am I the only one who finds the terminology “decimal odds” annoying? Let me see if I have it right:
2-to-1 and 3-for-1 are the same thing and denoted as “2/1 (fractional odds)” or “3.00 (decimal odds).” That’s European decimal odds as opposed to American" decimal odds which would be “2.00.”
If you take the opposite side of that bet, which would be 1-to-2 or 1/2 in “fractional odds,” the “decimal odds” quotation would be “1.50” and the “American” quotation would be “-2.00.” Note that 1.00 = -1.00 in the “American” system. :smack:
I find it annoying as well, but that may be just because I’m more accustomed to other methods. One nice thing about UK style decimal odds is that the implied probability is the the reciprocal of the odds listed.
I’m just trying to show the OP how to maximize expected value so that he can see for himself that it is generally a bad idea to bet against oneself. Some day we may wish to cross the bridge above, but we can’t even see the bridge from where we are.
These calculations are not trivial so I will not include them in this post, but the counterintuitive result is that we should wager approximately 12.4% of our bankroll on A and 0% on B. The high volatility of bet B counters its higher expectation. In this case the low risk low reward of A is the way to go.
OP raises an interesting question which was not fully and correctly answered upthread. In this post I correct the misconception and state a little-known but elegant theorem: In a well-defined “zero-vigorish” situation, you should partition your entire bankroll into bets (p, q, r, … ) proportional to the probabilities of the associated outcomes. This partitioning does not depend on the odds you are offered (as long as they are a feasible zero-vigorish odds).
In the sequel I demonstrate the theorem by example, but first prove a simple corollary. First, problem statement and notation:
In our scenario, a gambler chooses from betting on two mutually exclusive results. He can wager on A at payoff of a-for-1; A occurs with probability p. He can also wager on B at payoff of b-for-1; B occurs with probability q.
0 < p,q < p+q < 1
We suppose both bets are favorable with A more favorable; that is
ap > bq > 1
It is convenient to write r = q/p; Thus
a > br > 1/p > 1
We suppose the gambler has bankroll 1 and bets some fixed amount k
0 < k < 1
We want to solve for x, the amount he wagers on A
0 <= x <= k
He will wager k-x on B.
Thus there are three outcomes
With probability 1-p-q, his new bankroll (B[sub]new[/sub]) = 1 - k
With probability p, B[sub]new[/sub] = 1 - k + ax
With probability q, B[sub]new[/sub] = 1 + (b-1)*k - bx
By Kelly’s theorem we seek x which maximizes
f(x) = (1-p-q) log(1-k) + p log(1 - k + ax) + q log(1 + (b-1)k - bx)
Solving df/dx = 0, and recalling r = q/p yields
a / (1-k+ax) = br/(1 + (b-1)k - bx)
x(1+r) = (1/b + k - k/b - r/a + rk/a)
Solving this equation will produce x > k when k is sufficiently small. In that case the gambler should wager only on the more favorable bet, A. But optimal x is given by the following when it is less than k.
x[sub]Optimal[/sub] = ( (1-k)/b + k - (1-k)/(ar) ) / (1+r)
Now a positive “hedge bet” on B of $(k - x[sub]Optimal[/sub]) will be appropriate.
x[sub]Optimal[/sub] > 0
Since k,b > 0
kb > 0
kb + (1-k) - (1-k) > 0
k + (1-k)/b - (1-k)r/(br) > 0
Recalling a > br
k + (1-k)/b - (1-k)r/a > 0
(1/b + k - k/b - r/a + rk/a) ) > 0
But the left-hand side was shown above to be x[sub]Optimal[/sub](1+r), so
x[sub]Optimal[/sub](1+r) > 0
Since 1+r > 0, we have
x[sub]Optimal[/sub] > 0
~ ~ ~ ~ ~ ~
Before stating and demonstrating the Interesting Theorem, I replace our variables with simple constants. This will clarify and simplify. (We’ll be left without a general proof, but proofs can be found on the Internet)
I hope the simple numeric example will clarify the workings.
a = 5
b = 4
p = q = 1/3
r = p/q = 1
The condition for optimal x above
x = ( (1-k)/b + k - (1-k)/(ar) ) / (1+r)
x = (1 + 19k) / 40
With this substitution, the outcomes, each occurring with probability 1/3, are
The optimal k will maximize the product of these three equally likely outcomes
Maximize (1 - k)(11k + 9)^2
(1-k) * (242k + 198) - 121kk - 198 k - 81 = 0
k = (-154 + sqrt(154154 + 4363117)) / 726
k = 13/33
x = (1 + 19k) / 40 = (1 + 19*13/33) / 40 = 7/33 Therefore the optimal move in our scenario is to bet 7/33 of your bankroll on A and 6/33 on B. (Retain 20/33 of your bankroll unwagered so you can live to fight another wager should you be unlucky.)
The outcomes are now seen to be
case A: B[sub]new[/sub] = 1 - k + 5x = 55/33
case B: B[sub]new[/sub] = 1 + 3k - 4x = 44/33
case C: B[sub]new[/sub] = 1 - k = 20/33
C is defined as the remaining case, i.e. where both A and B fail.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Let us now consider a slightly different problem.
You and the oddsmaker have different notions of the true chances of outcomes A, B and C. Because the oddsmaker offers 5-for-1 and 4-for-1 he thinks the chances are 20% and 25% so C is 55%, so he offers 20-for-11 (the correct no-vigorish rate) on that. You are, let us suppose, required to bet your entire bankroll on outcomes A, B and C, in some proportion. (This stipulated bet is optimal for you even if not required.)
Interesting Theorem. In this situation, your best strategy is to bet a portion p of your bankroll where p is your judgment of A’s true probability, a portion q on B where q is again your judgment, and the remaining (1-p-q) on C.
In our example, p = q = 1-p-q = 1/3 so you bet 1/3 of your bankroll on each of the possible outcomes. Let us see what the results are.
In Case A, your 1/3 bet on A pays off 5-for-1 so you get $5/3 back or $55/33.
case A: B[sub]new[/sub] = 55/33
Similarly for Case B, and in Case C your 1/3 bet is paid off 20-for-11
case B: B[sub]new[/sub] = 44/33
case C: B[sub]new[/sub] = 20/33
These numbers for bets (1/3, 1/3, 1/3) are identical to the results derived earlier for (7/33, 6/33, 0). The two betting regimes are equivalent (recall that we’ve stipulated your counter-party is taking zero vigorish based on the probabilities he believes in.) But the latter was proven to be optimal. Therefore the former is as well; the Theorem is demonstrated for this case.
Wrong. You must have made an error in your calculation.
See Remark 1 above.
septimus, you have simplified away the interesting part of the OP. The OP wants to explore making two very specific kinds of bets. The first bet has a nonzero probability of pushing. This can’t be wished away because pushing the first bet is a necessary but not sufficient condition for winning the the second bet. There are four possible outcomes of this combination bet so your solution, while it appears solid mathematically, is not sophisticated enough to provide a solution to this problem.
Remark 1 does not apply for reasons given above. I have given the correct solution to the problem that I stated in post 16. It looks like you have given the correct solution to some other superficially similar problem.