Nightmare Math / maths (betting question)

This is really twisting my melon.

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 :slight_smile:

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?

Are these bets independent where P(Win A and B) = 15%? What happens here?

Or are they mutually exclusive where we have:
P(Win A) = 50%?
P(Win B) = 30%?
P(Lose A and B) = 20%?

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

One way to look at this is to divide it into mutually exclusive outcomes and multiply the probability of each outcome by the amount won in each case.

I think we have four outcomes to examine here:

  1. The team bet on in A wins.
  2. There is a 1-1 draw.
  3. There is a draw other than 1-1.
  4. The team bet on in A loses.

Call the probabilities of these outcomes p_1 … p_4 respectively. They will add up to 1.

Assume we bet X on a and 1 - X on B. X will be the proportion of the total amount we wish to bet on A.

Then we have:

E(X) = p_1(1.1 X - (1 - X)) + p_2(5(1 - X)) + p_3(-(1 - X)) + p_4(-X - (1-X))

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.

Let me know if this helps.

I will let you know Lance but do give me a while to ponder your thoughts :slight_smile:

there is also:

  1. Not a draw (which means the stake on Bet B is lost, but bet A may/not win (50/50))

Without knowing why I think that’s why I mentioned the probability of H, D, A:

I think p_4 is probably in there somewhere

p_3 is minimal at 3%

p_5 is approx 66%

Thanks :slight_smile:

Outcomes 1 and 4 are both not a draw and cover both possible not draw outcomes. There is no reason for outcome 5.

I’ll spell it out a little more clearly. Let H be the the the number of goals scored by the team bet on in bet A and V be the number of goals scored by their opponent.

There are four outcomes we need to consider.

  1. H > V. Result: Bet A wins, bet B loses.
  2. H = V = 1. Result: Bet A pushes, bet B wins.
  3. H = V != 1. Result: Bet A pushes, bet B loses.
  4. H < V. Result: Bet A loses, bet B loses.

Lance, could you help me with a couple of clarifications. I love the beauty of this btw.

what does 1.1 represent, and do I assume it is multiplied by X?
There is something similar in the p_2 equation with the ‘5’?

I’m sorry, I don’t know what X represents - stake, I guess? If so, I assume 100 units. Additional information - odds are decimal not fractional:

p_1 = .5 (odds: 2.09)
p_2 = .3 (odds: 7.2)
p_3 = .03 (odds: 34.0)
p_4 = .5 (odds: 1.89)

The odds on offer seems really key information …

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.

People, you are over thinking the mathematics.

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.

Okay, now we’re cooking:

X = Stake (100 units)
D = decimal odds
When we get into the equation we use (decimal odds minus 1, or D-1).
Thank you for that. I’ll work on the above and comment on below:

The original odds were slightly random - I was lost in the thinking, I’m sorry for that. The second set is accurate, I used them this week.

A question on ‘mutually exclusive’. The bet itself confuses me:
If bet A wins, bet B must lose
If bet A loses, bet B loses
If bet B wins, bet A loses
If bet B loses, bet A could win or lose

I don’t know enough to state if those are mutually exclusive or not.

Question on the four equaling 1:

Are you saying this because the market must equal 1. If so that is the nub of my confusion - I am taking odds from two different markets.

To add context, the stated probabilities are not a product of all matches, this is a slice, a niche of the general market - the general/expected set of results/math is skewered.

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:

Is that right?

OP asks about “betting against himself.” Let’s distinguish three cases.
(a) Both bets are unfavorable. Do not bet.
(b) One bet is favorable, one unfavorable. Place only the favorable bet unless a loss would impact your bankroll materially. If the bet is large, you may wish to hedge – place money on the unfavorable bet – but to compute the optimal bet/hedge ratio, we’d need your bankroll size and/or risk adversity.
© Both bets are favorable. If your total wager is smallish, then, as Lance Turbo implied upthread, you should just make the more favorable bet. If your bet is large enough to affect you materially, see (b).

That is confusing. :slight_smile: I think you’re saying B wins only when A pushes. Obviously you need to provide a complete schedule of outcomes, probabilities, and payoffs before anyone can deduce best strategy.

X is not the stake. X is the fraction of the stake we are wagering on bet A or X = (amount bet on A)/(amount bet on A + amount bet on B)

These are not mutually exclusive. “If bet A wins, bet B must lose” is completely contained in “If bet B loses, bet A could win or lose” for example.

Think of it this way. Bet A has three mutually exclusive outcomes. Win, lose, and push. The probability that exactly one of things happen is exactly 1.

Bet B cannot push so it has two mutually exclusive outcomes. Win and lose.

To find the mutually exclusive outcomes for a combination of bet A and bet B take the Cartesian product of these lists:

  1. Win A, Win B
  2. Win A, Lose B
  3. Lose A, Win B
  4. Lose A, Lose B
  5. Push A, Win B
  6. Push A, Lose B

However, two of these outcomes are impossible (probability zero) so they can be thrown out:

  1. Win A, Win B
  2. Lose A, Win B

That leaves us with four that we’re concerned with. I will reorder them to match previous posts:

  1. Win A, Lose B
  2. Push A, Win B
  3. Push A, Lose B
  4. Lose A, Lose B

These are the outcomes we need to determine probabilities for to answer your original question.

Can you please explain precisely what these four bets are and where the probabilities came from? I do not believe they correspond exactly to the four outcomes listed above.

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.

Let’s concretize this a real example:

We have Chelsea v. Sunderland this weekend. Chelsea draw no bet is being offered at 1.09 (A) and 1-1 draw is being offered at 11.5 (B).

In my estimation these are the probabilities for the four outcomes I care about:
Chelsea wins -> 77%
Sunderland wins -> 6%
1-1 draw -> 9%
any other draw -> 8%

These are a ballpark correct estimation of the true probabilities that I have tweaked a bit to illustrate a point.

Whatever the case, when we throw all these into the formula given earlier we get:
E(X) = -0.0257 X + 0.035

We would like to find the value of X taken from [0,1] that maximizes E(X), but this is easy since E(X) is a line with a negative slope so the maximum is achieved at X = 0.

This means we should put 100% of the money we wish to wager on bet B.

But wait, I’ve just heard that Chelsea’s coach has been fired. This causes me (for reasons unexplained) to reevaluate the probabilities above to the following:

Chelsea wins -> 77%
Sunderland wins -> 6%
1-1 draw -> 8.7%
any other draw -> 8.3%

Now I calculate:
E(X) = 0.0088 X + 0.0005

Since the slope is now positive, E(X) achieves its max at X = 1. This means I should put 100% of the money I’m willing to bet on bet A.

Hope this helps.

Finally, I will address **septimus **'s concern.

Say we have large bankroll and wish to know what fraction of said bankroll to wager on A and B to optimize bankroll growth given…

Chelsea v. Sunderland. Chelsea draw no bet is being offered at 1.09 (A) and 1-1 draw is being offered at 11.5 (B).

With probabilities…
Chelsea wins -> 77%
Sunderland wins -> 6%
1-1 draw -> 9%
any other draw -> 8%

Both bets are favorable and the expectation of B is slightly higher than that of A (3.5% versus 0.93%).

To do this we need to maximize the weighted geometric mean of possible outcomes.

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.

That’s a whole chunk of information for which this dyslexic s hugely grateful! Unlike Jose Mourinho I have a busy weekend ahead but will certainly make the time to digest.

Lance, I don’t know if you know of this staking formula, it certainly has credibility among those who seem to know:

http://www.albionresearch.com/kelly/

Kelly betting is logically equivalent to maximizing the geometric mean. I referenced it indirectly in post 16.

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.

Remark 1
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
or
(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
Q.E.D.

~ ~ ~ ~ ~ ~

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.
Simple example:
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)
becomes
x = (1 + 19k) / 40
With this substitution, the outcomes, each occurring with probability 1/3, are

B[sub]new[/sub] = 1 - k
B[sub]new[/sub] = 1 - k + 5x = (11k + 9)/8
B[sub]new[/sub] = 1 + 3k - 4x = (11k + 9)/10

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(154
154 + 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.
Q.E.D.

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.