[nixbits]

what offer would minimize the payout by the banker on the average in a two box {a,b} deal or no deal game?

[RollandGarros]

Depends on the last two amounts, surely?

[CLS94]

It depends on what he judges the likelihood of the player accepting to be. In the long term, any accepted offers below the average of the remaining boxes will reduces his average payout, so it becomes a matter of how far below the average he can go without the offer being rejected. This is normally dependent on the precise situation (mainly the risk, a player is more likely to gamble when there is only a small difference between the remaining boxes) and the circumstances and personality of the player which he will have judged throughout the game and even before it (whether they are a gambler, have a dependant family, their financial security etc.)

[MMoor18]

I'd probably go with a formula along the lines of (A+B)/4

so if it was 100k and £50 an offer of £25,012.50, would be hard to say no to but would still have that chance of getting 100k.

[Cll_ws]

(Original post by MMoor18)
I'd probably go with a formula along the lines of (A+B)/4

so if it was 100k and £50 an offer of £25,012.50, would be hard to say no to but would still have that chance of getting 100k.

That wouldn't work in some cases though would it? What if they had 100k and 250k remaining? with that formula the offer would be below the lowest amount they could win!

[MMoor18]

(Original post by Cll_ws)
That wouldn't work in some cases though would it? What if they had 100k and 250k remaining? with that formula the offer would be below the lowest amount they could win!

ah ye that's true..... so maybe A being lowest B being highest

B/4 + A

[mikeyd85]

High box (H) - Low box (L) = Difference (D)

L + (D/100*55)

[MMoor18]

(Original post by mikeyd85)
High box (H) - Low box (L) = Difference (D)

L + (D/100*55)

55% of diffrence + lower box? that's too high

It would never get rejected.

again with 250 and 100k it would be 55% of 150 (82.5k) +100k

182.5k is a huge offer and no one would reject it as your betting 82.5k against 67.5k, poor odds that don't make sense

[mikeyd85]

(Original post by MMoor18)
55% of diffrence + lower box? that's too high

It would never get rejected.

again with 250 and 100k it would be 55% of 150 (82.5k) +100k

182.5k is a huge offer and no one would reject it as your betting 82.5k against 67.5k, poor odds that don't make sense

I wouldn't deal for anything less in a high value two box game.

[MMoor18]

(Original post by mikeyd85)
I wouldn't deal for anything less in a high value two box game.

I would, in that scenario you're betting 87.5k against 62.5k in a 50/50 bet.

You'd be better off taking the money, betting the same 87.5k on black on roulett with the same 50/50 odds and getting 175k instead of 150k........


If the reward is lower than the stake in a 50/50 bet then the risk isn't worth taking. Simple maths

[CLS94]

(Original post by mikeyd85)
High box (H) - Low box (L) = Difference (D)

L + (D/100*55)

In addition, that would be likely to make the banker lose money because the amount the deal would be done at would exceed his avaerage payout were the deal to be rejected.

[mikeyd85]

(Original post by CLS94)
In addition, that would be likely to make the banker lose money because the amount the deal would be done at would exceed his avaerage payout were the deal to be rejected.

What can I say? It takes the extra 5% to stop me gambling

[samiz20891]

Well assuming the only two boxes were the 100k and 250k boxes in an one-shot game. Then the banker would offer the average, 175k, minus whatever he feels the players risk aversion is, so that the more risk averse a player is, the lower the offer from the banker as the player is against risk. However, if the player is risk loving, then the banker cannot really offer more than 175k as any more than the average payoff will lead to a loss over time assuming a 50/50 chance.
So it depends on risk preferences of the player, what the optimum offer for that player is, and this will generally be looked at during the game.

[nixbits]

No offer should be larger than the mean E= (a+b)/2 as on the average if no deals were made this is what would be payed out per game.

Obviously the offer should be larger than the lower value b so if x is the offer
b< x < E

as the offer gets bigger the probability of dealing will increase. Call prob of dealing for offer x = Pd(x).

The average ammount the banker pays out for this game = A(x)

A(x)= x.Pd(x) + E.(1-Pd(x))

To find the offer which minimises the payout calculate dA/dx = 0.

To do this we need a model of how the probability of making a deal varies with the deal ammount. We could get this empirically by seeing how a population of players actually behaved. But a simple model may give an idea of the optimum ammount. Eg suppose the probability increased linearly from 0 for an offer of b, to some value D for the offer E= (a+b)/2. D would be the fraction of the population who would gamble on a 50/50 bet rather than take the certainty of the offer.

so Pd(x)= {D/(E-b)} (x-b) is this linear model using this in the A(x) and rearranging gives

so A(x) = {D/(E-b)}(x.x -(E+b).x +Eb) + E

so dA/dx = 0 = {D/(E-b)}(2x - (E+b))

so x = (E + b) /2 = (a + 3b)/4 as the best offer to minimise payout


The model maybe a bit unrealistic, i would expect more of an s shaped curve with low probability of deal until the amount the player risks loosing = (X-b), by not accepting the offer is comparable to the expected gain = (a- X) where it will increase rapidly toward some plateau . see for example a function like

{exp(cx) -1}/{exp(cx) +1} which runs from 0 to 1 as x goes from 0 to infinity

[nixbits]

A plausible non linear, s shaped probability curve for the problem is:

P(z) = k.[exp(cz)-1]/[k.exp(cz)+1] where 0< z < 1 z=(x-b)/(a-b)

where x is the offer and (a,b) the two box values with a>b . k and c are constants. (I'll post some pics of the way this function behaves as c or k changes when i capture them from my spreadsheet).

A reasonable model of the behaviour of contestants playing the (250k,100k) game occurs for c=10 and k=0.02. Using this model to calculate the average ammount payed out if the offer is x shows that the

optimum offer X= 0.35 a + 0.65b ie a little more than (a+2b)/3.

This is a larger offer than predicted by the linear model.

This s shaped curve model is appropriate for all the cases when a and b are both large values or both small. If b is much smaller than a, the contestants will be more likely to accept lower offers rather than risk getting the b value as prize. We can model this behaviour by changing c and k (i.e we could allow both to be functions of b/a). In the case of the 2 box game {250K,1p} appropriate values of the parameters are c=15 k=0.1. This gives a probability of accepting offers which rise rapidly with the offer and reach a plateau. With such a model

optimum offer X= 0.23a + 0.77b i.e a bit smaller than (a+3b)/4 which was the

best offer for the linear model.

So we can say the offer will roughly be 0.23a + 0.77b < X < 0.35a + 0.65b

the lower bound ocuring when b/a is negligible the upper when 1/20< b/a .

[nixbits]

Here are the promised pics of the s shaped probability model for accepting an offer and the expected pay out by the banker given the model applies:



The constraint on the probability model was that the probability of accepting an offer equal to b ,the lower value box, is zero ; of accepting an offer equal to the higher value box, equal to a, is 1 (certainty) and that the probability of accepting an offer equal to the mean , E, is some value D which characterises the average behavior of the population towards gambling on a 50:50 bet. I have estimated that given an offer equal to the mean value (a+b)/2 about 80% of the population will take the certainty of the offer rather than gamble on getting a and risking getting b ie P(X=b) =0 , P(X=a)=1, P(X=E) ~ 0.8.

The 4 curves represent different models of population behaviour. curve 1 and 3 are applicable to the cases where the 2 box values are similar ( (b/a)~1) while 2 and 4 apply to the case where b is much smaller than a ((b/a)<<<1).

The payout curves show the average amount the banker will have to pay out in a specific 2 box game {a b} if he makes the offer z . Each curve (1-4)corresponds to the probability curve with the same number. The optimum offer the banker should make to minimise his pyout is obviously the z at the minimum of the payout curves = Zmin

The size of the best offer = a . Zmin + b . (1-Zmin).

so 0.23a + 0.77b < X < 0.44a + 0.66b
case b<<a approx (a+3b)/4 case b~a approx (2a+3b)/5 ; as rough rule of thumb

These agree quite well with data on the offers made in the game.

[nixbits]

some data from actual games comparing offer and predicted offer:

case 1 (value of b similar to a)

{250000,75000} offer made 150000
(2a+3b)/5 =145000; 0.44a + 0.66b= 159,500
using rule of thumb (2a+3b)/5:
{250000,20000} offer 110000 predict 112000
{35000, 3000} offer 15000 predict 15800
{3000,500} offer 1500 predicted 1500
{20000, 15000} offer 17200 predicted 17000
{100000,10000} offer 45000 predicted 46000
{250000,100000} offer 168000 predicted 160000

case 2 b much smaller than a

{10000,0.01} offer 22000 predict (a+3b)/4 =23000
{35000, 50} offer 8000 predict 8787


{

[Mark85]

(Original post by MMoor18)
I would, in that scenario you're betting 87.5k against 62.5k in a 50/50 bet.

You'd be better off taking the money, betting the same 87.5k on black on roulett with the same 50/50 odds and getting 175k instead of 150k........


If the reward is lower than the stake in a 50/50 bet then the risk isn't worth taking. Simple maths

It isn't as simple as you make out. For example, £200k isn't worth twice £100k to me and the same applies to most people. So you have to factor in the law of diminishing returns, the appetite for risk of the contestent and also their personal circumstances (if someone has £40k left to pay on their mortgage, then offered £45k when £20k and £100k are left might seem more appealing)

[MMoor18]

(Original post by Mark85)
It isn't as simple as you make out. For example, £200k isn't worth twice £100k to me and the same applies to most people. So you have to factor in the law of diminishing returns, the appetite for risk of the contestent and also their personal circumstances (if someone has £40k left to pay on their mortgage, then offered £45k when £20k and £100k are left might seem more appealing)

no I'm not saying don't deal, if you want to deal that's fine, that's the game.

Infact I said the opposite, in that scenario I put dealing is the only viable option. Even if you'd rather risk it for the 250k you're financially better off dealing and putting the excess over 100k on black, because the chance of winning is still 50/50 but the returns are better. Anyone who would bet 187.5k (or but the 87.5k at least as the 100 is garunteed) on a 50/50 when they pay out is less than the stake then they deserve to lose the money.

[nixbits]

I agree with the last comment but in actual games of deal/no deal the offer for the 2 box game {a,b} is always less than the box average (a+b)/2 , often much less at about (a+3b)/4, some times even lower, so the gain/stake can be as high as 4 on this 50/50 bet (gain/stake=(1-Zmin)/Zmin where the bankers offer was a.Zmin + b.(1-Zmin). In which case no deal is the better option from a purely betting perspective.