Valuation question

[Bill]

I have a sort of math/business question--say I have 3 offers that won't expire, that are worth:

$9.3m, but has a 75% chance of not being approved
$7m, but has a 50% chance of not being approved
$6m, which has a 100% chance of being approved

And let's say I'm negotiating a fourth offer and want to decide on my resistance point for it.

Now, my understanding is that the correct way to value the first offer, if it were the only one available, would be $2.325m (25% of $9.3m), but what about if all three available? Since the offers don't expire I can "roll the dice" on the first one, and if it fails take the second, and so on.

So with the fourth offer, what is the lowest I should accept before walking away and taking, or attempting to take, one of the original 3 offers?

[SpoofyChop]

I think you want the "expected value" of the deals.

It seems like you have the first deal's value backwards.

1) 75% chance of getting $9.3m which is $6.975m
2) 50% chance of getting $7m which is $3.5m
3) 100% chance of $6m

I think this means your best bet from a mathematical point of view is #1 and you would only bother with offer #4 if the expected value was greater than $6.975m

I'm not a math wizard though so don't take my advice.

[Orinoco]

I assume you mean 25% chance of being approved for the $9.3M? Otherwise it doesn't make sense.

[Rimbo]

"$9.3m, but has a 25% chance of not being approved" (if I have that correctly) does not mean a value of 3/4ths of $9.3m. It means it has a 3/4ths probability of being $9.3M and a 1/4ths probability of being $0. It only makes sense to multiply by 3/4ths if you get 4 tries to get the money and can get the money more than once.

[Bill]

I assume you mean 25% chance of being approved for the $9.3M? Otherwise it doesn't make sense.

You are correct, I mispoke. First post fixed now.

[tiohn]

Spoofy has it right. If I am understanding Rimbo correctly, he is incorrect.

[Wisbechlad]

The expected value of your deal is:
7.2m

You have a 25% chance of 9.3m
You have 37.5% chance of 7m (0.5 x 0.75, because 25% of the time, the 9.3m has bben approved, and you took that)
You have 37.5% chance of 6m (if above two deals fall through)

So expected value is .25*9.3+.375*7+.375*6= 7.2

[tiohn]

Crap. I didn't read the last couple lines of the OP.

[Sidd_Budd]

So with the fourth offer, what is the lowest I should accept before walking away and taking, or attempting to take, one of the original 3 offers?
Assuming you have a neutral risk tolerance, $7.2m, just as Wisbechlad found.

It gets a bit more complicated if you are risk-averse, however, because you are comparing a certain amount to an uncertain outcome (the series of 3 original offers).

For example, assume the fourth offer is $6.8m. This is lower than the expected value of the uncertain 3 offer series, so a person whose only consideration was expected outcome would reject the fourth offer. A conservative person might take it anyway, because he can't stand the thought that if he rejects the fourth offer, 38% of the time he ends up with only $6m (the worst of the original 3 offers).

[Bill]

Thanks Sidd_Budd and Wisbechlad, that's basically what I was thinking but was unsure about how to work the math.

Btw, this was part of a case in a negotiation class, where I had negotiate the fourth offer but I was unsure what my minimum walkaway value (resistant point) should be. But it all worked out ok because I negotiated for $9.4m :)