Interesting game theory question (1 Viewer)

Leonard

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From a friend:

Since you are the math major and odds expert, can you solve this problem?

How Badly Can A Car Salesman Swindle You?

Source: http://fivethirtyeight.com/features/how-badly-can-a-car-salesman-swindle-you/



My reply so far:
First thoughts.
You have to take the first two cards. If they are 400 apart then your strategy is clear: take the lowest card if it was second, otherwise wait for the lowest +100. If the first two cards are less than 400 apart different strategies will be needed

I'm at work and unable to put any time into the question but I thought it would make an interesting discussion
 
Well we are talking about a car salesman - - so every price will be $2,000+ correct price no matter that you think the spread is $100 - $400 because he lied about the conditions of the game. And then there are the special charges, prep fee, dealers fee, county tax fee, asparagus fee, nitrogen for your tires, bed liners, undercoating, xm radio up charge . . . .

You'd be doing well only to get skinned for $3,000

DrStrange
 
I'm inclined to say the expected average would be N+250. Not running any math here, just my natural gut calculation - which is usually off when calculating my odds of catching a draw.
 
Because those aren't the five cards......

he takes five index cards and writes down a number on each of them: N, N+100, N+200, N+300 and N+400.
 
If you stop at the first card, your avg purchase price will be N+200:

N+0
N+100
N+200
N+300
N+400
------
5N+1000 / 5 = N+200
 
Without getting into it too deep, I'd expect that optimum play would be to:

a) if second card is lower than first card, stop and pay that amount (worst case, purchase price = N=300)
b) if second card is higher than first card, continue drawing until a card hits that is lower -- unless the first two cards are N and N+400 in that order, then use Leonard's strategy (continue drawing until N+100 appears)

If the first four cards in b) are all incrementally higher than the previous one, then the fifth card is either N or N+400. Staying with the fourth card makes the purchase price either N+300 or N+400 (N+350 avg). Drawing the fifth card makes the purchase price either N or N+400 (N+200 avg), so drawing the fifth card is correct.

If the cards get higher but not in set increments, optimum strategy should play out pretty easily.
 
Last edited:
First, the average is N+200.

Second, I think you can beat that by playing the game correctly.

Draw two cards:
Without getting into it too deep, I'd expect that optimum play would be to:

a) if second card is lower than first card, stop and pay that amount
b) if second card is higher than first card, continue drawing until a card hits that is lower

You can improve on that by saying in case b, if the two cards are not adjacent (n and n+100) keep drawing until you get <n OR exactly n+100. You know n+100 is coming and it is your worst case scenario.

L
 
Ah, yes... thanks BG. I revise my initial assumption to N+150. I guess that I felt that N+0 was not even a possibility.
 
Ah, yes... thanks BG. I revise my initial assumption to N+150. I guess that I felt that N+0 was not even a possibility.

It is not an option at a car dealer in real life, but it is an option in the question as asked.
 
OK, I solved it :)

I believe the answer should be $96.67 (or $96 & 2/3 to be precise).

Here's how I solved it...

Given: We must take at least 2 cards to gain information, otherwise we can expect to pay the EV of a random card, which is N+$200

There are then 8 possible outcomes, each with different odds of occurring. These outcomes are that our 2nd card is a delta of the following values from the 1st card: +400, +300, +200, +100, -100, -200, -300, -400. We must weight each outcome according to it's respective probability within our decision tree.

I then calculate the EV for decisions within each branch of the tree, and come up with the following strategy as being GTO (game theory optimal).

Case +400: STOP! which yields paying N
Case +300: STOP! which yields paying N, N+100
Case +200: STOP! which yeilds paying N, N+100, & N+200
Case +100: DRAW which yields paying an average of N+108.33, 4x
Case -100: DRAW which yields paying an average of N+108.33, 4x
Case -200: DRAW which yields paying an average of N+88.89, 3x
Case -300: DRAW which yields paying N+100, & N+100
Case -400: DRAW which yields paying N+200

So... when you add it all up and divide by 20, you get an expected value payout of N+$96.67

I have it all on chicken scratch notebook paper if you really want to see it, but it's kinda all over the place and not easy to follow :)
 
My guess.

If the gap between 1st card and 2nd card is negative or +100 you pay the amont of the 2nd card
If the gap between 1st card and 2nd card is +200 or +300, it will depend on the 3rd card
If the gap between 1st card and 2nd card is +400 you pay +100


1st,2nd ; Gap ; You pay

+0,+400 ; +400 ; +100
+0,+300 ; +300 ; +100
+0,+200 ; +200 ; +100
+0,+100 ; +100 ; +100

+100,+0 ; -100 ; +0
+100,+400 ; +300 ; +0+200
+100,+300 ; +200 ; +0+200
+100,+200 ; +100 ; +200

+200,+0 ; -200 ; +0
+200,+100 ; -100 ; +100
+200,+300 ; +100 ; +300
+200,+400 ; +200 ; +0+100+300

+300,+0 ; -300 ; +0
+300,+100 ; -200 ; +100
+300,+200 ; -100 ; +200
+300,+400 ; +100 ; +400

+400,+0 ; -400 ; +0
+400,+100 ; -300 ; +100
+400,+200 ; -200 ; +200
+400,+300 ; -100 ; +300
--------------------------
Avg. +3100/24 = +129.16
 
Here's a twist. If you were the salesman what order would you present the cards to get the best price?
 
RM.jpg
 
I suppose all of this assumes we know the pattern of the cards. If we don't, we cannot play optimally.
 
OK, I solved it :)

I believe the answer should be $96.67 (or $96 & 2/3 to be precise).

Here's how I solved it...

Given: We must take at least 2 cards to gain information, otherwise we can expect to pay the EV of a random card, which is N+$200

There are then 8 possible outcomes, each with different odds of occurring. These outcomes are that our 2nd card is a delta of the following values from the 1st card: +400, +300, +200, +100, -100, -200, -300, -400. We must weight each outcome according to it's respective probability within our decision tree.

I then calculate the EV for decisions within each branch of the tree, and come up with the following strategy as being GTO (game theory optimal).

Case +400: STOP! which yields paying N
Case +300: STOP! which yields paying N, N+100
Case +200: STOP! which yeilds paying N, N+100, & N+200
Case +100: DRAW which yields paying an average of N+108.33, 4x
Case -100: DRAW which yields paying an average of N+108.33, 4x
Case -200: DRAW which yields paying an average of N+88.89, 3x
Case -300: DRAW which yields paying N+100, & N+100
Case -400: DRAW which yields paying N+200

So... when you add it all up and divide by 20, you get an expected value payout of N+$96.67

I have it all on chicken scratch notebook paper if you really want to see it, but it's kinda all over the place and not easy to follow :)

Damn, I made a miscalculation somewhere. The correct answer was $90, not $96.67 as I thought it was... Sorry guys, I'll do better next time :)
 

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