Price is Right - Game Theory Problem

[Scotty.]

I was reading an NVG thread, and a very similar game theory problem came up regarding one of The Price is Right games. It wasn't (I don't think) accurately solved, and I'm curious to know how the real answer comes out.

The game is Come on Down. There are 4 contestants in this game. They are each shown an object to bid on, we will say a dining room set in this case. Each contestant must make an open bid (not simultaneous) on what they think the dining room set is worth. The winner of the game will be the person with the bid closest to the actual price without going over. Everyone estimates the set to be worth somewhere in between $900 and $1000. The actual item is worth something in this range, uniformly distributed with equal probability for each value.

What is the optimal bidding strategy?

(if I screwed up the explanation, someone please correct me)

[hmkpoker]

My guess: (in white below)

<font color="white">
1st $975

2nd $950

3rd $925

4th $900 (or $1) </font>

[DougShrapnel]

If everyone overbids, Does the bidding begin again?

[Max Weinberg]

[ QUOTE ]
If everyone overbids, Does the bidding begin again?

[/ QUOTE ]
Some ass always bets $1. Price is Right standard.

[MathEconomist]

I haven't really thought about it or checked, but I think the equilibrium is the first bidder chooses 975, second 950, third 925 and forth 900. If anyone tries to deviate, one of the later bidders can screw them. This assumes that they don't actually have private signals but simply know the distribution of the item. If they have private information which is better then the solution probably changes.

[amirite]

Wouldn't the last person be better guessing $926, $951, or $976? This seems it would carry the added bonus of being able to guess again if the answer falls in the $900-$924 range without hurting his probability of winning.

I don't know much about the specifics of game theory, but this at least seems right in my opinion.

[MathEconomist]

The problem doesn't specify what happens in that case, but I guess if a rebid occurs after everyone overbids, then the equilibrium I proposed doesn't work. It shouldn't be that hard to solve this simply by working backwards in either case.

[Scotty.]

[ QUOTE ]
If everyone overbids, Does the bidding begin again?

[/ QUOTE ]

I have no idea how it works in reality, but for the purposes of this question I will say yes.

[DougShrapnel]

1st $975

2nd $950

3rd $925

4th $900 (or $1)

If the bidding begins again wouldn't 4 do better to screw one of the others and bid at random either 976, 951, 926

if the first 3 go 975 950 925, 4 can give up 1% to gain 8.3%

4 has a 32% chace to win
while the remaing 3 have a 22%

[amirite]

The best part is that 95% of the people on that show don't know what the hell they're doing so if contestant four is one of us (us being the majority of the SMP readers) our likelyhood of winning is excellent.

[hmkpoker]

oof, good point. I didn't think about that getting to bid again thing.

[ZeeJustin]

Thinking about this as I go so bear with the fumbling...

It seems like player 1 should bid 1000-x, player 2 should bid 1000-2x and player 3 should be 1000-3x where x is the largest number possible that still leaves player 4's optimal bid to be $900.

If x is 25, as we have already determined, 4 is not best off bidding $900, so X &lt; 25.

If x is 20, we have bids of 980, 960, 940. Player 4 can then take a 40% chance of winning by bidding 900, or he can screw someone winning approxmiately 32% of the time. He'd clearly be better off just bidding $1 (or $900, same thing).

25 &gt; x &gt; 20

(guess and check sucks)

Actually, I don't need to use guess and check since player 4 should win 32% of the time.
Final bids shoud be
4- 900
3 - 933
2 - 933+x
1 - 933+2x

1000 - 933 = 67, / 3 = 22.3
so final bids should be
1 - 977.7
2 - 955.4
3 - 933.1
4 - 900

There was a little bit of rounding in here, so the numbers might be off.

With this set of strategies, only the last player has a poisitonal advantage. Let's see if with the same bidding, player 3 can have some advantage...

1 - 977.7
2 - 955.4

If player 3 sticks with the plan, he gets a 22.3% chance of winning. If he bids less than 933.1, player 4 will just 1 up him, and if he bids more, hes winning less than 22.3% of the time, so I think the above is optimal for all 4 players.

[atrifix]

[ QUOTE ]
If x is 20, we have bids of 980, 960, 940. Player 4 can then take a 40% chance of winning by bidding 900, or he can screw someone winning approxmiately 32% of the time. He'd clearly be better off just bidding $1.

[/ QUOTE ]Why does he win 32% of the time? Doesn't he win with 19% + .4*probability of winning the rebid?

I'm not sure how to answer this question because I'm not clear on the rebid rules. Do they simply rebid in the same order if everyone overbids?

[ZeeJustin]

[ QUOTE ]
Why does he win 32% of the time? Doesn't he win with 19% + .4*probability of winning the rebid?

[/ QUOTE ]

If he steals someones bid, essentially there will be 3 people with even bids and 1 guy sitting out and they keep bidding uuntil one of the 3 guys wins. It's 32% rather than 33.333% because he will have to give up $1 in his bid to do this. 32% isn't exact, it depends on the other bids, but it's a good estimate.

I was assuming rebids happen in the same order.