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Imagine a situation where 4 players are playing in an auction game. There are 4 letters for sale A, B, C and D. A player "wins" if they are the first player to buy 3 of the same letter. Each player starts with the same amount of money. The player with the highest bid for each item pays their bid and the letter is added to the collection. If nobody bids then nobody gets the letter.

The number of each letter in the auction is known as well as the order in which they would be auctioned. For example 40 letter A's 30 letter B's 20 letter C's and 10 letter D's.

Can anyone think of a good strategy for winning this game?

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  • $\begingroup$ What strategy would you suggest if there were one letter being auctioned with two players participating? $\endgroup$ – Ethan Bolker Apr 7 '18 at 14:23
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    $\begingroup$ How many of each letter are being auctioned? How much money do the players have. The best strategy for winning the game as you have described it is to start with the biggest bankroll. $\endgroup$ – saulspatz Apr 7 '18 at 14:27
  • $\begingroup$ Games like this are sensitive to collusion. Three players can make sure the fourth does not win. You need to specify how the auction is held. Is it sealed bid? Around the table with each person having to raise the bid or pass? $\endgroup$ – Ross Millikan Apr 7 '18 at 15:09
  • $\begingroup$ If we assume its a sealed bid auction and that each player starts with the same amount of money. I'm trying to compute the valuations of each letter based on their occurence rates. Would it be fair to assume that if there are 40 letter A's that their valuation would be twice that of letter C's which occur half as often? How would I compute the utility in case of a winning bid. The valuations are not currency in this case so would it still make sense to have utility = valuation - bid? $\endgroup$ – Simon Wilson Apr 8 '18 at 0:49
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There are a few pieces of information that I am missing when trying to come up with a strategy, so I have made the following assumptions;

  1. Given there are 40 A's, 30 B's, 20 C's and 10 D's, I am assuming that the letters are dealt in a repeating order of AAAABBBCCD, and;
  2. I am assuming this auction takes the style of an "English auction".

I cannot personally think of a mathematical strategy that guarantees victory, but I can produce a strategy that gives you a much better probability of victory.

It is clear to me that you have 3 main objectives in this game, and they are as follows (1 is the most important, 2 the second most, 3 the third most);

  1. Stop any of the other players from getting 3 of the same letter (by bidding against them if they).
  2. Get 3 of the same letter for yourself.
  3. Protect your money relative to the other players.

As the first 4 letters are all A, the game could theoretically be won in the first 3 sales. Many people would assume the first and second A are the most important letters, because if you get either of these, you can theoretically win without waiting for the second round of sales. Therefore, the best option is not to bid on either of these letters, as they are likely to sell for a high price.

Of course, the below advice is useless if a player has 2 of the same letter and is about to win the bid for a third of the same letter, so in this case, the advice would be simply to beat their bid.

Given the perceived importance of the letters, in all likelihood, the first 2 A's will go to different players (we will call these opponents 1 and 2). If this is the case, the 2 players will be competing for the third A, because if they get this, the victory is still possible if they get the final A, so this is once again likely to sell for a high price, and in all likelihood will go to either opponent 1 or opponent 2. The final A could either go to the opponent who lost out in the previous bid, or the third opponent may choose to try to purchase the letter, given it will be at a much lower price than the previous 3 A's.

The third most important letter is the first B, because whoever wins this will still have a chance of victory without waiting for another round. The advantage you have at this point though, is at best you have the joint-highest amount of money, and at least 2 of your opponents have spent some of their money. The best idea for the B's is to use them to reduce your opponents' money. There are a few options for this, and they depend on your opponents' behaviour. The options are as follows;

  1. Assuming all three opponents either have a similar amount of money, or at least two of them have enough to still compete for the B's, then you can once again step back and allow your opponents to dwindle their money as they compete for the B's, and you only step in if absolutely necessary (i.e. someone is on the brink of getting all 3 B's).
  2. Assuming the second richest opponent is trying to get the B's, you could start a bidding war for the first B, and then drop out once the price reaches a suitable level, thereby costing your opponent a sizeable amount. This could be repeated for the second B, and then, given their reduced funds, you can purchase the third B for a relatively small amount simply to stop them from taking a victory.

Now, I would not bid for the C's or D's.

It is on the second round that you start trying to get 3 of the same letter.

Given that if you get the first or second A, you could take victory in this round, they have quite a high value to you, but similarly, your opponents will want to spend money stopping you from getting them. Therefore, you should bid for these, and only bail-out if both of these below statements are true;

  1. The lead bidder doesn't already have 2 A's, and;
  2. The bidding price will use up a sizeable chunk of the lead bidder's remaining money.

If you have won at least 1 A, then you should follow the same strategy as the first 2 A's for the third A. If you don't currently have an A, then clearly each of your opponents have 2 A's, so the advice is win the bid at all costs.

If you have two A's already, then you should go all-in for the fourth A. This will either result in victory, or force any opponent who can beat you to spend a large chunk of their money. If you only have 1 A, you must beat all of your opponents to this letter, regardless of how much they bid.

For the second round of B's, you should out-bid your opponents for each B, unless the cost means you will not be the richest player. Once again, ignore this if the lead bidder already has 2 B's, as you must stop this whatever the cost.

If you still don't have 3 letters after the B's, the winner will be the person who gets the A in the next round (assuming no-one else had bid on the C's or D's), as each player has 2 A's. Therefore, as long as you are the richest competitor, you are guaranteed to win, as you can outbid your competitors for the next A.

Summary

You will have the best chance of success if you only bid when absolutely necessary, as it should leave you with more money than your competitors, meaning you can beat them in the bidding during the latter part of the game.

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