The Breakfast Game

Post Reply New Topic

I eat breakfast occasionally with a few gentlemen who play an interesting game at the end of each breakfast. The "winner" of this game must pay for breakfast, while the "losers" leave the tip.

The game is played as follows:

1. Every morning a number is chosen in advance by Alan. The number is a secret and is between 0-1000 inclusively.
2. The person (Barry) to Alan's left must choose a number, and Alan will tell them if they are high or low.
3. The guessing continues by the next person to Barry's left and continues in a clockwise motion until someone picks "the number". If you pick the number, you "win".
4. Everyone except for Alan gets to guess their own number. Alan is assigned a number in the following manner. If the person to Alan's right guesses a number and it's low, Alan is assigned the next highest number. If the person to Alan's right guesses a number and it's high, Alan is assigned the next lowest number.

Here's an example of a typical day. There are four people at breakfast including Alan. Alan secretly chooses the number 454. The person to Alans right picks 499, and Alan says high. The next person must pick a number lower than 499. So the next person picks 492, and Alan says low. The next person must pick a number higher than 492. They pick 493, and Alan says low. Now Alan gets assigned the next highest number which is 494 and he is the "winner" and must pay for breakfast.

Of course, it normally doesn't happen this quickly, this was just an example. But here are my questions:

If it were just you and Alan, what is the likelihood you will pay for breakfast?

If you showed up to breakfast and there were 3 people (including yourself), where would you want to sit to minimize your chances of paying?
If you showed up to breakfast and there were 4 people (including yourself), where would you want to sit to minimize your chances of paying?
If you showed up to breakfast and there were 5 people (including yourself), where would you want to sit to minimize your chances of paying?
If you showed up to breakfast and there were N people (including yourself), where would you want to sit to minimize your chances of paying?

An interesting problem. I think your description is a little unclear, at one point.

To clarify, I hope, all bar Alan get a free choice of what number to nominate, provided that that number is still possible, and has not yet been chosen.

When Alan takes his turn, he is obliged to chose the still-possible number adjacent to the last player's guess - which may, of course, cause him to chose the number he initially selected, and hence "win" (as you put it).

========

Other than that, I wonder if you have fully thought through the problem. E.g. the initial number that Alan chooses... if he employs strategy, He will optimally chose a number according to a non-uniform distribution across the range. Ditto for each player - their "guesses" should employ strategy.

E.g. take the simplest non-trivial form of the problem: two players and a choice of just three numbers: 0, 1 or 2.

The best strategy for Alan is to choose 1 with probability 1/2 and either of 0 or 2 with probability 1/4 each. When Barry then adopts exactly the same probabilities for his choice of first guess, the outcome turns out to be a purely fair game: both Alan and Barry "win" with probability 1/2. Neither player can improve on these strategies - unless they know that the other player is stupid .

Strategy! Well that's a horse of a different color.

I would suggest that in a two player game (me and Alan), I could employ a strategy where I am expected to pick the number only 1/3 of the time. Think about it for a while, and it should come to you.

*********************************

Just to clarify, Alan is unwilling to change the way the game is played. He chooses the number secretly at the beginning, and is assigned a number as described above. Period.

*********************************
Years ago, I suggested (and still believe) that assuming everyone guesses at random. Alan has the highest probability of paying for breakfast when there are 8 or fewer people at the table. I have come to this conclusion through simulations.

We have 5 years of actual data. Buried in this data are each players individual strategies at work. The bottomline is that Alan has paid more than his fair share for years (and he is ok with that. He loves the company).

This is correct. I may try to rewrite the game description.

If you are stating that you will (as player B) choose each of 0, 1 and 2 with equal probability (i.e. 1/3), then I, as Player A will never choose 1. Indeed, I could just always choose 0, say. The probability that you pay for breakfast will be 2/3.


Yes.


Why should I chose numbers with equal probabilities, when I can significantly improve my chances of avoiding paying for breakfast, by employing an optimal non-uniform distribution? See above.
If I am assured that all other players will chose at random, that means I can chose a strategy that is far more successful.
If I assume that all other players adopt optimal strategies, I'm quite certain there are variations on the expected "payment avoidability", depending on what position around the table you sit, and the range of numbers in use. Sit far enough round the table (for numbers 0..1,000, sit any where after position 1,001), and the game can never reach you, and your chance of paying is zero.
The neat, fair result for three numbers and two players is not the usual case. I haven't evaluated any other combinations.


I don't know what this means. It is a game. It is, moreover, a fairly simple game. The optimal strategy is quite easy to find. If the other players are announcing that they will use non-optimal strategies, you can adopt a variant strategy to significant increase your gains (as above).

As for player A tending to pay more often that his fair share... I would say that was an expected result - as that player has no control over his strategy, beyond the distribution they use for their initial choice. Of course, if there are many people at the table, it becomes unlikely that he will pay at all, because, ultimately, the game can never reach back to him.

In summary, the first players have good chances of avoiding payment, as the range of numbers available is wide. Later players have smaller ranges remaining, and are the likeliest to pay. Eventually, player are again unlikely to pay, because the chances are that the game has already reached a conclusion before reaching them.

With the 0..1,000 range, players loosely attempting a binary chop, in order to avoid the play going all round and back to them, would imply that people around the 10th position in sequence might have the least enviable seats. This, however, is pure guesswork.