## HN. Barbette

In which we discuss still more 2012 facts—Matt Zinno points out that we are emerging from a spell of years with repeated digits, and in fact this is just about the longest run in the last 1000 years!  (So, folks, enjoy working out other long spells!)

Ben Anderman shares his online Princess-and-Suitor app.

And Kyle and I discuss some bar bets, including the great Barbette, shown here in a photo by Man Ray:

The challenge this week is to work out a strategy for the following game, that works 50% of the time on average:

The Victim believes you will lose twice as often as you win, so in order to make money, you should somehow get The Victim to bet a little bit more than you do, say \$1.50 for each \$1 you put up.

The Victim writes any three numbers on three pieces of paper, turns them over, and mixes them up.

One by one you flip over a card, and then either stop, selecting that one, or discard it and move on. If you select the highest number overall, you win!

We discussed this in more generality a long time ago, but this version has the merit that it’s simple enough to demonstrate quickly, work out precisely why it works and on top of all that, of all the cases has the single highest probability of winning per round.

1. ### Scott said,

February 27, 2012 at 9:41 pm

If you change your bar bet so that the the opponent must choose natural numbers, and you are looking for the smallest one, then you can actually win more than half the time.

2. ### Harry Kaplan said,

February 28, 2012 at 7:46 pm

It seems to me that the desired strategy for the 3-number game and the famous Monty Hall Problem are cousins.  In both you use information revealed, maybe subtly, by the process itself to up your odds.

3. ### Mark said,

February 29, 2012 at 11:27 pm

Regarding the two-heads-in-a-row problem: I analyzed this problem two different ways and got two different answers. The apparently wrong, simpler, analysis is to imagine all 20 flips done at once and then take the results two at a time, #1 & #2, #2 & #3, #3 & #4, …, #19 & #20. The probability that any consecutive two are heads is .25. The probability that every pair is not HH is 1-(1-.25)^19 = 0.995772 or about 4 losses per thousand tries. (This would result in a winning bet, by the way.)
The other analysis looks something like recursion, or calculating the Fibonacci sequence. It’s really tough to explain only in words (my wife, who very long ago, was a math teacher, didn’t even want to hear it).
It involves imagining the flips sequentially, walking down the possibility tree to the first Tail flip. Then working up a recursion formula for calclulating the probability of two heads following any tail flip, starting from the last flip (0), to the next to last (0), to the next to next to last (.25), etc. That yields 0.983109, or about 16 losses per thousand tries. (A losing bet, the way you describe the problem, losing \$600 every thousand games.)
Empirically running the test in Excel yields 12-24 losses per thousand tries, vindicating the more complicated analysis.
Why is the first analysis wrong? Thanks, Mark

4. ### Byon said,

March 1, 2012 at 12:17 pm

Here are my thoughts…
[spoiler]
The playing algorithm is flip one, discard it, then flip another, and keep it if it is greater than the first, otherwise discard it and flip and keep the third.

The reason your odds are 50% can be found by considering which of the 3 in your selection order is the greatest.  There is a 1/3 chance it is the first, in which case you lose because you always discard it.  There is 1/3 chance it is the second, and you win, and in the 1/3 case it is the last, you win that 1/2 the time, depending on if 1 was greater than 2 or not.

Does it matter if the 3 numbers may not be unique?

I like Scott’s modification, too!
[/spoiler]

5. ### Shawn said,

March 6, 2012 at 10:42 pm

Here’s an idea I had: [spoiler] You flip over the first piece of paper, read and remember the number, and discard it. Then you flip over the second piece of paper and select it if it is higher than the first number and discard it otherwise. It appears that in that instance, you would lose if the first piece of paper had the highest number or if the numbers were in ascending order, and win otherwise. Let’s assume that the numbers are three different real numbers. In this case, we call refer to them as the lowest number (L), the middle number (M), and the highest number (H), L < M < H. You would lose if the papers were ordered L-M-H, H-L-M, or H-M-L, but you would win if they were ordered M-H-L, L-H-M, or M-L-H. [/spoiler]

6. ### Shawn said,

March 22, 2012 at 3:10 am

Mark,
That’s an interesting question. I’d like to know the same thing. I’m no math expert; the only idea I could come up with is that perhaps the nineteen different pairs of coins are not completely independent since each coin is a member of two pairs. For example, if you were trying to figure out the probability that #5 and #6 were not both heads, that might be affected by what you know about whether #4 and #5 are both heads and whether #6 and #7 are both heads. If you know that #4 and #5 are both heads as well as that #6 and #7 are both heads, then probability is 1 that #5 and #6 are both heads. For each pair, TT, TH, and HT will result in not having a consecutive pair of heads, but HT cannot be preceded by TH and TH cannot be followed by HT. In that way, each pair doesn’t only have to worry about itself not being HH, but also that any H that does appear is not adjacent to another H. Each pair of TT can be followed by HT, TH or TT, but each pair of HT can only be followed by TH or TT.

7. ### Marcos said,

February 15, 2013 at 9:44 pm

I thought of a way to “ruin” your bet: instead of writting numbers directly, I write numerical expressions so difficult to compute or compare that the choice you make is effectively random. Later, I show you which number is the highest :)