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Ask the Wizard #289

The Michigan Lottery has a three-player game with the following rules:

  1. Players play one at a time while players yet to play are kept off stage and are given no information about how any previous player did.
  2. There are 100 cards, numbered 1 to 100.
  3. A player starts by choosing any card.
  4. After looking at it, the player may keep it or switch for a new card.
  5. All cards are dealt with replacement. In other words, old cards are put back in the deck, including after discarding.
  6. The player who draws the highest card wins.


Is there any positional advantage to going last in this game? What is the optimal strategy for each player?

Here is a YouTube video showing the game.

Dween

First, there is no positional advantage to acting last. Since the players are kept in a sound-proof booth while any previous players play, order doesn't matter.

Second, there must be a Nash Equilibrium to the game where a strategy to stand with a score of at least x points will be superior to any other strategy. The question is finding x.

What I did was ask myself what would be the strategy if instead of a card numbered 1 to 100, each player got a random number uniformly distributed between 0 and 1 and look for the point x where a perfect logician would be indifferent between standing and switching. With that answer, it is easy to apply the answer to a discrete distribution from 1 to 100.

I'll stop talking at this point and let my readers enjoy the problem. See the links below for the answer and solution.

Answer for a continuous distribution from 0 to 1.

Answer for a discrete distribution from 1 to 100.

For my solution, please click here (PDF).

This question was raised and discussed in my forum at Wizard of Vegas.

With regulation almost non-existent now in Atlantic City, I have noticed some casinos (a.k.a. Caesars) have been flouting the rules. When I played craps there this past weekend, I noticed a huge amount of sevens being thrown, despite the randomness of the throwers and on multiple tables too. I'm well aware of variance and everything, but when I see sevens or craps thrown on over 40% of the throws (no exaggeration) I have to wonder. This was further confirmed upon inspecting the dice as the edges were worn down (they were whitish instead of the blue color of the dice).

Thundershock

Thank you for the warning, but you'll forgive me if I'm skeptical.

My advice to you if you find dice that land on a total of seven or craps (2, 3, or 12) 40% of the time is to keep your mouth shut and bet on those totals. If you bet one unit each on the 1-6, 2-5, and 3-4 easy hops, which each pay 15 to 1, and two units on any craps, which pays 7 to 1, then you will have a 40% chance of winning 11 units and a 60% chance of losing 5 units. That is a player advantage of (0.4×11 + 0.6×-5)/5 = 28%! Compare that to an approximate 1% advantage that card counters struggle for.

If you get rebuffed by the table maximum, you can also bet the any seven, and hop the 1&1, 1&2, and 6&6.

Don't get too greedy by winning too fast. If you slow play it you should be able to walk in with $100 and be able to buy an Atlantic City casino before you know it.

This question is raised and discussed in my forum at Wizard of Vegas.

In the NFL is it a good bet to take over 50 at one sports book and under 52.5 at another?

Rob from Las Vegas

You can use my alternative totals in the NFL for this kind of question. That will show the probability and fair line for getting extra points on a total bet.

Let's assume that the fair over/under line is 51. The first table shows that the probability of winning betting under 52.5 is 54.4%, assuming no tie. The third table shows that the probability of winning betting over is 53.5%.

To beat the 10% juice, you need to have a probability of winning of 11/21 = 52.38%. The average of these two bets is 53.95%, so, yes, play that middle. Your advantage is 3.0%.

According to the article Customer Stunned After Spending $3,750 For Wine Bottle, a patron at a Borgota restaurant asked the waitress for a suggestion on a bottle of wine. He professed no knowledge of wine prior to this and simply didn't know what to pick. The waitress suggested one for "thirty-seven fifty." He assumed this meant $37.50, but was shocked to get charged for $3,750 on the bill. After he protested the restaurant dropped the price to $2,200.

What is your opinion about what should have happened?

MrV

I'm with the consumer on this one. There is a well-established notion that in the event of a contract dispute the benefit of any ambiguity should go to the party that didn't write the contract. In this case, it was the waitress/restaurant that presented the offer to sell the bottle of wine, so the burden is on them to make sure the agreement is clear.

In this case, the waitress didn't make it clear whether the price was $37.50 or $3,750. I think it is reasonable to conclude that a recreational wine drinker would assume she meant $37.50. It seems unethical and unreasonable to suggest a $3,750 bottle to somebody unless he clearly indicated that was his price range.

I tend to think the patron and his guests were under duress to accept the $2,200 offer. In my opinion, the restaurant should refund him $2,162.50, the different between what he paid and $37.50.

This question was raised and discussed in my forum at Wizard of Vegas.