Five-Card Battle

Logic Level 1

You are playing a game against a casino called Five-Card Battle.

For each round, both you and the dealer have the five cards above (aces are worth 1). You and the dealer each choose a card secretly, and then reveal the choices simultaneously. If the sum of the cards is even, then the casino wins $1. If the sum of the cards is odd, then you win $1. Note that 13 out of the 25 possible sums are even, so if both you and the dealer pick randomly the casino has an advantage in the long run.

Suppose you know the dealer's protocol is to randomly and independently pick a card every round. Is there a strategy for you to have an advantage in the long run?

Yes No

4 solutions

Michael Mendrin
Jan 24, 2017

Since the dealer is likely to pick an odd number 3 out of 5 times, pick an even number.

If the protocol were that both you and the dealer were to choose cards to gain the advantage, then how would the Battle end? I would suspect the guessing game would go in circles, so you would both end up going back to random picks, giving back the edge to the casino. The House always wins in the end ...

Brian Charlesworth - 4 years, 4 months ago

It's interesting to see how the casino would develop a winning protocol in this game. It has three alternatives:

1) Pick randomly, hoping that "average" customers will also pick randomly

2) Pick only even cards, hoping that "smart" customers would be picking even cards, expecting the dealer to pick randomly

3) Pick only odd cards, hoping that "super smart" customers would be picking even cards, expecting the dealer to pick even cards (see 2 above)

And so on? This sounds like the game of rock, scissors, and paper. But what the casino will have to do is to guess the mix, or distribution, of "average", "smart", and "super smart" customers would be---and then that will create even higher levels of customers anticipating this move. This is a lot like what's happening in computerized stock market trading today. A very interesting problem!

Michael Mendrin - 4 years, 4 months ago

Good breakdown. I guess the difference with RPS is that its participants are on an even playing field, while this game throws the Nash equilibrium off balance. The casino could lean on data gathered over time on how best to "deal", (pun intended), with each type of player once identified, so I would suggest the best strategy for a player is to be super-super smart are change things up so that the casino's algorithm can't easily identify your type. To stay disciplined, you could use some difficult-to-detect pattern, (say the unit digits of the Fibonacci sequence, or if you have a good memory the digits of $\pi$ ), for the number of times you use the average/smart/super smart approaches in sequence. I'm not sure if this would yield a better outcome long-term than any single approach, but it couldn't be any worse ... or could it? I think one could base a thesis on this problem. Anyway, Here is an interesting analysis of how human psychology can be exploited to develop optimal strategies in the RPS game.

Brian Charlesworth - 4 years, 4 months ago

@Brian Charlesworth – I agree, this problem is an idea for a thesis, if it hasn't been done already. Maybe somebody do a wiki on this?

Michael Mendrin - 4 years, 4 months ago

@Michael Mendrin – See K-level thinking

Michael Mendrin - 4 years, 4 months ago

It's actually no different than what was going on before computers were involved. Sussing out high-value information and making predictions of how that information might be utilized by different levels of traders has been going on for hundreds of years. Bucket shop operators and brokers created elaborate signalling networks in the early 1900s. And the public screamed foul then. That is the exact basis of HFT only on a shorter time frame. Nothing new under the sun.

Jack Bowie - 3 years, 4 months ago

A Former Brilliant Member
Feb 3, 2018

Quite simple!!! Whosoever takes the first chance, suppose the other person gets even numbered card, then probability of mine selecting odd (i.e. winning) is 3/4 which is reduced to 1/2 if odd is first selected. So main aim is to get one even numbered card out.

Jack Bowie
Jan 28, 2018

What is it with the need to rely on obscure/obtuse language with so many problems? How did this approach take hold? Why does it persist? When someone actively avoids specificity so as to make use of the resultant waves of uncertainty for what otherwise would be a straightforward solution had only a couple more words been added, it wreaks of desperation and certainly not cleverness. Knock it off.

Bob Loesch
Jan 25, 2018

The dealer has a probability of $\frac{3}{5}$ of drawing an odd card and a probability of $\frac{2}{5}$ of drawing an even card. If you draw an even card each time, then the probability of the sum being odd is $\frac{3}{5}$ whereas the probability of the sum being even is $\frac{2}{5}$ . You now have the longterm advantage over the house.

Five-Card Battle

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