Betting on red

[bonanova]

Here's a variation on a previous puzzle that let you make multiple bets while choosing the color of the next card.

I shuffle an ordinary deck of playing cards and then turn over the top card sequentially so that you can see it. At any time you may ask me to stop and place a $1 bet that the next card to be exposed will be red. If you never ask me to stop, you will automatically bet on the last card. To summarize:

1. You can bet on only one card.
2. You don't get to choose the color.
3. You must bet the card will be red.

What is your best strategy?

How much better than even can you do?

[bushindo]

Interesting puzzle. I'm getting some non-intuitive results...

Let S₀ be the strategy where we do not choose any card (default choice is the 52nd card).

There are 52! different possible ways that the deck can be arranged. Let this entire set of 52! card arrangement be denoted as D.

Let S be any strategy where we divide D into two sets: a set B where we do not choose any card (default choice is the 52nd card), and a set A where for each card arrangement we bet on some card before the 52nd card.

Let's consider set A. Suppose for some arrangement a_i in A we bet on the c_i-th card, where c_i < 52. By symmetry, the chance of the c_i-th card being red is the same as the last card (52nd) card being red. Therefore for all arrangements in A we would have the same winning chance if we default to the 52nd card.

For instance, suppose we opened 30 cards and they turned out to be 18 blacks and 12 red, and our strategy calls for betting on the 31 card. The 31st card has 14/22 chance of being red, but the 52nd card also has a 14/22 chance of being red. So you might as well as not choose the 31st card at all and let the game default to the 52nd card.

Consequently, any such strategy S would have the same winning chance as S₀, which is 50%.

[phil1882]

based on computer sim

you should bet on red after 3 red cards, or 4 black cards, have revealed themselves.

the odds of winning are very slightly better than 50/50.

[dgreening]

based on computer sim

you should bet on red after 3 red cards, or 4 black cards, have revealed themselves.

the odds of winning are very slightly better than 50/50.

Very interesting strategy.

I was working on something like bet after so many [3 or 4] black in a row. The addition of a red component seems very counter intuitive - although in my small number of simulations, the red strategy paid off much better than the black strategy.

Overall in 60 iterations [Excel sheet that is very labor intensive] I had the following results:

• The red component successfully bet 22 times
• The black component successfully bet 2 times
  • Note: no instance went to the last card
• Total number of successful bets - 24 out of 60
• The actual random return [no bet] for the cases I ran were 28

I reanalyzed the data using just the black component of the strategy

• The black component successfully bet 26 times
• 4 instances wne to the end of the sequence
  • 3 of these ended in Red
  • 1 ended in Black
• Thus using just the Black component of this strategy, resulted in 29 wins out of 60 [vs 28 based on no bets at all.

Since your sim is probably much more automated, it might be interesting to see what happens if you use only the black component of your strategy.

[harey]

A priori, the p of any card being red is the same, so it does not matter on which card you bet. (Therefore the best strategy is to bet on the first card because you are not losing time.)

Roulette gamblers will not agree with this, they will argue that if there remain 3 red cards and 1 black card, p=3/4.

Is there an easy way to calculate the p that R red cards and B black cards remain?