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Ask the Wizard #342

Supposed I had a 48 card deck, four suits, twelve cards each suit. If I drew 15 cards, what is the probability that I will draw at least one card in each suit?

SignGuyDino

Let's start with 100% and subtract out the probabilities that result in less than four suits.

What is the probability the 48 cards have no hearts, for example? There are 36 cards that are not a heart. The number of ways to choose 15 cards out of 36 is combin(36,15) = 5,567,902,560. The number of ways to choose 15 cards out of all 48 is 1,093,260,079,344. So the probability of the 15 cards without a heart is 5,567,902,560 / 1,093,260,079,344 = 0.005093.

Next, let's multiply that by four, to get the probability of missing any suit, not just hearts: 4 × combin(36,15)/combin(48,15) = 0.02037174.

However, this double-counts some situations. Consider getting 15 black cards. That would omit both hearts and diamonds. We would have double counted that situation. So, we need to correct for that. There are combin(4,2) = 6 ways to choose two suits out of four. The probability of all 15 cards being any two specific suits is combin(24,15)/combin(48,15) = 1307504/1,093,260,079,344 = 0.00000120. As mentioned, there are six ways to choose two suits out of four, so the number of ways all the cards will be of two suits is 6 × combin(24,15)/combin(48,15) = 0.00000718.

Subracting out what we double counted, we get a probability of two or three suits being represented of 0.02037174 - 0.00000718 = 0.02036456.

Note that we don't need to worry about one suit being represented, because it's impossible to choose 15 cards out of 12.

As a final step, subtract the probability of 2 or 3 suits from 100% to get the probability of all four suits being represented: 1.00000000 - 0.02037174 = 0.97963544.

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

What is your opinion of the Comp Killer roulette strategy, as covered in this video?

joedol

It's easy to see the purpose of that system is to cover most numbers so it's a fairly low risk way of playing roulette. Here is what to bet on each spin:

  • $5 each on 3, 16, 24, 28, and 33.
  • Make a corner bet on each of these sets of numbers: 2/3/5/6, 7/8/10/11, 14/15/17/18, 19/20/22/23, 26/27/29/30, 31/32/34/35.

Note that this does not cover the following nine numbers: 0, 00, 4, 9, 12, 13, 21, 25, and 36.

The following return table shows the probability and contribution to the return of all possible outcomes.

Comp Killer

Event Net Win Combinations Probability Return
Straight up win 5 5 0.131579 0.657895
Corner win 50 24 0.631579 31.578947
All other -175 9 0.236842 -41.447368
Total 38 1.000000 -9.210526

The lower right cell shows an expected loss of $9.21 per spin. The total amount bet per spin is $175. This results in a house edge of $9.21/$175 = 5.26%, the house edge in double-zero roulette.

I would like to add that you will probably lose more than you get back in comps with this strategy, or any roulette strategy. The rule of thumb is casinos will give you back about 1/3 of your expected loss in comps. There are ways of fooling the casinos that your expected loss is more than it really is, but playing this strategy isn't one of them.

Spot It is a kids game. Pretty addictive actually. I believe it’s called Dobble in the UK. There are 55 circular cards with 8 images chosen from 57 possible images. Cards are such that each card has exactly one match with each other card (no more and no less). Each person starts with a card and you flip over a third in the center. Whoever “spots” the match with their own card takes center card, and flips a new card.

My question, is what is the most cards one can have, given there are a total of 57 possible images and eight pictures per card?

unJon

I hope you're happy (said humorously); I've spent hours on this problem and still can't offer a solution.

However, the answer can found found in the article The Mind-Bending Math Behind Spot It!, the Beloved Family Card Game. For the the situation of n symbols, where any two overlap exactly once, the maximum number of cards is n^2 - n + 1. In this case, n=8, so the most number of cards is 8^2 - 8 + 1 = 57. The actual games uses 55. I imagine they arbitrarily chose to remove two of the possible combinations.

Personally, I have not arrived at why the n^2 -n + 1 formula is true.

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