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Ask the Wizard #302

In a two-player game of Texas Hold 'Em, which hand has the best odds against pocket aces of unknown suit?

Mike B.

Assuming both hands go to the end, I show the best competing hand is 5-6 suited. If the suit is not represented in the pair of aces the possible outcomes are:

  • Win: 22.87%
  • Tie: 0.37%
  • Lose: 76.76%


If the suit is represented in the pair of aces (lowering the chance of a flush), the possible outcomes are:

  • Win: 21.71%
  • Tie: 0.46%
  • Lose: 77.83%


Overall, the possible outcomes are:

  • Win: 22.290%
  • Tie: 0.415%
  • Lose: 77.295%


The Hot Roll bonus round on slot machines awards the player the following number of coins according to the total of two dice. The player keeps collecting until he rolls a total of seven, which ends the bonus. If he rolls a seven on the first roll, he gets a consolation prize of 70 coins. Following are the prizes for all other totals besides seven:

  • 2 or 12: 1,000
  • 3 or 11: 600
  • 4 or 10: 400
  • 5 or 9: 300
  • 6 or 8: 200


My question is what is average bonus win?

Anonymous

Click the following button for the answer.

The answer is 1983.33.

Click the following button for the solution.

Let x be the answer. As long as the player doesn't roll a seven he can always expect future wins to be x, in addition to all previous wins. In other words, there is a memory-less property to throwing dice in that no matter how many rolls you have already thrown you are no closer to a seven than you were when you started.

I won't go into the basics of dice probabilities but just say the probability of each total is as follows:
  • 2: 1/36
  • 3: 2/36
  • 4: 3/36
  • 5: 4/36
  • 6: 5/36
  • 7: 6/36
  • 8: 5/36
  • 9: 4/36
  • 10: 3/36
  • 11: 2/36
  • 12: 1/36


Before considering the consolation prize, the value of x can be expressed as:

x = (1/36)*(1000 + x) + (2/36)*(600 + x) + (3/36)*(400 + x) + (4/36)*(300 + x) + (5/36)*(200 + x) + (5/36)*(200 + x) + (4/36)*(300 + x) + (3/36)*(400 + x) + (2/36)*(600 + x) + (1/36)*(1000 + x)

Next, multiply both sides by 36:

36x = (1000 + x) + 2*(600 + x) + 3*(400 + x) + 4*(300 + x) + 5*(200 + x) + 5*(200 + x) + 4*(300 + x) + 3*(400 + x) + 2*(600 + x) + (1000 + x)

36x = 11,200 + 30x

6x = 11,200

x = 11,200/6 = 1866.67.

Next, the value of the consolation prize is 700*(6/36) = 116.67.

Thus, the average win of the bonus is 1866.67 + 116.67 = 1983.33.

What is the expected number of random numbers drawn from the uniform distribution between 0 and 1 for the sum to exceed 1?

Anonymous

Answer:

e=2.718281828...
Solution:
Here is solution.

How can I calculate the number of combinations of each win for slots with "Multiway wins"? You can assume that I have the reel strips.

James from Vermont

For the benefit of other readers, slots with "Multiway" wins cover all possible pay-lines. However, the game will pay only once for each way combination through winning symbols. Once a reel is reached with no winning symbols, the pay-lines end there.

Let's look at an example based on a game with five reels and three visible rows. All wins are left aligned only. Suppose the player had a winning symbol on reels 1, 2, 3, and 5. The player would be paid only once for three of that symbol. It does not matter that there are 9 ways to run through reels 4 and 5, because in this example the pay-lines end with reel 3.

Next suppose the player had the same winning symbol this many times on each reel:

  • Reel 1: 2
  • Reel 2: 1
  • Reel 3: 3
  • Reel 4: 2
  • Reel 5: 1


The player would be paid for 2×1×3×2×1 = 12 pay-lines.

If the player covered the entire screen with the same winning symbol, he would be paid for 35=243 pay-lines.

Next, let's move onto the answer. Let's assume there are wins for 3 to 5 symbols only.

Let's define some terminology:

  • tx = total reel stops on reel x.
  • nx = total count of the winning symbol on reel x.
  • px = positions on reel strips x where there is no winning symbol visible on the reel.


  • For reel 3 the answer is 33 × n1 × n2 × n3 × p4 × t5.

    For reel 4 the answer is 34 × n1 × n2 × n3 × n4 × p5.

    For reel 5 the answer is 35 × n1 × n2 × n3 × n4 × n5.