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Ask The Wizard #421

Assuming all players are equally skilled, what the is the probability the team that serves first in pickleball wins the whole game? 

anonymous

Let me remind the rest of the readers the rules of scoring in pickleball.

  1. The first team to 11 points and wins by at least two points, wins the game.
  2. Each team has two players, I'll refer to as player 1 and player 2. I'll refer to the two teams as A and B, with A serving first.
  3. Player 2 on team A serves.
  4. If team A from step 3 wins the rally, they get a point and the same person serves again. This continues until team B wins the rally.
  5. Player 1 on team B serves.
  6. If team B from step 5 wins the rally, they get a point and the same person serves again. This continues until team A wins the rally.
  7. Player 2 on team B serves.
  8. If team B from step 5 wins the rally, they get a point and the same person serves again. This continues until team A wins the rally.
  9. Player 1 on team A serves.
  10. If team A from step 7 wins the rally, they get a point and the same person serves again. This continues until team B wins the rally.
  11. Go back to rule 3.

Please note the receiving team cannot win points. They are playing to get back the serve.

A shorter way to explain it is the same person serves and earns a point for each won rally until the other team wins a rally. The receiving team does not earn points. When the turn to serves switches from one team to the other, both players on the serving team get a chance to serve. To make the odds more fair, play starts with the second player on one of the teams serving. This continues until either team has at least 11 points and a margin of victory of at least 2.

That said, my answer is the probability the serving team wins is 0.499999997522. This was solved using a Markov Chain.

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

Suppose we have a lottery in which 6 balls are randomly picked in the range of 1 to 54. What is the probability that, in 50 draws, at least 1 ball is not drawn?

KevinAA

The answer is 0.140150159777671.

 

A start is to ask what is the probability any specific number is not called in 50 draws. That answer is (combin(53,6)/combin(54,6))50 = (8/9)50 = 0.002769325.

For the probability any number is not called in 50 draws, multiply the figure above by 54: 54 × 0.002769325 = 0.149543533246569.

However, that is double-counting situations where two numbers are not called in 50 games. The probability two specific numbers are not called in 50 games is (combin(52,6)/combin(54,6))50 = 0.78826050 = 0.00000681512. There are combin(54,2)=1431 ways to choose any two balls out of 54. So, the probability any two balls are not called in 50 games is 1431 × (combin(52,6)/combin(54,6))50 = 0.009752432.

So, now we're at 0.149543533246569 - 0.009752431939662 = 0.139791101306907.

However, the double-counting adjustment above double-counts situations where three numbers are not called in 50 games. That probability is combin(54,3)*(combin(51,6)/combin(54,6))50 = 0.000367891216781.

So, now we're at 0.149543533246569 - 0.009752431939662 + 0.000367891216781 = 0.140158992523688.

We keep doing this, alternating between adding and subtracting. Excel can handle only about 15 significant digits, so we need to only do this through eight missing numbers to be correct within those 15 significant digits.

In the end, the probability comes to 0.140150159777671.

 

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

Suppose fair odds for an event happening are 6.3 to 1. A sports book offers this bet at 6 to 1. If they wanted to offer a bet on the favored team winning with the same house edge as the underdog, what odds should it offer?

anonymous

If fair odds are 6.3 to 1, then the probability of winning is 1/7.3.

A bet on the underdog pays 6 to 1, which is the same as 7 for 1. That makes the expected win = 7/7.3 = 70/73 = 0.958904.

The probability the favorite wins is 6.3/7.3 = 63/73.

Let's call the odds, on a "for one" basis, that the favorite wins f.

Solve for f, such that:

(63/73) × f = 70/73.

Multiply both sides by 73:

63f = 70

f = 70/63 = 10/9

To convert that to a "to one" basis, subtract 1. So, the odds that the favorite wins should be set to 1 to 9.