Ask The Wizard #360

In the movie Casino Royale, in the final hand of a poker tournament, the four players have the following hands:

  • Flush
  • Full house
  • Full house (of different value than first one)
  • Straight flush

What is the probability of that?


I had to run a simulation for this one. In my simulation, I assume nobody ever folds. In running almost 2.2 billion rounds, this happened 312 times. That equates to a probability of about 1 in seven million.

On Roll to Win craps games the player may make both lay and place-to-lose bets. Here are the odds offered on the place-to-lose bets:

  • 4 and 10: 5 to 11
  • 5 and 9: 5 to 8
  • 6 and 8: 4 to 5

The lay bets pay fair odds, except the player must pay a 5% commission, based on the win amount, if he wins.

My question is which type of bet offers the better odds?

John Cokos

The following table shows the house edge both ways by the number bet on. You can see the house edge is lower on the lay bets for all points except the 6 and 8.

House Edge on Place to Lose and Lay Bets

Number Place to Lose Lay
4 or 10 3.03% 1.67%
5 or 9 2.50% 2.00%
6 or 8 1.82% 2.27%

The following question comes to us courtesy of the Riddler Express.

Let's assume NFL rules. Consider the following situation:

  • The Red team is down 14 points late in the game
  • The Red team will have two more possessions
  • The Blue team will have zero more possessions
  • Let's ignore field goals and safeties, as the Red team must score two touchdowns to have a chance at winning
  • If the game goes into overtime, each team will have a 50% chance to win. The game cannot end in a tie.
  • The probability of making a one-point kick after a touchdown is 100%.
  • The probability of making a two-point conversion is p.

At what value of p should the red team be indifferent to kicking and going for a two-point conversion after a first touchdown (now down by 8)?


(3-sqrt(2))/2 = apx. 0.381966011250105


Let p = indifference point between going for the two-point conversion and the kick.

If the first two-point conversion attempt is successful, then the red team can kick the ball the second time and win.

If the first two-point conversion attempt is unsuccessful, then the red team must try again after the second touchdown and then win the game in overtime.

The probability of winning, going for the two-point conversion after the first touchdown is p + (1-p)*p/2. We equate this to the 50% chance of winning by kicking after the first touchdown and solve for p.

p + (1-p)*p/2 = 1/2
2p + (1-p)*p = 1
3p - p^2 = 1
p^2 - 3p + 1 = 0

Using the quadratic formula, solve for p:

p = (3 +/- sqrt(5))/2

We take the negative option, to keep p between 0 and 1, to get p = (3-sqrt(2))/2 = apx. 0.381966011250105

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