## Wizard Recommends

What is the probability of getting three to a royal flush on the deal and then completing it on the draw TWICE in a span of ten hands and in the same suit?

Daknight0721

For the first royal, the probability of getting three to a royal on the deal, in any suit, is 4*combin(5,3)*combin(47,2)/combin(52,5) = 0.01663742. The probability of completing the royal on the draw is 1/combin(47,2) = 0.00092507. So the probability of both events is 0.01663742 * 0.00092507 = 0.00001539, or 1 in 64,974.

The probability of getting any two royals, in any two suits, this way in ten hands is combin(10,2) * 0.000015392 (1-0.00001539)8 = 0.00000001065810. You also specified the two royals must be in the same suit. The probability the second royal matches the first is 1/4, so divide the previous probability by 4 to get 0.00000000266453, which is 1 in 375,301,378.

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

Consider a game show with two contestants who are both selfish and perfect logicians. Here are the rules.

1. The host places \$1,000,000 on a table between the two contestants.
2. Contestant A is asked to make a suggestion on how to divide the money between the two contestants.
3. Contestant B will be asked to accept or reject the suggestion.
4. If contestant B accepts the suggestion, then they divide the money that way and the game is over.
5. If contestant B rejects the suggestion, then the host will remove 10% of the amount currently on the table.
6. The host will then ask contestant B to make a suggestion and contestant A will have the same chance to accept or reject it.
7. If contestant A accepts the suggestion, they split it that way and the game is over. If he rejects it, then the host rakes another 10% of the remaining amount on the table. Then go back to step 2 and keep repeating until a suggestion is accepted.

The question is how should contestant A suggest dividing the money on his initial turn?

"Anonymous" .

He should suggest keeping 10/19 of the money for himself, less one penny, and offer B 9/19 of the money, plus a penny. In other words:

A: \$526,315.78
B: \$473,684.22

The key is A should put B as close as possible to an indifference point.

Let's call the ratio of the pot to the other player r. If B accepts the offer, he gets r×\$1,000,000.

If B rejects the offer, then the host rakes out 10%. After which, B will have a position advantage and would offer contestant A a share of r and keep 1-r for himself.

Solving for r...

r×\$1,000,000 = (1-r)×\$900,000.
r×\$1,900,000 = \$900,000.
r = \$900,000/\$1,900,000 = 9/19.

A does not want B to be completely indifferent, lest a chose randomly and stand a chance of the host raking the pot. So, A should throw in the extra penny to B and offer him (9/19) × \$1,000,000 + \$0.01 = \$473,684.22.

A: \$526,315.78
B: \$473,684.22

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

BetMGM sometimes offers what they call a "Risk Free Bet," although it's not risk free. I think a better term would be a "second chance" bet. Here are the rules.

1. The player makes a bet, subject to a maximum amount, on any event (no parlays, teasers, etc.)
2. If the bet wins, it wins and the player is paid normally.
3. If the bet loses, the player is given a promotional bet equal to the amount he lost.
4. The promotional bet may also bet bet on any one event.
5. If the promotional bet wins, the player is paid the winnings. If the promotional bet loses, then the player gets nothing. Either way, the promotional bet is taken away.

Here are my questions:

1. What would be the value of a \$100 Risk Free Bet if played against the spread at -110 odds?
2. What strategy do you recommend?

odiousgambit

First, let's look at betting against the spread at -110 odds. Let's assume a 50% chance of winning each bet.

• There is a 50% chance you win the original bet and profit \$90.91.
• There is a 25% chance you lose the original bet and win the second one. Here you will have lost \$100 and won \$90.91, for a net win of -\$9.09.
• There is a 25% chance you lose both bets for a loss of \$100.

The expected value of this promotional bet is 0.5×\$90.91 + 0.25×-9.09 + 0.25×-100 = \$18.18.

Second, what do I recommend? I suggest betting on the biggest longshot you can find. At the time you asked this question, the biggest longshot I could find was this college football game:

Miami (FL) +575
Alabama -1000

Assuming the house edge is the same on both bets, the probability of Miami winning is 14.01%. This would result in a house edge of 5.41% both ways.

Let's assume if the player loses he will find another game at the same odds to use his second chance on. That said, here are the possible outcomes:

• There is a 14.01% chance you win the original bet and profit \$575.00.
• There is a 12.05% chance you lose the original bet and win the second one. Here you will have lost \$100 and won \$575, for a net win of \$475.
• There is a 25% chance you lose both bets for a loss of \$100.

The expected value of this promotional bet is 0.1401×\$575 + 0.1205×\$475 + 0.7394×-\$100 = \$63.87.

Bottom line is to throw a Hail Mary both times. This advice is true of "use once" promotional chips in general. Unfortunately, such chips are usually restricted to even money bets.

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