Ask The Wizard #317

No, I don't.

I will have to take on faith their statistic that for an amateur golfer, the probability of a hole-in-one are 1 in 12,500 per hole.

However, almost all hole-in-ones are achieved on par-3 holes. The typical course will have four such holes. So, the probability of a hole-in-one on a par 3 hole is (1/12500) × (4/18) = 1 in 2,778.

That said, the probability that exactly two golfers out of four get a hole-in-one on a par 3 hole is combin(4,2) × (1/12500)² × (12449/12500)² = 1 in 1,286,935.

Assuming four par-3 holes in a round, the probability of two hole-in-ones in the same hole are 4 × (1/1,286,935) = 1 in 321,734.

Where Golf Digest seems to have gone wrong is in assuming every hole has the same probability of a hole in one.

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

With any free-roll like this, you should shoot for the moon. If they allow parlays, then parlay bet as much as they allow under this promotion and parlay enough events to get up to whatever their maximum payout is, which in Las Vegas is typically around $100,000.

If you have to bet on a single event, then find the biggest long-shot you can.

As an example of why this strategy is best, consider the same promotion in double-zero roulette with a $100 bet. If you make an even-money bet, then the expected amount you'll get back is (18/38)×$200 + (20/38)×$100 = $147.37. However, if you bet on a single number, then you can expect to get back (1/38)×$3600 + (37/38)×$100 = $192.11.

Let's assume six decks. The probability of the up card being an ace is 1/13. Next, the probability the hole card is a 10-point card is 96/311. So, this opportunity will occur 2.37% of hands.

The insurance bet I assume will be half a unit and insurance pays 2 to 1. So, every time this happens, you can expect to gain a unit. Since it happens with probability 2.37% of the time, that is what it is worth to you. Subtract the house edge under whatever the blackjack rules are to get the overall player advantage of the game.

There are 42 ways the dealer could break down $10. Here they are:

Mathematicians call these partitions. Here are the number of partitions for starting quantities up to 405, which is as high as my computer can calculate (2^64).

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