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In this YouTube video, Matt Damon says John McCain has a 1 in 3 chance of not surviving his first term. Is he right?

Lisa from San Antonio, TX

No. Using this actuarial table from the CDC (Centers for Disease Control), the probability of a 72 year old white male making it to age 76 is 85.63%. That is about a 1 in 7 chance of death. The survival rate can be found by dividing the birth cohort at age 76 of 57,985, by birth cohort at age 72 of 67,719, from the white male table on page 14. The table used is a called a "period life table," which assumes 2003 rates of mortality will not change in the future, and is the most commonly used kind of actuarial table. A perfectionist might want to use a 1936 cohort life table, but I don’t think it would make much difference.

p.s. After posting this answer I have received several comments that my response did not take into consideration John McCain’s individual health situation. Working against him is being a cancer survivor. Working in his favor is access to the best medical care money can buy, he is obviously still in good shape mentally and physically for a 72-year old, and longevity, as evidenced by the fact that his mother is still alive. However, I never intended to factor in this information. It was Matt Damon who quoted actuarial tables, which is what I was referring to. All I am saying is that for the average 72-year old white male, the probability of surviving four more years is 86%. If forced, I would predict John McCain’s odds are even better than that.

A reader asked about a slot tournament at the Wynn. The cost to enter was \$25,000, and the average prize was \$30,000. You said that you need a bankroll of about three million to enter, according to the Kelly Criterion. I have two questions:

1. Does this take into account the unknown house edge on the slot machines?

2. What would be the playing strategy for the best overall return? Could you just sit back and not gamble, and hope that the other 49 players all end up behind, while you break even and take the grand prize of \$1,000,000?

Gray C. from Silicon Valley, CA

Slot tournaments are always held on dedicated tournament machines. Usually these machines don’t accept bets, so your balance will either stay even or go up, after each play. So it doesn’t make any difference what the return is; the more you play, the more you can expect your balance to go up. Even if you had to play conventional slot machines, I would still bet as fast as possible, stopping only if I got a jackpot large enough to likely win the tournament. The reason is that it is very unlikely that 49 out of 49 players would be negative.

Interestingly, there was once a slot tournament at Caesars Palace where they gave a prize to the person who finished last. However, they didn’t announce this rule until the award ceremony. If you somehow knew of such a rule, indeed, it might be best to not bet.

Why is it that dealers at craps tables are very reluctant to color up your chip stack, unless you are leaving the table? Although I’ve never had a dealer outright refuse to do so, they often begrudgingly comply with a wise or snide comment, as though I were asking too much of them.

Ron from State College

This is true of all table games, not just craps. The policy against coloring up, except when leaving, comes down from management, so don’t blame the dealers. A good dealer is supposed to keep the player well armed with chips at the level he is betting. Coloring up goes against this purpose. It will lead to chip shortages, causing the player to ask to break down the big chips, which wastes time. There may also be an unstated purpose that the player will be unlikely to bet the big chip.

In slots or video poker, when playing with a double or nothing feature, how many times should I try to double?

Joseph R. from Manila, Philippines

It depends on your reason for playing. If you are trying to achieve some winning goal, like doubling your bankroll, then you should keep doubling until you reach your goal, or you reach the maximum number of doubles allowed. If you are trying to play as long as possible on a given bankroll, then I would double only on small wins, and then only once. If you have some combination of both goals, then I would have a mixed strategy. The more important winning is to you, the more aggressive you should be doubling. The more important “time on device” is to you, the less you should be.

In baccarat, the cut card is placed in front of the last 13 cards in the shoe, and one hand is dealt following the hand the cut card came out on. If the cut card came out after the first player card was dealt, and both the player and banker draw a card, only 8 cards will remain in the shoe for the last hand. If you are tracking the cards, and know the last 8 cards are all 0-value cards, a table max bet on tie would net you a huge profit. My question is, what are the odds that the last 8, 9, or 10 cards in an 8 deck shoe are all 10 value? Also if you knew exactly what the last 8 cards were, could you use a formula or program to figure the odds that the next hand will be banker, player, or tie?

Mike S. from Michigan City

To answer your first question, the probability that the last 8 cards in an 8-deck shoe are all 0-valued cards is combin(128,8)/combin(416,8) = 0.0000687746. So, it isn't something to wait around for. I know of no easy formula for what to bet in other situations. If you could find a casino that would allow you to use a computer, the advantages would sometimes be huge towards the end of the shoe, especially on the tie.

In 180 consecutive rolls of the dice, how many times can I expect to see the following:

Two sevens in a row?
Three sevens in a row?
Four sevens in a row?

Melanie D. from Elizabeth City, NC

I can’t think of any useful reason to know this information, but I get asked this kind of thing a lot, so I’ll humor you.

It is a little easier getting a specified sequence of sevens starting with the first roll, or ending with the last, because the sequence is bounded on one side. Specifically, the probability of getting a sequence of s sevens, starting with the first roll, or ending with the last, is (1/6)s × (5/6). The 5/6 term is because you have to get a non-7 at the open end of the sequence.

The probability of starting a sequence of s sevens at any point in the middle of the sequence is (1/6)s × (5/6)2. We square the 5/6 term, because the player must get a non-7 on both ends of the sequence.

If there are r rolls, there will be 2 places for an inside sequence, and r-n-1 places for a run of n sevens. Putting these equations in a table, here is the expected number of runs of sevens, from 1 to 10. The "inside" column is 2*(5/6)*(1/6)r, and the "outside" column is (179-r)*(5/6)2*(1/6)r, where r is the number of sevens in the run. So, we can expect 3.46 runs of two sevens, 0.57 runs of three sevens, and 0.10 runs of four sevens.

### Expected Runs of Sevens in 180 Rolls

 Run Inside Outside Total 1 0.277778 20.601852 20.87963 2 0.046296 3.414352 3.460648 3 0.007716 0.565844 0.57356 4 0.001286 0.093771 0.095057 5 0.000214 0.015539 0.015754 6 0.000036 0.002575 0.002611 7 0.000006 0.000427 0.000433 8 0.000001 0.000071 0.000072 9 0 0.000012 0.000012 10 0 0.000002 0.000002