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# Why strings of losses don't, yet do, better your chances

30 January 2007

Gambling gurus say no matter how often you take a shot at winning a bet and fail, the chance on your next try stays the same. But common sense says the more you play, the higher the likelihood you'll score. Just as 10 lottery tickets give you a better chance than one, so wagering on a single number at roulette on 10 spins gives you a greater chance at winning than going for only a single coup. And when going for broke at the slots, a thousand rounds somehow buy you more of a chance than a hundred.

These lines of reasoning aren't contradictory. The conclusions differ because the premises aren't equivalent.

Picture the toss of a balanced coin. Flip once and you can get heads (H) or tails (T) with equal probability. That's 50 percent each, and 2 x 50 = 100 percent, indicating all the possibilities.

Flipping twice, you could get HH, HT, TH, or TT. Each of the four possible results is equally likely, one out of four or 25 percent. Again, 4 x 25 = 100 percent. The alternate sets had two heads and two tails on the second toss. So, the chance of tails on the second toss was two out of four or 50 percent. It didn't matter whether the first toss landed heads or tails.

With three tosses, you could have HHH, HHT, HTH, THH, HTT, THT, TTH, or TTT – eight equiprobable results. The third toss had four each, heads and tails, so the chance is four out of eight or 50 percent regardless of the first two tosses. Tails on the third toss is no more or less likely after HH than TT, HT, or TH.

You could also ask the chance of success by flipping until you got heads once. Assume you could go for three coups before giving up. You'd win if you flipped H, TH, or TTH; you'd lose if the result was TTT. Using the figures above, the chance of success is 50 percent (H) plus 25 percent (TH) plus 12.5 percent (TTH). The total is 87.5 percent, the same as flipping three times regardless of result and obtaining at least one head.

In real life, solid citizens often ask these questions about longshots and numbers of trials that can't be easily jotted down and counted like three flips of a coin. Then math, which I'll spare you and just give some answers, flashes its fearsome fangs.

For instance, make believe you want to bet \$10 per spin on a particular number at double-zero roulette. The wheel has 38 positions so your chance of winning on any spin is one out of 38 – 2.6 percent. If you had a \$100 bankroll and your strategy was to play until you won, for up to 10 spins, your chance of joy would be 23.4 percent. With a \$200 bankroll, you'd have a 41.3 percent chance. Turn this around and find how many losing rounds you'd have to be prepared to endure to have 90 percent chance of success. It's 87 rounds, a downswing of \$870. And, alas, a hit at that point would still leave you \$520 in the hole.

The situation doesn't improve on the slots, where a jackpot could conceivably cover a career of losing spins. For simplicity, posit an all-or-nothing game with no house edge. You bet \$1 with a chance of one in 10,000 to win \$9,999. For 50-50 prospects, you'd need enough for 6,931 tries. To get to a 90 percent chance, your bankroll would have to be \$23,024 and hitting on the 23,024th try would still leave you \$13,024 down. As for a 100 percent chance, the coupleteer, Sumner A Ingmark, caught the conundrum like this:

When odds are long, and still you fight 'em,

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Alan Krigman

Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013.
Alan Krigman
Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013.