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This is a dice game where the host throws the dice and the players bet on 3 numbers - numbers below 7, 7 and numbers above 7. On numbers below and above 7 the players get double their stake and on 7 the players get 5 times their stake. For example if i bet 100 bucks on "below" and 7 comes, I will lose, but if 6 comes, then i win 200 bucks. Similarly if someone has betted 100 bucks on 7 and if 7 comes, he wins 500 bucks.

What is the best way to bet in this game?

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  • $\begingroup$ There are 2 dice. Sorry, missed that out. $\endgroup$
    – user7030
    Commented Feb 13, 2011 at 14:31
  • $\begingroup$ though it's not completing germane to your particular problem, you might be interested in Dubins and Savage, How to gamble if you must: Inequalities in stochastic processes. There are also some introductory treatments online and a section in Billingsley about it. The basic strategy is often referred to as bold play. $\endgroup$
    – cardinal
    Commented Feb 13, 2011 at 15:52

4 Answers 4

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As they say, "the only winning move is not to play." I assume there are two dice. In that case, the expected value of all three bets is the same, and it's negative: on average, you lose $\frac{1}{6}$ of what you bet.

Some details: when rolling $2$ dice the probabilities of getting a sum of $2, 3, 4, ... 12$ are $\frac{1}{36}, \frac{2}{36}, \frac{3}{36}$, and so on, increasing until $7$ and then decreasing. This is a nice exercise if you've never done it; you just need to count the number of ways a pair of numbers between $1$ and $6$ can add up to a given number. It follows that the probability of getting a sum of $7$ is $\frac{1}{6}$, and by symmetry the probability of getting above $7$ or below $7$ are each $\frac{5}{12}$.

It follows that if you bet $7$ then on average you multiply your stake by $\frac{5}{6}$, whereas if you bet either below or above then on average you also multiply your stake by $\frac{5}{6}$. So all three options are equally bad, in the long run.

Some meta-lessons to learn from this computation:

  • Generally the expected value of a game like this is negative, or the host would never bother to offer it. For example, the expected value of every game at a casino and the lottery has to be negative, or else the casino (resp. the institution running the lottery) would not make money. This is as much a lesson about incentives as it is about probability.
  • If one of the options was better than the others, at some point people would be aware of this and play only that option.
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choose one side of the three sides. Say $7$ up. Start your game by placing the minimum bet say Rs. $10$. If you lose then double your bets everytime you lose(remember only to double your bets if you lose) but put your money on the same side(i.e. $7$ up). Within three to four games of continuous losses you will surely win in fourth to fifth time. :) ;) Remember this is the trick of winning this game , the only thing you need to do is place your bets only on the same side (i.e. The side on which you were losing). You can change your side once you win. I hope I helped you.. ;) . I always follow this rule and the result is profits.

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    $\begingroup$ This is the classic Martingale system. It misses the point that a series of negative expectation bets cannot have positive expectation. It is true that your chances of winning are high, but the losses are so high when you hit the limit that on average you lose. $\endgroup$ Commented Nov 1, 2013 at 15:20
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The logic is to keep betting the same choice with increasing bet amount such that if you win that particular bet it at least covers loss of all previous bets. As per your setup lets say you want to win 100 bucks with the probability of 95%, then the strategy should be to start by betting 100 bucks in any of 7up or 7down, lets keep it 7up. Next if you lose, bet again but double the previous bet on the same choice i.e. 200 bucks on 7up, and keeping betting until you win : 100, 200, 400, 800, 1600, 3200, 5% chance that you will need to go futher. As per my calculation, the probability that your win takes at most 6 bets is more than 95% as needed. And thus you should be ready for at least for 6 bets for which in total you will need (2^6-1)*100 = 6300 bucks. Conclusion: If you have 6300 bucks, you can earn 100 bucks in 6 continuous bets with a probability of 95%. Loop this process if you wish and earn 100 bucks again.

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Dont worry about all these mathematics, make it simple, gambling is all about sheer luck. Doubling the loss bets isn't a good idea, I have lost many times doing that, even after 15 deals the side won't change, this can happen anytime, this is very risky attempt for someone playing with limited amount, and this idea will be successful for the one who have the money backed for the betting. Who knows how the cards are shuffled, then there is no point of permutations, combinations and probabilities. Please dont misguide people with this kind if advices. For me gambling is sheer luck to add some more money to our hard earned money what I say is accept both loss and profit. And do not gamble with commitments, if u do that chances of losing is more than 100%

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