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November 26

Martingale betting system

If I apply a Martingale betting strategy to a random process (dice, roulette, coins) I'll end up broke. Variations of it (not doubling after each loss, but halving after each loss, for example) won't make a difference.

However, what would happen if I apply it (or some variation of it) to a process where the results are not independent? That is, a process where there are tendencies, bad and good stretches? E.g. sports, politics, horses.

Would Martingale work here? --Hofhof (talk) 14:29, 26 November 2016 (UTC)[reply]

With a martingale betting system for coin tosses, you won't necessarily end up broke if you stop as soon as you are ahead. For example, if you have enough initial wealth to finance 6 tosses, then the chance of going broke is the chance of throwing 6 straight tails, or

1/2^{6}=1/64.

Now if the tosses are positively serially correlated, then one tail is more likely to be followed by another tail, so you still could go broke (I would think with a higher probability than before). On the other hand, if the tosses are negatively serially correlated, except in the extreme case where a tail is necessarily followed by a head, there is still a chance of a sufficiently long string of tails to make you go broke. Loraof (talk) 20:56, 26 November 2016 (UTC)[reply]

The problem is that "as soon as you are ahead" will almost surely happen with the "double when lose" martingale, but you will win only little. If you repeat until you either (say) double your initial account or go bankrupt, your success chances are not improved (they actually deteriorate, compared to betting your whole money at once, if the house takes a percentage on each toss). I would assume Hofhof is asking for a way to (say) play until they make double or bankrupt with a probability of winning > 0.5. Tigraan^{Click here to contact me} 21:13, 27 November 2016 (UTC)[reply]

It would depend on whether the odds you're given accurately reflect the correlation and whether you know the effects of the correlation. In an extreme case where a head is always followed by a tail and vice versa, if the person you're betting against always sets the odds at 1:1 based on the average, then you can bet according to the rule and win each time. On the other hand suppose that there is a deterministic rule for the next result, but it depends on the previous 100 results. In that case you'd need at least 2¹⁰⁰ results to figure out what the rule is and you'd be back where you started. In real life, if there was a way to figure the odds better than the person you're betting against then that person would soon go broke and would no longer accept bets. An example is card counting where you could, with a sufficiently good memory, win against the house in Blackjack. But the axiom "The house always wins," Dustin Hoffman aside, holds because if the house didn't win it would no longer be in business. --RDBury (talk) 00:32, 27 November 2016 (UTC)[reply]

Probability/statistics question: estimating indoor climbing safety

Does anyone have micromort figures for indoor climbing (as opposed to mountaineering) in terms of say, micromorts per climbing visit? The nearest I can find is this citing this paper, which suggests a really low figure. But is that figure statistically sound, given how rare an event a one-in-a-million chance is? Clearly, even though the study given recorded no deaths, the actual risk cannot be zero: see this report. Does anyone here have the relevant knowledge to work out what we can tell from these results using, say, a Bayesian framework? The Anome (talk) 18:28, 26 November 2016 (UTC)[reply]