Quantifying a non self weighing effect

[DrVanNostrin]

Mason Malmuth wrote:

Suppose you make a trip to Las Vegas, walk up to a craps table, and start betting $1 every time you roll the dice. You do this for 10,000 rolls, and then suddenly on the next roll you pull $1 million out of your pocket and put it all into action. Now how many times have you rolled the dice? Well, there are two answers, both correct. The mathematicians will say 10,001 times, but the statisticians will say one time simply because your results are all clustered around that one big bet. Notice, that during the first 10,000 rolls of the dice, we were looking at a self-weighting gambling experience. But after that $1 million bet came along, we had a very non-self-weighting gambling experience. Incidentally, and this is important, all successful gamblers are statisticians, not mathematicians.


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But in GTaOT there is no mention of a formula to quantify such effects. In the example given it's clear that the answer is that the dice has been rolled once statistically. But what if instead of putting $1M into play on roll #10,001, you put $10 or $100 into play. How many times would you statistically have rolled the dice then?

[Siegmund]

He's speaking somewhat loosely when he says "rolled the dice once" -- that is, almost your entire bankroll fluctuation for the trip is caused by that last roll.

Rather than "effective number of dice rolls," we usually just report the mean and standard deviation of a series of bets of differing sizes.

Simplifying from craps to betting on even-money coin flips:

10,000 independent fair $1 bets (standard deviation $1 each) add up to something that is, in effect, one fair bet with standard deviation $100.

Those 10,000 plus one million dollar bet make one fair bet with standard deviation $1000000.005.

Those 10,000 plus one $10 bet would have standard deviation $100.49, and those 10,000 plus one $100 bet would have standard deviation $141.42.

If you wish an "effective number," you can ask, "how many large bets could I have made, to get the same standard deviation in my outcome as I got by making the series of small bets followed by one large bet?" The answer is 1.00000001 million dollar bets, 2 $100 bets, or 101 $10 bets. In general, 1 + (numberof$1bets)/(sizeoflargebet^2).

[DrVanNostrin]

Thanks for the reply. That makes a lot of sense.