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[Albalaha]

http://albalaha.lefora.com/2013/08/20/how-izak-matatya-beats-15000-spins-of-american-rou/

[esoito]

A quote from your site: "...the penultimate winner number (last number that has just won in the last spin)..."


No. Penultimate does NOT mean 'the last' -- it means THE ONE BEFORE LAST.

This needs correcting on the site.


So which one exactly is Izak's selection?

[Tomla]

yes very confusing as always

[ADulay]

No. Penultimate does NOT mean 'the last' -- it means THE ONE BEFORE LAST.

Thank you for bringing that up.  It bothered me too.

AD

[Albalaha]

My mistake. He is betting the "last" number and not penultimate. Correcting on my blog too.[smiley]cps/nodding.gif[/smiley]

[esoito]

Very good.

Obviously there's no need to thank me for bringing it to your attention.

[Bayes]

The arithmetic seems suspect to me.

If you bet only on repeats you get 404 X 35 = 14,140 units versus 15,000 - 404 = 14,596, thus you would lose 14,596 - 14,140 = 456 units.

With the parlay, you get 12 X 1295 = 15,540 units.  You have been betting 15000 units with 1 unit on each spin, minus the 404 repeats that you don't lose and 12 parlays that you do not lose either, thus 15,000 - (404+12) = 14,584 units lost.

Thus net profit is 15,540 - 14,584 = 956 units.

If you're using a parlay and betting the last number spun, then you DO lose those times when you get a single repeat, because you're parlaying your winnings onto the next bet, which, if it loses (i.e., you get TWO hits of the same number, NOT three which is needed for the parlay to be successful) then you have lost your original stake.


Apart from that, these particular results could easily be a fluke; 15,000 spins isn't a lot of testing for a low probability bet like this.