Re: Card Sharks
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First, I think you switched the labels on these cases.
Second, I get 4*(27+31+35+39+43+47)=888 ways to win twice and 4*(23+19+15+11+7+3)=312 ways to win and then lose. As should be expected, these total 888+312 = 1200 = 6*4*50, since there are 6 ranks that win for you, 4 cards in each of those ranks, and 50 possible third cards given the first two.
Did you assume aces are low?
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Whoops! You're right. But my post sounded good. I used the same method, but somehow moved the nines on the second card from winners to losers. That's why I'm 4*27 = 108 short on the WW and 4*23 = 92 short on the WL; with those cases moved to LW and LL respectively.
You're also right that I mixed up the labels, of course to win twice after a seven you need a nine or higher on the second card, and the third card to be lower than the second card, not higher.
No, I assumed Aces were high.
In an attempt to salvage some credibility, here is a simple example of the principle I was defending. There is an urn with one white marble and two black ones. You have $1,000 and are allowed to bet on two draws from the urn. There is no replacement, once a ball is drawn it is discarded. The house pays $1.80 for each $1.00 bet if you win.
On your first bet you have 2 chances in 3 of winning if you bet on black. (2*$0.80 - 1*$1)/3 = $0.20 so this is a positive expected value bet. However, if you win the first bet the odds are even on the second bet, so you have a negative expectation and wouldn't bet. But if you lose the first bet you're guaranteed to win the second, so you would bet everything.
If you bet X on the first draw and lose, you end up with 1.8*(1,000 - X). If you win, you end up with 1,000 + 0.8*X. The first happens one time in three, for an expected value of 0.6*(1,000 - X) and the second happens two times out of three for an expected value of (2,000 + 1.6*X)/3. Add them together and you get (3,800 - 0.2*X)/3. So you should bet nothing on the first draw, despite the positive expected value.
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