I lost my credit card in the parking lot today. Some very nice lady found it and called the credit card company. I don’t know who she is, but thank you very much. There are still good people in the world. Now back to the Piranha Club.
I was at my local Staples and the old lady at the check out pissed me off. I was upset. That’s why I lost it. Turns out Visa cancelled and will send me a new card. Also my recurring payments will continue. I don’t have to change anything.
Bud Grace’s method of calculating the 52 card game odds is faulty. Consider the case of only four cards being left: his method would yield 1/4+1/3+1/2+1=2+1/12 as the odds total of the last for draws, which is impossible. The odds of the total any mutually exclusive events cannot be greater than one. Moreover when insisting on at least one successful outcome of any mutually exclusive events the odds will total exactly one.
So where does Grace’s method go wrong? There was only a 4/52 (=1/13) chance of arriving at the final four cards, and that history of 48 unsuccessful draws must be taken into account. The first draw does in fact have a successful chance of 1/52, but a failure chance of 51/52; the second draw is only possible because of the previous failure with its 51/52 odds. Therefore the odds of the second draw become 1/51 x 51/52 = 1/52. Third draw: 1/50 x 51/52 x 50/51 = 1/52, etc. So we see that the odds of a successful draw remain 1/52 for all 52 draws, with odds correctly totalling one.
Losing your credit card in a parking sounds like something I would do. You were lucky..
I was at my local Staples and the old lady at the check out pissed me off. I was upset. That’s why I lost it. Turns out Visa cancelled and will send me a new card. Also my recurring payments will continue. I don’t have to change anything.
Bud Grace’s method of calculating the 52 card game odds is faulty. Consider the case of only four cards being left: his method would yield 1/4+1/3+1/2+1=2+1/12 as the odds total of the last for draws, which is impossible. The odds of the total any mutually exclusive events cannot be greater than one. Moreover when insisting on at least one successful outcome of any mutually exclusive events the odds will total exactly one.
So where does Grace’s method go wrong? There was only a 4/52 (=1/13) chance of arriving at the final four cards, and that history of 48 unsuccessful draws must be taken into account. The first draw does in fact have a successful chance of 1/52, but a failure chance of 51/52; the second draw is only possible because of the previous failure with its 51/52 odds. Therefore the odds of the second draw become 1/51 x 51/52 = 1/52. Third draw: 1/50 x 51/52 x 50/51 = 1/52, etc. So we see that the odds of a successful draw remain 1/52 for all 52 draws, with odds correctly totalling one.
Geeze! I got to have a math genius checking me! My only excuse is I’ve been hors de combat for 55 years or so.