xxi: Please check my tournament math...

[xxi]

I'm trying to figure out the EV of my wild card chances for the $1 million Hilton tournament. I'd appreciate someone checking my math.

There were 1,613 total entrants. 191 of them qualified already, so there are 1,422 (1,613 - 191) entries in a wild card drawing for 9 remaining spaces. I have 3 entries, so my chance of being picked is:

9 x 3/1422 = 0.0190 (1.9%)

Assuming the prize is $1 million (ignoring the much lower runner-up amounts) and setting aside any judgement of relative skill, one's chances of winning are 1/200 or 0.5%.

So the EV of showing up at the wild card drawing (assuming everyone else showed up, too) is:

0.0190 x 0.005 x $1 million = $95

Not exactly worth a long trip.

If someone could confirm or correct my calculations, I'd appreciate it. Thanks.

[Don Schlesinger]

> I'm trying to figure out the EV of my wild
> card chances for the $1 million Hilton
> tournament. I'd appreciate someone checking
> my math.

> There were 1,613 total entrants. 191 of them
> qualified already, so there are 1,422 (1,613
> - 191) entries in a wild card drawing for 9
> remaining spaces. I have 3 entries, so my
> chance of being picked is:

> 9 x 3/1422 = 0.0190 (1.9%)

This assumes drawing without replacement, which isn't the case. I get 1.88%, which, of course, is too insignificant a difference to matter.

> Assuming the prize is $1 million (ignoring
> the much lower runner-up amounts) and
> setting aside any judgement of relative
> skill, one's chances of winning are 1/200 or
> 0.5%.

Correct.

> So the EV of showing up at the wild card
> drawing (assuming everyone else showed up,
> too) is:

> 0.0190 x 0.005 x $1 million = $95

Correct.

> Not exactly worth a long trip.

Only you can decide whether or not a shot at the million is worth the trip. Every day, millions of people play Lotto, even though their EV is infinitesimal. They do it because, should they win, their lives may be changed forever -- EV be damned.

> If someone could confirm or correct my
> calculations, I'd appreciate it. Thanks.

See above. You're welcome.

Don

[xxi]

> This assumes drawing without replacement,
> which isn't the case. I get 1.88%, which, of
> course, is too insignificant a difference to
> matter.

Thanks, Don. It's a pleasure to be advised with one of the true giants of the game.

I'm not sure what you mean by "drawing w/o replacement". My guess is that the odds on each draw change slightly because there is one fewer entry left in the pool. So I'd have to add the discrete odds for each draw, rather than simply multiple.

I.e., the first draw my odds are 3/1422. My odds on the next (assuming I didn't win, in which case I don't care about the next!) are 3/1422. And so forth for the next 7 drawings. I should add the foregoing for the total probability.

Is that correct, or is it something else? Thanks again.

[xxi]

[Don Schlesinger]

> Thanks, Don. It's a pleasure to be advised
> with one of the true giants of the game.

> I'm not sure what you mean by "drawing
> w/o replacement". My guess is that the
> odds on each draw change slightly because
> there is one fewer entry left in the pool.
> So I'd have to add the discrete odds for
> each draw, rather than simply multiple.

> I.e., the first draw my odds are 3/1422. My
> odds on the next (assuming I didn't win, in
> which case I don't care about the next!) are
> 3/1422. And so forth for the next 7
> drawings. I should add the foregoing for the
> total probability.

> Is that correct, or is it something else?
> Thanks again.

Yes, you have the idea, but in problems such as these, it's always much easier to work with the complement, or one minus the number you're interested in. Then, once you have the probability that the event won't happen, you once again use one minus the value you've come up with to find the chance that it will happen.

In the above problem, just remove the three chances that can make you successful. So, on the first draw, you have 1-(3/1422) = 1419/1422 probability that you won't be picked. On the next draw, you have 1418/1421 chance to not be picked. Multiply all the way down to the ninth drawing, where the odds of missing out are now 1411/1414. Once you have this product, you subtract your result from one, and the answer is the probability that you WILL get picked somewhere in the nine drawings.

Clear?

Don

[Susan]

There is more to the story.

1. Two players have qualified twice increasing the original eight open spots to ten (not nine).

2. There will be some ?no show? of qualified players additionally increasing number of drawings to (make a guess) 12-20.

3. 1613 entries include players who qualified and entered the tournament more then once. Perhaps only around 1200 entries will be eligible for the drawing.

4. The 1200 entries belong to some 400-600 ?unique? players.

5. Many players, especially those with only one or two entries, will not attend the drawing. I guess that there will be less than 400 eligible players (with about 1000 entries) participating in the drawing.

6. This is not significant (and rather difficult to calculate without knowing the exact distribution) but players with more entries will have statistically better chance of being drawn. For example, if there are 350 players with 1050 entries and there are fifteen names drawn, xxi (with three entries) would have 1/350 chance during the first drawing. The next fourteen lucky guys would invalidate about sixty entries, giving xxi 1/330 chance to succeed during the last name drawing. In this scenario xxi would have more then four percent chance of being drawn.

7. The median player in Hilton?s April tournament has just slightly more then half the chance of the average player to win the tournament (and a ?ten percent from the top? player is about twice the favorite to win over the average player). Luck is very important but skill factor can not be dismissed.

8. I have a $50 over/under bet about the number of ?no show? at plus 15. I am sharing my winnings with a friend who drives a Hummer. Be careful when crossing street going to Hilton?s drawing.

Good luck,
Susan