How long on average before winning a Stars doubleshootout?

[M.B.E.]

I'm not taking a survey here, just curious about the precise mathematical answer. I think it's an application of binomial theorem but I've forgotten how to do that.

Assume that the player can win the first table with probability 16% and, after winning the first table, will have a 14% probability of winning the final table, a 76% probability of finishing 2nd through 8th, and a 10% probability of finishing 9th.

It's easy to calculate EV. Cost is $160 to enter, first place wins a WSOP package worth (let's say) $12,000; second through eighth get $160, and ninth gets $30.

So EV per attempt is:

(0.0238 x $12,000) + (0.1292 x $160) + (0.017 x $30) - $160

= $146.78

Hmmm, a 92% ROI seems extremely high. Are my assumptions totally out of whack? Keep in mind that competition is quite soft because many of the players won their seats in a $5 rebuy satellite.

If you think my assumptions are way too optimistic, then substitute your own (assuming the player is good but not great). Otherwise use my figures set out above.

The questions are:

(1) On average, how many attempts does it take to win the WSOP seat?

(2) Taking account of the $30 and $160 prizes along the way, how much will the player have lost on average before winning the WSOP seat?

(3) What is the probability that the player will not have won a WSOP seat after the 60th attempt?

(4) What is the probability that the player will have lost $12,000 (net) without winning a seat?

[PrayingMantis]

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Assume that the player can win the first table with probability 16% and, after winning the first table, will have a 14% probability of winning the final table,

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Without getting into the math part at all, I just think that it's >16% to win the first, and <14% to win the second, for what you would call a good player. Generally, the final table is MUCH tougher than the first one. I'd say that sometimes it's like a jump from $10 SNG to $200.

[durron597]

One reason that ROI number is so high is because you are giving yourself an extra $1000 of overlay (or Stars is) since Stars only deducts $11000 of the prize pool to be part of your seat.

[shaniac]

Why are you valuing the package at $12,000?

Edit: I've played a lot of shootouts, and won 3.5 of them--I just emailed stars for the records, will post when I get them

[Che]

You can use this calculator to figure out all your answers if you invest a little time.

If I'm interpreting it correctly, the answer to Q3 is 25.7%. I don't have time to do the others now, but I'm interested in the results.

Later,
Che

edit: And Q1 is between 44 and 45. [img]/images/graemlins/grin.gif[/img]

[M.B.E.]

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Why are you valuing the package at $12,000?

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$10K seat, $1K cash, plus either 9 days hotel or an additional $1K cash.

Of course that's for the first seat. After you have one, the second should probably be valued at $10,200 or so (assuming you can sell the 10K W$ for $9,200 cash).

[M.B.E.]

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You can use this calculator to figure out all your answers if you invest a little time.

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Thanks, that's a cool page.

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If I'm interpreting it correctly, the answer to Q3 is 25.7%.

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Wouldn't it be 23.6%? That's the probability of not winning, to the power of 60.

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edit: And Q1 is between 44 and 45. [img]/images/graemlins/grin.gif[/img]

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Not quite sure where you got that?

I think the answer to Q1 should be:

(1 x .0238) + (2 x .9762 x .0238) + (3 x [.9762]^2 x .0238) + ... + (n x [.9762]^[n-1] x .0238) + ...

Does that look right? It's an infinite series which should be a weighted average, e.g. the 58th term is 58 multiplied by the probability that you win your first seat on the 58th attempt.

If I've set up my spreadsheet correctly, it looks like the series converges to something a bit larger than 42.

[M.B.E.]

Just realized that 1/.0238 is 42.0168, so maybe that's the answer to Q1?

[Che]

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Wouldn't it be 23.6%? That's the probability of not winning, to the power of 60.

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Your answer is correct if you use 2.38% as the probability of winning. I was using (.14)(.16)=2.24% based on your original assumptions.

To get the answer to question 1, I entered my percentage (2.24%) and then used trial and error to determine which number of trials yielded a "mean number of events" closest to 1. The answer is slightly less than 45.

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Hmmm, a 92% ROI seems extremely high. Are my assumptions totally out of whack?

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I would say no. The reason the ROI is so high is because winner-take-all formats (which DS's effectively are) reward the better players more than flat payouts that spread the money out more (so that weaker players get a little). Winning just 1% more often than the "average" player yields a large ROI because you are winning an extra 1% of a prize ~72x your buyin. (That's not a very good explanation, but it makes sense to me. [img]/images/graemlins/crazy.gif[/img])

Later,
Che

[M.B.E.]

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Wouldn't it be 23.6%? That's the probability of not winning, to the power of 60.

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Your answer is correct if you use 2.38% as the probability of winning. I was using (.14)(.16)=2.24% based on your original assumptions.

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Oh yeah, you're right. That was a mistake in my original post. Originally I had (.14)(.17)=2.38% and I forgot to change that part.

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Hmmm, a 92% ROI seems extremely high. Are my assumptions totally out of whack?

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I would say no. The reason the ROI is so high is because winner-take-all formats (which DS's effectively are) reward the better players more than flat payouts that spread the money out more (so that weaker players get a little).

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Yes, good explanation.