50 years ago in college I could do this....

[gillenr]

....but not now.
The #'s 1 to 100 contain one winning #. If randomly you select one at a time, what are the odds against the winner being the one hundredth picked?
What is the formula?
Thanks

[Hoofless_Wonder]

About the same as me picking the daily double at Pimlico. :(

Well, I'm no expert at permutations, combinations and factorials, but since there's a full moon and it's late, I'll take a shot. I believe the odds are 1%, or, the same as picking the winning number first and not last.

Formula would be something like:

99/100 x 98/99 x 97/98 ..... 2/3 x 1/2 = 0.01

or

99!/100!

You start out with 99 out of 100 numbers as losers, then 98 out 99 as losers, and so on until you get down to the last two numbers in which you'd then have a 50-50 shot of picking the loser.

Now, if you want to program that, here's a link that might help:

http://www.programmingsimplified.com/c-program-find-factorial

[maddog42]

About the same as me picking the daily double at Pimlico. :(

Well, I'm no expert at permutations, combinations and factorials, but since there's a full moon and it's late, I'll take a shot. I believe the odds are 1%, or, the same as picking the winning number first and not last.

Formula would be something like:

99/100 x 98/99 x 97/98 ..... 2/3 x 1/2 = 0.01

or

99!/100!


You start out with 99 out of 100 numbers as losers, then 98 out 99 as losers, and so on until you get down to the last two numbers in which you'd then have a 50-50 shot of picking the loser.

Now, if you want to program that, here's a link that might help:

http://www.programmingsimplified.com/c-program-find-factorial

If the odds of picking start at 1 percent that you will pick the ball/number then on the next pick they are 1 in 99 or 1.0101 %, then after 2 picks they are already 2.01 % that you would have picked the ball. On the third pick odds would be 1.03092 % etcetera. The odds of the last ball being THE ball are absolutely huge.
When you get down to 50 balls the odds would improve to 2% and then 2.04081 % with the 49 th pick etcetera.
At the very least the odds would be 99.9999% against picking the ball/number on the last pick.

[Robert Goren]

The odds of selecting the winning ball remain the same whether it the first ball drawn or the 50th or the last 1 in 100. It sounds more complicated than it actual is.

[maddog42]

The odds of selecting the winning ball remain the same whether it the first ball drawn or the 50th or the last 1 in 100. It sounds more complicated than it actual is.

I assumed a number was removed with each pick.

[maddog42]

....but not now.
The #'s 1 to 100 contain one winning #. If randomly you select one at a time, what are the odds against the winner being the one hundredth picked?
What is the formula?
Thanks

The odds of the number being the correct pick if no balls are removed, are certainly 1 in 100. Trick question or real math problem?

[DeltaLover]

If each number is not removed from the pool then the probability of picking 99 consequtive losers will be:

(99 / 100) ^ 99

which is:

0.36972963764972644

This means that after 99 draws we expect not to pick the winner approximately 37% of the time


This means that the fair odds for this event are 1.7 - 1


The probability to pick the winner in the 100th is goining to be:

0.36972963764972644 * (1 / 100)

which is :

0.0036972963764972644

This means that the proper odds are 270 - 1

[Dave Schwartz]

DeltaLover is precisely right.

1st draw chances of NOT drawing are 99/100.

2nd draw is 98/99.

Thus, chances of NOT drawing in the 1st two are 0.98.

3rd draw is 97/98 x 0.98=.97

4th draw is 96/97 x 0.97 =.96

Notice a pattern?


Works like 1-spot keno.

[DeltaLover]

The case of removal was correctly answered by Wonder and the odds are 99-1

[Robert Goren]

I assumed a number was removed with each pick.then answer would be 100% because in order to get to the 100th pick all the other balls would be gone. If this is a trick question 100% is the right answer.

[gillenr]

I guess I should have worded it differently.
Yes, each selection is removed.
Prior to any selections, what are the odds 99 straight losers will be picked?
Thanks.

[raybo]

I guess I should have worded it differently.
Yes, each selection is removed.
Prior to any selections, what are the odds 99 straight losers will be picked?
Thanks.

.01/1?

[DeltaLover]

I guess I should have worded it differently.
Yes, each selection is removed.
Prior to any selections, what are the odds 99 straight losers will be picked?
Thanks.


This was answered already, the odds are exactly 99-1.

Exactly the same are the odds for picking the winner at any position.

To understand it try this:


The odds to pick the winner first are:
99 - 1

since the probability is: 1 / 100

To pick it second : again 99 -1 since the probability is:

(99 / 100) X ( 1 / 99)

To pick it third : again 99 -1 since the probability is:

(99 / 100) X ( 98 / 99) X ( 1 / 98)

etc

I think the problem is very simple and clean

[raybo]

.01/1?

Oops, that would be the chances of picking 99 straight winners. if the odds are correct that is.

[DeltaLover]

Oops, that would be the chances of picking 99 straight winners. if the odds are correct that is.

With removal is impossible to pick more than 1 straight winner.

[raybo]

With removal is impossible to pick more than 1 straight winner.

You're correct of course. The chances of picking 99 straight losers would be 1 chance of doing it and 99 chances of not doing it, so a 1 in 99 probability of picking 99 straight losers. Or 99/1 odds.

[Hoofless_Wonder]

Now that the moon is no longer full, I'll take another shot here.

I assumed that a number would be removed with my first answer, and gillenr has confirmed that.

I believe the series is best represented by 99!/100!, which is the only way to mathematically demonstrate the changing odds as the series progresses. So, picking the winner goes from:

1st draw: .0100 (1 out of 100)
2nd draw: .0101 (1 out of 99)
3rd draw: .0102 (1 out of 98)
...
...
98th draw: 0.3333 (1 out of 3)
99th draw: 0.5000 (1 out of 2)
100th draw: 1.0000 (1 out of 1)

Picking the winner on the 1st draw or exactly 100 draws = .01 overall, which is the same as dividing 99! by 100!.

Picking the winner anywhere else in the series is better than .01, from .0101 on the second draw to 50 percent on the 99th draw, to 100% on the last "draw".

Now, the more important question might be why does gillenr want to know this, and how does it apply to picking horses?

[raybo]

Don't know, but he sure got a lot of answers about probabilities and odds.

[highnote]

I guess I should have worded it differently.
Yes, each selection is removed.
Prior to any selections, what are the odds 99 straight losers will be picked?
Thanks.


Another way of describing it is "random sampling without replacement".

[Actor]

About the same as me picking the daily double at Pimlico. :(

Well, I'm no expert at permutations, combinations and factorials, but since there's a full moon and it's late, I'll take a shot. I believe the odds are 1%, or, the same as picking the winning number first and not last.

Formula would be something like:

99/100 x 98/99 x 97/98 ..... 2/3 x 1/2 = 0.01

or

99!/100!

You start out with 99 out of 100 numbers as losers, then 98 out 99 as losers, and so on until you get down to the last two numbers in which you'd then have a 50-50 shot of picking the loser.

Now, if you want to program that, here's a link that might help:

http://www.programmingsimplified.com/c-program-find-factorialThis is the correct answer.

Note that in the summation

(99/100)(98/99)(97/98) ..... (3/4)(2/3 )(1/2)

The numerator in the first term (99) cancels the denominator in the second term (99). In general the numerator in any term cancels the denominator in the following term. After all cancellations what is left over is 1/100 = 0.01 :)

[gillenr]

If the odds on the first pick are 99-1, second pick 98-1 , etc. shouldn't the answer be either 99+98+97 etc., or 99!?

[DeltaLover]

If the odds on the first pick are 99-1, second pick 98-1 , etc. shouldn't the answer be either 99+98+97 etc., or 99!?

The odds for the second pick are not 98-1. As we said before they are again 99-1. They are given by the following formula:

p(2) = (99/100 ) * (1/99) = 1/ 100

thus odds are 99-1

[Hoofless_Wonder]

Delta, I don't think you're taking into account that a number is removed every time. Your formula does sort of look like 99!/100! though, but I'm not sure how it shows the odds progression through each draw.

The odds are 1 out of 98 of picking the winner on the second draw, or .0101. The progression is NOT simply 1 out of 99 for 100 draws in a row.

What we're looking to find is the odds of selecting the winning number LAST. Not first or second, but last and ONLY last. Therefore you have to multiply the previous draw's odds together to determine that. It starts off as 1 out of 100 on the first draw, to one out of 2 on the second to last draw. Multiply all the together and you get .01, which are the same as drawing it on the first and only the first draw.

gillenr, you brought up an interesting question, but since a number is removed each time, the final result would be .01, as the odds increase to pick the winner with each losing draw. I think 99!, which is quite a large number, simply represents the total number of combinations of drawing the balls, all losers and the one winner, in any random series.

Sort of like the cost of a 10 cent super in a 99 horse field, in which you'd be required to select the exact order of finish for all 99 horses.

Is that the application you're looking for?

[raybo]

The OP asked what the odds of picking 99 losers in a row, in 100 events. That is 1 chance of picking 99 losers in a row, and 99 chances of not picking 99 losers in a row. So, 99/1 odds for picking 99 losers in a row, and .01/1 odds for not picking 99 losers in a row.

Now, if he changes the original post to include making odds after each play, and removing that play from the total then the odds change with each play, and he ends up with 1 chance of losing the last play, and 1 chance of winning the last play, 1/1 odds or 50% chance of winning or losing that last play.

But then that wasn't what he asked, was it? He asked what were the odds of losing 99 plays in a row, out of 100 plays. The only way he can possibly lose 99 plays in a row, is either, 1) he wins the first play and loses the next 99, or 2) he loses the first 99 plays.

[Hoofless_Wonder]

Well Raybo, I have to admit your outlook on this problem seems simpler than mine.

What I'm trying to wrap my head around is that like the OP, intuitively it seems that odds of picking 99 losers in a row would be far higher than 99-1, or .01. Only by looking at the odds progression along the series can you make sense of that, or at least the math of it all.

Which is why I have a better chance of picking 99 losers in a row rather than the daily double at Pimlico....

[raybo]

Well Raybo, I have to admit your outlook on this problem seems simpler than mine.

What I'm trying to wrap my head around is that like the OP, intuitively it seems that odds of picking 99 losers in a row would be far higher than 99-1, or .01. Only by looking at the odds progression along the series can you make sense of that, or at least the math of it all.

Which is why I have a better chance of picking 99 losers in a row rather than the daily double at Pimlico....

Aren't we talking about "natural odds" here? In a 10 horse race, for example, the natural odds for any of the runners winning is 1 chance of winning and 9 chances of not winning, 9/1 odds of winning. So with 100 runners in that race, the natural odds of any runner winning would be 1 chance of winning and 99 chances of not winning, 99/1 odds of winning. And the converse for losing, 99 chances of any horse losing and 1 chance of any horse not losing, 1/99 odds of any horse losing, or .01010101/1 odds (rounded to .01/1).

[DeltaLover]

Delta, I don't think you're taking into account that a number is removed every time. Your formula does sort of look like 99!/100! though, but I'm not sure how it shows the odds progression through each draw.

The odds are 1 out of 98 of picking the winner on the second draw, or .0101. The progression is NOT simply 1 out of 99 for 100 draws in a row.

What we're looking to find is the odds of selecting the winning number LAST. Not first or second, but last and ONLY last. Therefore you have to multiply the previous draw's odds together to determine that. It starts off as 1 out of 100 on the first draw, to one out of 2 on the second to last draw. Multiply all the together and you get .01, which are the same as drawing it on the first and only the first draw.

gillenr, you brought up an interesting question, but since a number is removed each time, the final result would be .01, as the odds increase to pick the winner with each losing draw. I think 99!, which is quite a large number, simply represents the total number of combinations of drawing the balls, all losers and the one winner, in any random series.

Sort of like the cost of a 10 cent super in a 99 horse field, in which you'd be required to select the exact order of finish for all 99 horses.

Is that the application you're looking for?


Wonder,

the probability of picking the winner in the second draw are the following:

P(Loosing the First) * P(winning the second) = (99 / 100) * 1 / 99 = 1/ 100

same applies for any other position. For example to pick it third:


P(Loosing the First)* P(Loosing the second) * P(winning the third) =

(99/100) * (98 / 99 ) * (1/ 98) = 1 / 100

[Hoofless_Wonder]

Aren't we talking about "natural odds" here? In a 10 horse race, for example, the natural odds for any of the runners winning is 1 chance of winning and 9 chances of not winning, 9/1 odds of winning. So with 100 runners in that race, the natural odds of any runner winning would be 1 chance of winning and 99 chances of not winning, 99/1 odds of winning. And the converse for losing, 99 chances of any horse losing and 1 chance of any horse not losing, 1/99 odds of any horse losing, or .01010101/1 odds (rounded to .01/1).

At the risk of bringing up formal definitions of some terms here, I'd argue that:

In a 10 horse race, the "odds" of winning are 1 in 10, or a probability of .10. When you reference it as natural odds of 9-1, that infers a profit of 9-1, like payouts at the track, which really mean getting back $10 ($9 profit, $1 wagered). But really, most people would look at that as 10-1 (like in Australia), at least if we're just asking "what are the odds"?

In the 100 horse field, any horse would have exactly a .01 chance of winning (1 out of 100, not 1 out of 99), and exactly a .99 chance of losing (99 out of 100).

Not that it's critical, but I simply look at the odds as probability, without subtracting my original wager from the "odds".

[Hoofless_Wonder]

Wonder,

the probability of picking the winner in the second draw are the following:

P(Loosing the First) * P(winning the second) = (99 / 100) * 1 / 99 = 1/ 100

same applies for any other position. For example to pick it third:


P(Loosing the First)* P(Loosing the second) * P(winning the third) =

(99/100) * (98 / 99 ) * (1/ 98) = 1 / 100

Delta, that is a slick way to show the odds of picking the winner anywhere along the series are the same as the first draw, as each number is removed.

Assuming the winning number is drawn last, then your formula results in 99!/100!. I pictured it as drawing numbers from a hat, with each draw an individual event, and seeing if I could avoid drawing the winner along the way. If there's a more complicated way to think about it, that's the way I usually go.

30 years since I took graduate level stats in college, and I sort of remembered a few things....