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**jasonbosche****Member**- Registered: 2013-01-07
- Posts: 4

Should someone enter into a competition where he has a 1 in 1000 chance of winning, then as a winner gets entered into a second competition where he has a 1 in 1000 chance of winning what would the probability of him winning the second competition?

Is it as simple as 1000 * 1000? Or is it 2 in 2000?

I am lost please help!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi jasonbosche;

Welcome to the forum.

Should someone enter into a competition where he has a 1 in 1000 chance of winning.

That question can only be answered in terms of expectation.

You are asking the chance of winning 2 tournaments both at 1 / 1000?

It is simply

( 1 / 1000 ) (1 / 1000 ) = 1 / 1000000

His chance of winning the second tournament is just 1 / 1000.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**jasonbosche****Member**- Registered: 2013-01-07
- Posts: 4

Ok great thanks.

So his chance is 1 in a million to win two in a row, so if we had to add this onto the question.

The chances he has of winning 3 competitions in a row where each of them he has a 1 in 1000 chance of winning.

This would be 1000* 1000*1000 = 1 a in a billion chance?

arent there some other factors that make his chances higher, i mean if winning the lotto is a one in a 175 million chance i would think winning three in row where theer are 1 in 1000 chances would give him better chances... If you know what i mean..

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

arent there some other factors that make his chances higher,

It depends on how sure you are of the 1 chance in a 1000. If that is an estimated probability ( subjective probability ) then there is some room for error. If that is an exact probability ( objective probability ) like the odds of getting a six when you throw a die then

( 1 / 1000 ) ( 1 / 1000 ) ( 1 / 1000 ) = 1/1000000000

is the chance of winning all three tournaments.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**jasonbosche****Member**- Registered: 2013-01-07
- Posts: 4

I would say its subjective probability because there is some skill in the game.

Does that change anything?

Lets say the game is 20% skill

*Last edited by jasonbosche (2013-01-07 05:27:00)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Sure does change things, the question is much more complicated now. What do you mean by 20% skill, please be exact.

Do you feel that player has a 20% chance of victory?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**jasonbosche****Member**- Registered: 2013-01-07
- Posts: 4

Lets say the game is mostly chance but there is a certain aspect of skill, like 20% of the game would be skill and 80% absolute game of chance.

like blackjack for example, some players do the wrong things and then the bank has a bigger chance of winning, normally their advantage is 20% but with a bad player its 35% advantage.

If that makes sense... There is a certain aspect of skill but not a lot only 20%

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

hI jasonbosche;

bigger chance of winning, normally their advantage is 20% but with a bad player its 35% advantage.

The house edge against a good player in blackjack is very small, less than 1%. Against a good card counter they have a negative percentage depending on the game and the bet spread. Who knows how much of a disadvatage a bad player is at.

Basically the question you have to ask is, is the player playing with a positive expectation or a negative one? That is the only stat that counts.

For instance, the house has about a 2% advantage on the average player on a blackjack table. That means for every dollar bet the house wins 51 cents and player win 49 cents, regardless of who wins any particular bet. This expectation stays the same. The player loses 2 cents on every dollar bet.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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