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What are the odds of losing a bet 10 times in a row when I have a 25% chance of winning?

which of these fair gambles gives your the greatest chance of winning?What are the odds of not being selected 2 years in a row when you have a 71% chance of winning each year?What are the odds of losing a bet 20 times in a row when I have a 49.95% chance of winning?What are the odds that the pattern, win lose lose, will happen 23 times in a row (69 rounds)?Roulette odds of winning using specific strategywhat's the odds of winning 3 times my money on the first try when I have a 50% chance of winning every bet?what are the chances of losing a 31.5% chance win bet 14 times in a row?PROBLEM: Probability and net sum from samplesWhat is the probablity of flipping 1024 fair coins without ever getting 11 of the same side in a row?Bingo-alike game Probability, $frac14$ chance… $20$ times losing in a row.

0

$begingroup$

I have been looking for an online calculator to solve the below 2 mathematical problems, but I haven't yet found such a calculator. Thus, can someone answer these questions for me?

1. What are the odds of losing a bet 10 times in a row when I have a 25% chance of winning?


2. What are the odds of losing a bet 20 times in a row when I have a 25% chance of winning?

edit: I mistakenly forgot to mention that I would like the reply answer to be in the equation of "1 in 1000" or "1 in 12,448" (or whatever the real answer is). I apologize for forgetting this important part.

I will appreciate any help I can get with the above 2 math questions.

probability

share|cite|improve this question

edited Oct 24 '17 at 5:10

user495020

asked Oct 24 '17 at 4:18

user495020user495020

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ten: $$(1-0.25)^10 = 0.75^10 approx 0.0563 = boxed5.63%$$ How would you do twenty, given this?
  $endgroup$
  – Zubin Mukerjee
  Oct 24 '17 at 4:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Stattrek binomial calculator is such a calculator
  $endgroup$
  – pjs36
  Oct 24 '17 at 4:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. But I really like the answer to be in the equation such as: "1 in 1000 chance" or "1 in 74,226 chance" (or whatever the answer is).
  $endgroup$
  – user495020
  Oct 24 '17 at 5:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $5.63%$ is equivalent to "5.63 in 100 chance" then dividing this ratio by $5.63$ on both sides gives you a "1 in 17.76 chance" or, rounding to the nearest integer a "1 in 18 chance". Try doing it for the 20 games now using the same method as I did.
  $endgroup$
  – thesmallprint
  Oct 24 '17 at 5:43
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I have been looking for an online calculator to solve the below 2 mathematical problems, but I haven't yet found such a calculator. Thus, can someone answer these questions for me?

1. What are the odds of losing a bet 10 times in a row when I have a 25% chance of winning?


2. What are the odds of losing a bet 20 times in a row when I have a 25% chance of winning?

edit: I mistakenly forgot to mention that I would like the reply answer to be in the equation of "1 in 1000" or "1 in 12,448" (or whatever the real answer is). I apologize for forgetting this important part.

I will appreciate any help I can get with the above 2 math questions.

probability

share|cite|improve this question

edited Oct 24 '17 at 5:10

user495020

asked Oct 24 '17 at 4:18

user495020user495020

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ten: $$(1-0.25)^10 = 0.75^10 approx 0.0563 = boxed5.63%$$ How would you do twenty, given this?
  $endgroup$
  – Zubin Mukerjee
  Oct 24 '17 at 4:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Stattrek binomial calculator is such a calculator
  $endgroup$
  – pjs36
  Oct 24 '17 at 4:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. But I really like the answer to be in the equation such as: "1 in 1000 chance" or "1 in 74,226 chance" (or whatever the answer is).
  $endgroup$
  – user495020
  Oct 24 '17 at 5:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $5.63%$ is equivalent to "5.63 in 100 chance" then dividing this ratio by $5.63$ on both sides gives you a "1 in 17.76 chance" or, rounding to the nearest integer a "1 in 18 chance". Try doing it for the 20 games now using the same method as I did.
  $endgroup$
  – thesmallprint
  Oct 24 '17 at 5:43
  
  

  
  
  
  

add a comment | 

$begingroup$

I have been looking for an online calculator to solve the below 2 mathematical problems, but I haven't yet found such a calculator. Thus, can someone answer these questions for me?

1. What are the odds of losing a bet 10 times in a row when I have a 25% chance of winning?


2. What are the odds of losing a bet 20 times in a row when I have a 25% chance of winning?

edit: I mistakenly forgot to mention that I would like the reply answer to be in the equation of "1 in 1000" or "1 in 12,448" (or whatever the real answer is). I apologize for forgetting this important part.

I will appreciate any help I can get with the above 2 math questions.

probability

share|cite|improve this question

edited Oct 24 '17 at 5:10

user495020

asked Oct 24 '17 at 4:18

user495020user495020

11

$endgroup$

share|cite|improve this question

edited Oct 24 '17 at 5:10

user495020

asked Oct 24 '17 at 4:18

user495020user495020

11

share|cite|improve this question

edited Oct 24 '17 at 5:10

user495020

asked Oct 24 '17 at 4:18

user495020user495020

11

asked Oct 24 '17 at 4:18

user495020user495020

11

asked Oct 24 '17 at 4:18

user495020user495020

11

1 Answer
1

active

oldest

votes

1

$begingroup$

First part:

Probability of losing 10 times in a row = $(1-0.25)^10 = (0.75)^10$

$$textodds = fractextfavorable outcometextunfavorable outcome$$

Here the probability of losing $10$ times in a row is favorable outcome, so

$textodds = frac0.75^101-0.75^10$

Similarly, for second part:

$textodds = frac0.75^201-0.75^20$

share|cite|improve this answer

edited Oct 24 '17 at 10:29

Andrew Tawfeek

1,8041722

answered Oct 24 '17 at 4:26

Sagar ChandSagar Chand

1,2731520

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sagar Chand, I have to admit I only understand very basic math. Thus, I do not understand the equations you have given. What I really like is for you (or anyone else) give an answer such as: "With a 25% chance of winning, your chances of losing 10 in a row is 1 in 35,782" (or whatever the real answer is). And the same applies to my question about losing 20 in a row with a 25% chance of winning.
  $endgroup$
  – user495020
  Oct 24 '17 at 5:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  probability of losing once is 0.75,so probability of losing 2 times is 0.75*0.75. Similarly proba of losing 10 times in a row is $0.75^10$ Is this clear?
  $endgroup$
  – Sagar Chand
  Oct 24 '17 at 5:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But can you can you specifically tell me if it is "1 in 1000 chance" or "1 in 25,837 chance" or whatever?
  $endgroup$
  – user495020
  Oct 24 '17 at 5:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1 in $frac10.75^10$ chance for part a
  $endgroup$
  – Sagar Chand
  Oct 24 '17 at 5:18
  

  
  
  
  

add a comment | 

Your Answer

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1

$begingroup$

First part:

Probability of losing 10 times in a row = $(1-0.25)^10 = (0.75)^10$

$$textodds = fractextfavorable outcometextunfavorable outcome$$

Here the probability of losing $10$ times in a row is favorable outcome, so

$textodds = frac0.75^101-0.75^10$

Similarly, for second part:

$textodds = frac0.75^201-0.75^20$

share|cite|improve this answer

edited Oct 24 '17 at 10:29

Andrew Tawfeek

1,8041722

answered Oct 24 '17 at 4:26

Sagar ChandSagar Chand

1,2731520

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sagar Chand, I have to admit I only understand very basic math. Thus, I do not understand the equations you have given. What I really like is for you (or anyone else) give an answer such as: "With a 25% chance of winning, your chances of losing 10 in a row is 1 in 35,782" (or whatever the real answer is). And the same applies to my question about losing 20 in a row with a 25% chance of winning.
  $endgroup$
  – user495020
  Oct 24 '17 at 5:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  probability of losing once is 0.75,so probability of losing 2 times is 0.75*0.75. Similarly proba of losing 10 times in a row is $0.75^10$ Is this clear?
  $endgroup$
  – Sagar Chand
  Oct 24 '17 at 5:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But can you can you specifically tell me if it is "1 in 1000 chance" or "1 in 25,837 chance" or whatever?
  $endgroup$
  – user495020
  Oct 24 '17 at 5:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1 in $frac10.75^10$ chance for part a
  $endgroup$
  – Sagar Chand
  Oct 24 '17 at 5:18
  

  
  
  
  

add a comment | 

1

$begingroup$

First part:

Probability of losing 10 times in a row = $(1-0.25)^10 = (0.75)^10$

$$textodds = fractextfavorable outcometextunfavorable outcome$$

Here the probability of losing $10$ times in a row is favorable outcome, so

$textodds = frac0.75^101-0.75^10$

Similarly, for second part:

$textodds = frac0.75^201-0.75^20$

share|cite|improve this answer

edited Oct 24 '17 at 10:29

Andrew Tawfeek

1,8041722

answered Oct 24 '17 at 4:26

Sagar ChandSagar Chand

1,2731520

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sagar Chand, I have to admit I only understand very basic math. Thus, I do not understand the equations you have given. What I really like is for you (or anyone else) give an answer such as: "With a 25% chance of winning, your chances of losing 10 in a row is 1 in 35,782" (or whatever the real answer is). And the same applies to my question about losing 20 in a row with a 25% chance of winning.
  $endgroup$
  – user495020
  Oct 24 '17 at 5:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  probability of losing once is 0.75,so probability of losing 2 times is 0.75*0.75. Similarly proba of losing 10 times in a row is $0.75^10$ Is this clear?
  $endgroup$
  – Sagar Chand
  Oct 24 '17 at 5:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But can you can you specifically tell me if it is "1 in 1000 chance" or "1 in 25,837 chance" or whatever?
  $endgroup$
  – user495020
  Oct 24 '17 at 5:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1 in $frac10.75^10$ chance for part a
  $endgroup$
  – Sagar Chand
  Oct 24 '17 at 5:18
  

  
  
  
  

add a comment | 

$begingroup$

First part:

Probability of losing 10 times in a row = $(1-0.25)^10 = (0.75)^10$

$$textodds = fractextfavorable outcometextunfavorable outcome$$

Here the probability of losing $10$ times in a row is favorable outcome, so

$textodds = frac0.75^101-0.75^10$

Similarly, for second part:

$textodds = frac0.75^201-0.75^20$

share|cite|improve this answer

edited Oct 24 '17 at 10:29

Andrew Tawfeek

1,8041722

answered Oct 24 '17 at 4:26

Sagar ChandSagar Chand

1,2731520

$endgroup$

share|cite|improve this answer

edited Oct 24 '17 at 10:29

Andrew Tawfeek

1,8041722

answered Oct 24 '17 at 4:26

Sagar ChandSagar Chand

1,2731520

edited Oct 24 '17 at 10:29

Andrew Tawfeek

1,8041722

edited Oct 24 '17 at 10:29

Andrew Tawfeek

1,8041722

answered Oct 24 '17 at 4:26

Sagar ChandSagar Chand

1,2731520

answered Oct 24 '17 at 4:26

Sagar ChandSagar Chand

1,2731520