Binomial Distribution / Consecutive Odds Question

jschroeder5

New Member
Hi everyone,

Looking to solve a 3 part problem. My answers to the first 2 parts agrees with the textbook answers, but I'm confused by the 3rd part.

Here is the problem.

The chance of winning a bet on 00 at roulette is 1/38 = 0.026315. In 325 bets on 00 at roulette, the chance of six wins is 0.104840. Use this fact, and consideration of odds ratios, to answer the following questions without long calculations.

A) What is the most likely number of wins in 325 bets on 00, and what is its probability?

Correct Answer: 8 (with a prob of .1387, calculated by successive substitutions into the consecutive odds ratio formula)

B) Find the chance of 10 wins in 325 bets on 00

Correct Answer: 0.1128 (calculated by successive substitutions into the consecutive odds ratio formula)

C) Find the chance of 10 wins in 326 bets on 00

Correct Answer according to the text is 0.1133

I figured that I could calculate based on knowing the following info, which I had already calculated in order to answer the first 2 questions...

p of 9 wins in 325 games (.132059)
p of 10 wins in 325 games (.1127856)

I figured there are 2 paths to arrive at 10 wins in 326 games. They are...

9 wins in the first 325 games, combined with a win in the 326th game

10 wins in the first 325 games, combined with a loss in the 326th game.

So I calculated as P = 1/38 (.132059) + 37/38 (.1127856).

Shouldn't this work? My answer is not agreeing with the text. Does anybody have any other suggestions for figuring out this problem based on the odds ratio formula alone?

Thanks in advance for any help!
John

dianaforbes

New Member
You can simply use the binomial theorem. N=326 and p(success)=1/38 so (326 C 10)*(1/38)^10*(37/38)^316 is approximately 0.1133. Have a good day.